1. The unit vectors \(\vec{i}, \vec{j}, \vec{k}\) represent the directions along the \(x, y, and z\) axes respectively in a right-handed coordinate system.


2. The cross product of unit vectors follows a cyclic rule: \(\vec{i} \times \vec{j} = \vec{k}\), \(\vec{j} \times \vec{k} = \vec{i}\), and \(\vec{k} \times \vec{i} = \vec{j}\).


3. Moving in the clockwise (reverse) direction of this cycle results in a negative sign: \(\vec{j} \times \vec{i} = -\vec{k}\), \(\vec{k} \times \vec{j} = -\vec{i}\), and \(\vec{i} \times \vec{k} = -\vec{j}\).


4. Since the question asks for \(\vec{i} \times \vec{k}\), we apply the anti-commutative property of the cross product: \(\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})\).


5. Therefore, \(\vec{i} \times \vec{k} = -(\vec{k} \times \vec{i}) = -\vec{j}\).
Quick Tip: Visualize the cyclic order: \(i \rightarrow j \rightarrow k \rightarrow i\). Going in the direction of the arrow gives a positive result. Going against the arrow (like \(i\) to \(k\)) results in the third vector with a negative sign.

1. First, evaluate the expression inside the parentheses: \((\vec{a} \times \vec{a})\).


2. By the definition of the cross product, \(\vec{A} \times \vec{B} = |\vec{A}||\vec{B}|\sin\theta \hat{n}\), where \(\theta\) is the angle between the vectors.


3. Since any vector is parallel to itself, the angle \(\theta\) between \(\vec{a}\) and \(\vec{a}\) is \(0^\circ\).


4. Because \(\sin(0^\circ) = 0\), the cross product of a vector with itself is always the zero vector: \(\vec{a} \times \vec{a} = \vec{0}\).


5. Now substitute this result into the original expression: \(\vec{a} \cdot \vec{0}\).


6. The dot product of any vector with the zero vector is the scalar value 0.
Quick Tip: In a scalar triple product \([\vec{a} \vec{b} \vec{c}]\), if any two vectors are identical or parallel, the result is always zero because the volume of the parallelepiped formed by them is zero.

1. Let \(\vec{u} = 3\vec{i} + \vec{j} - 2\vec{k}\) and \(\vec{v} = \vec{i} + \lambda\vec{j} - 3\vec{k}\).


2. Two non-zero vectors are perpendicular if and only if their dot product is zero: \(\vec{u} \cdot \vec{v} = 0\).


3. Compute the dot product by summing the products of corresponding scalar components: \((3)(1) + (1)(\lambda) + (-2)(-3) = 0\).


4. Simplify the arithmetic terms: \(3 + \lambda + 6 = 0\).


5. Combine the constants: \(\lambda + 9 = 0\).


6. Solve for \(\lambda\): \(\lambda = -9\).
Quick Tip: For perpendicular vectors \(\vec{a}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}\) and \(\vec{b}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k}\), always set \(a_1b_1 + a_2b_2 + a_3b_3 = 0\). This is the most efficient way to solve for unknown coefficients.

1. Let vector \(\vec{A} = \vec{i} + 3\vec{j} + 7\vec{k}\) and vector \(\vec{B} = 2\vec{i} - 3\vec{j} + 6\vec{k}\).


2. The formula for the scalar projection of vector \(\vec{A}\) onto vector \(\vec{B}\) is: \(Projection = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}\).


3. First, calculate the dot product \(\vec{A} \cdot \vec{B}\): \((1)(2) + (3)(-3) + (7)(6) = 2 - 9 + 42 = 35\).


4. Second, calculate the magnitude of vector \(\vec{B}\): \(|\vec{B}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\).


5. Divide the dot product by the magnitude: \(Projection = \frac{35}{7} = 5\).
Quick Tip: Remember the denominator in the projection formula is always the magnitude of the vector that you are projecting "onto".

1. Use the vector identity relating magnitude and dot product: \(|\vec{u}|^2 = \vec{u} \cdot \vec{u}\).


2. Thus, \(|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})\).


3. Expand using the distributive property: \(|\vec{a} - \vec{b}|^2 = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}\).


4. Since dot product is commutative, this becomes: \(|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})\).


5. Substitute the given values (\(|\vec{a}|=2, |\vec{b}|=3, \vec{a} \cdot \vec{b}=4\)): \(|\vec{a} - \vec{b}|^2 = (2)^2 + (3)^2 - 2(4)\).


6. Calculate the sum: \(|\vec{a} - \vec{b}|^2 = 4 + 9 - 8 = 5\).


7. Solve for the magnitude by taking the square root: \(|\vec{a} - \vec{b}| = \sqrt{5}\).
Quick Tip: This formula is essentially the Law of Cosines for vectors. Squaring a binomial magnitude like \(|\vec{a} \pm \vec{b}|\) is a standard technique for solving vector geometry problems.

1. First, find the cross product \(\vec{a} \times \vec{b}\) using a determinant.


2. \(\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}
1 & -1 & 2
2 & 3 & -4 \end{vmatrix}\).


3. Expand the determinant: \(\vec{i}((-1)(-4) - (2)(3)) - \vec{j}((1)(-4) - (2)(2)) + \vec{k}((1)(3) - (-1)(2))\).


4. Calculate the components: \(\vec{i}(4 - 6) - \vec{j}(-4 - 4) + \vec{k}(3 + 2) = -2\vec{i} + 8\vec{j} + 5\vec{k}\).


5. Now find the magnitude of the resulting vector: \(|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + 8^2 + 5^2}\).


6. Simplify inside the square root: \(\sqrt{4 + 64 + 25} = \sqrt{93}\).
Quick Tip: Be very careful with the negative sign in front of the \(j\) component when expanding the determinant. Many errors in cross products occur at that specific step.

1. Start by squaring both sides of the given equation: \(|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2\).


2. Expand both sides using the dot product identity: \(|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})\).


3. Subtract \(|\vec{a}|^2\) and \(|\vec{b}|^2\) from both sides to simplify: \(2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})\).


4. Move all terms to one side: \(4(\vec{a} \cdot \vec{b}) = 0\).


5. This implies \(\vec{a} \cdot \vec{b} = 0\).


6. Since the dot product of two vectors is zero, they are perpendicular (\(\vec{a} \perp \vec{b}\)).
Quick Tip: Geometrically, if the diagonals of a parallelogram (\(|\vec{a}+\vec{b}|\) and \(|\vec{a}-\vec{b}|\)) are equal, the parallelogram is a rectangle, and the adjacent sides are perpendicular.

1. The expression involves the dot product of two vectors: \(2\vec{j}\) and \(-3\vec{k}\).


2. Scalar multiplication allows us to factor out the constants: \(2\vec{j} \cdot (-3\vec{k}) = (2 \times -3) (\vec{j} \cdot \vec{k})\).


3. Simplify the product of constants: \(-6 (\vec{j} \cdot \vec{k})\).


4. In an orthonormal coordinate system, the unit vectors \(\vec{j}\) and \(\vec{k}\) are mutually perpendicular.


5. The dot product of any two distinct unit basis vectors (\(\vec{i}, \vec{j}, \vec{k}\)) is zero.


6. Therefore, \(\vec{j} \cdot \vec{k} = 0\).


7. The final result is \(-6 \times 0 = 0\).
Quick Tip: Dot products of different basis vectors (\(\vec{i}\cdot\vec{j}, \vec{j}\cdot\vec{k}, \vec{k}\cdot\vec{i}\)) are always 0. Cross products of different basis vectors (\(\vec{i}\times\vec{j}\), etc.) yield a third vector. Don't confuse the two operations.

1. Let the direction ratios be \(a = 2\), \(b = 6\), and \(c = -3\).


2. The direction cosines (\(l, m, n\)) are calculated by dividing each direction ratio by the square root of the sum of their squares: \(\sqrt{a^2 + b^2 + c^2}\).


3. Calculate the normalization factor: \(\sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\).


4. Now divide each ratio by this factor:

\(l = \frac{a}{7} = \frac{2}{7}\)

\(m = \frac{b}{7} = \frac{6}{7}\)

\(n = \frac{c}{7} = \frac{-3}{7}\)


5. The resulting direction cosines are \((\frac{2}{7}, \frac{6}{7}, \frac{-3}{7})\).
Quick Tip: Direction cosines always satisfy the property \(l^2 + m^2 + n^2 = 1\). You can use this to check your final answer.

1. The direction cosines of a line are defined as \(l = \cos \alpha\), \(m = \cos \beta\), and \(n = \cos \gamma\), where \(\alpha, \beta, \gamma\) are the angles with the axes.


2. For any line in 3D space, the sum of the squares of its direction cosines is always equal to unity.


3. This is derived from the fact that if \((l, m, n)\) are components of a unit vector along the line, then its magnitude must be 1.


4. Applying the magnitude formula: \(\sqrt{l^2 + m^2 + n^2} = 1 \implies l^2 + m^2 + n^2 = 1\).


5. Substituting the cosine terms back in: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).
Quick Tip: An alternative form of this identity is \(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2\). It is derived using \(\cos^2 \theta = 1 - \sin^2 \theta\). Option (A) is the fundamental standard form.

1. Observe the denominator: \(1 + x^8\) can be written as \(1 + (x^4)^2\).


2. Perform a substitution: Let \(u = x^4\). Then, taking the differential gives \(du = 4x^3 dx\), or \(\frac{1}{4} du = x^3 dx\).


3. Change the limits of integration: When \(x=0\), \(u=0^4=0\). When \(x=1\), \(u=1^4=1\).


4. Rewrite the integral in terms of \(u\): \(\int_0^1 \frac{\frac{1}{4} du}{1 + u^2} = \frac{1}{4} \int_0^1 \frac{1}{1+u^2} du\).


5. Evaluate the integral using the standard formula \(\int \frac{1}{1+x^2} dx = \tan^{-1} x\):

\(\frac{1}{4} [\tan^{-1} u]_0^1 = \frac{1}{4} (\tan^{-1} 1 - \tan^{-1} 0)\).


6. Substitute known values: \(\frac{1}{4} (\frac{\pi}{4} - 0) = \frac{\pi}{16}\).
Quick Tip: When the power of \(x\) in the numerator is one less than a power in the denominator (like \(x^3\) and \(x^8\)), substitution is usually the fastest method.

1. Let \(u = \log x\). Then, \(du = \frac{1}{x} dx\).


2. Change the limits of integration: When \(x=1\), \(u = \log 1 = 0\). When \(x=e\), \(u = \log e = 1\).


3. Substitute \(u\) and \(du\) into the integral: \(\int_0^1 u^2 du\).


4. Evaluate the basic power rule integral: \([\frac{u^3}{3}]_0^1\).


5. Apply the upper and lower limits: \(\frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}\).
Quick Tip: For integrals involving both \(\log x\) and \(1/x\), the substitution \(u = \log x\) will almost always simplify the expression into a standard power rule form.

1. The matrix \(A = [4 2 3]\) is a row matrix of order \(1 \times 3\).


2. The notation \(A'\) (or \(A^T\)) represents the transpose of matrix \(A\).


3. By definition, the transpose of a matrix is obtained by swapping its rows and columns.


4. A row matrix of order \(1 \times 3\) becomes a column matrix of order \(3 \times 1\) upon transposition.


5. Therefore, \(A' = \begin{bmatrix} 4
2
3 \end{bmatrix}\).
Quick Tip: Transposing doesn't change the values or their relative sequence, only their orientation (horizontal to vertical or vice-versa).

1. Rewrite the integral by noting that \(\frac{1}{\cos^2 x} = \sec^2 x\): \(\int_0^{\pi/4} e^{\tan x} \cdot \sec^2 x dx\).


2. Let \(u = \tan x\). Then, \(du = \sec^2 x dx\).


3. Change the limits: When \(x=0\), \(u = \tan 0 = 0\). When \(x = \pi/4\), \(u = \tan(\pi/4) = 1\).


4. Substitute into the integral: \(\int_0^1 e^u du\).


5. Evaluate: \([e^u]_0^1 = e^1 - e^0\).


6. Since \(e^1 = e\) and \(e^0 = 1\), the result is \(e - 1\).
Quick Tip: Recognizing trigonometric identities like \(1/\cos^2 x = \sec^2 x\) is often the first step in simplifying complex-looking integrals.

1. Use the trigonometric identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\).


2. Substitute this into the integral: \(\int_0^{\pi/2} \frac{1 + \cos 2x}{2} dx = \frac{1}{2} \int_0^{\pi/2} (1 + \cos 2x) dx\).


3. Evaluate the antiderivative: \(\frac{1}{2} [x + \frac{\sin 2x}{2}]_0^{\pi/2}\).


4. Apply the limits: \(\frac{1}{2} [(\frac{\pi}{2} + \frac{\sin \pi}{2}) - (0 + \frac{\sin 0}{2})]\).


5. Since \(\sin \pi = 0\) and \(\sin 0 = 0\): \(\frac{1}{2} [\frac{\pi}{2} + 0 - 0] = \frac{\pi}{4}\).
Quick Tip: Wallis' Formula for \(\int_0^{\pi/2} \sin^n x dx\) or \(\int_0^{\pi/2} \cos^n x dx\): for \(n=2\), the result is \(\frac{n-1}{n} \cdot \frac{\pi}{2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}\). This is a great time-saver.

1. Use the identity: \(\cos 3x = 4 \cos^3 x - 3 \cos x \implies \cos^3 x = \frac{1}{4} ( \cos 3x + 3 \cos x)\).


2. Substitute into the integral: \(\frac{1}{4} \int_0^{\pi/3} ( \cos 3x + 3 \cos x) dx\).


3. Integrate each term: \(\frac{1}{4} [\frac{\sin 3x}{3} + 3 \sin x]_0^{\pi/3}\).


4. Evaluate at the upper limit: \(\frac{1}{4} [ \frac{\sin \pi}{3} + 3 \sin(\frac{\pi}{3}) ] = \frac{1}{4} [ 0 + 3 \cdot \frac{\sqrt{3}}{2} ] = \frac{3\sqrt{3}}{8}\).


5. The lower limit evaluation yields 0 as \(\sin 0 = 0\). So the final result is \(\frac{3\sqrt{3}}{8}\).
Quick Tip: Powers of trig functions like \(\sin^3 x\) or \(\cos^3 x\) are best integrated by converting them to multiple angle formulas.

1. The function \(|\sin x|\) is periodic with a period of \(\pi\).


2. Thus, \(\int_0^{2\pi} |\sin x| dx = 2 \int_0^\pi |\sin x| dx\).


3. In the interval \([0, \pi]\), \(\sin x\) is non-negative, so \(|\sin x| = \sin x\).


4. The integral becomes: \(2 \int_0^\pi \sin x dx\).


5. Evaluate: \(2 [-\cos x]_0^\pi = 2 [-\cos \pi - (-\cos 0)]\).


6. Substitute values: \(2 [-(-1) + 1] = 2 [1 + 1] = 4\).
Quick Tip: The area of one "loop" of a sine or cosine curve (from one zero to the next) is exactly 2 square units. In \(2\pi\), there are two such loops for the absolute value function.

1. Let \(I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\).


2. Apply the property \(\int_0^a f(x) dx = \int_0^a f(a - x) dx\).


3. \(I = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} dx = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx\).


4. Add the two expressions for \(I\): \(2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} dx\).


5. Simplify: \(2I = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \frac{\pi}{2}\).


6. Solve for \(I\): \(I = \frac{\pi}{4}\).
Quick Tip: This property is a classic trick for symmetric integrals over \([0, \pi/2]\). Whenever you see \(\sin x\) and \(\cos x\) in a fraction like this, \(2I\) will often simplify to \(\int 1 dx\).

1. Check the symmetry of the integrand \(f(x) = \sin^5 x\).


2. Evaluate \(f(-x)\): \(f(-x) = \sin^5(-x) = (\sin(-x))^5 = (-\sin x)^5 = -\sin^5 x\).


3. Since \(f(-x) = -f(x)\), the function is an odd function.


4. For any odd function \(f(x)\), the definite integral over a symmetric interval \([-a, a]\) is always zero: \(\int_{-a}^a f(x) dx = 0\).


5. Therefore, \(\int_{-\pi}^\pi \sin^5 x dx = 0\).
Quick Tip: Before doing any calculation for a definite integral over a symmetric range like \([-a, a]\), always check if the function is odd. It can save minutes of tedious calculation.

1. Let \(I = \int_0^a \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} dx\).


2. Apply the property \(\int_0^a f(x) dx = \int_0^a f(a - x) dx\).


3. \(I = \int_0^a \frac{\sqrt{a - x}}{\sqrt{a - (a - x)} + \sqrt{a - x}} dx = \int_0^a \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} dx\).


4. Add the two expressions for \(I\): \(2I = \int_0^a \frac{\sqrt{x} + \sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} dx\).


5. Simplify the integrand: \(2I = \int_0^a 1 dx = [x]_0^a = a\).


6. Solve for \(I\): \(I = \frac{a}{2}\).
Quick Tip: For any integral of the form \(\int_0^a \frac{f(x)}{f(x) + f(a-x)} dx\), the answer is always \(a/2\). This pattern is extremely common in multiple-choice exams.

1. Let \(\theta = \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right)\).


2. By the definition of inverse trigonometric functions, \(\tan \theta = -\frac{1}{\sqrt{3}}\).


3. We know that the principal value range of \(\tan^{-1}x\) is \(\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)\).


4. We also know that \(\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}\).


5. Using the identity \(\tan(-\theta) = -\tan \theta\), we can write \(\tan \left( -\frac{\pi}{6} \right) = -\frac{1}{\sqrt{3}}\).


6. Since \(-\frac{\pi}{6}\) lies within the principal value range, \(\theta = -\frac{\pi}{6}\).
Quick Tip: For inverse functions like \(\tan^{-1}(-x)\), \(\sin^{-1}(-x)\), and \(\csc^{-1}(-x)\), you can directly pull the negative sign out: \(f^{-1}(-x) = -f^{-1}(x)\).

1. Apply the identity for \(2\tan^{-1}x\): \(2\tan^{-1}x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)\), provided \(|x| < 1\).


2. Here, \(x = \frac{1}{3}\). Since \(|\frac{1}{3}| < 1\), the formula is valid.


3. Substitute \(x = \frac{1}{3}\) into the formula: \(\tan^{-1} \left( \frac{2(1/3)}{1 - (1/3)^2} \right)\).


4. Simplify the numerator: \(\frac{2}{3}\).


5. Simplify the denominator: \(1 - \frac{1}{9} = \frac{8}{9}\).


6. The expression becomes: \(\tan^{-1} \left( \frac{2/3}{8/9} \right) = \tan^{-1} \left( \frac{2}{3} \times \frac{9}{8} \right) = \tan^{-1} \left( \frac{18}{24} \right)\).


7. Reducing the fraction: \(\frac{18}{24} = \frac{3}{4}\). Thus, the final value is \(\tan^{-1} \frac{3}{4}\).
Quick Tip: This formula is derived from the double-angle formula \(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}\). Use it whenever you need to convert a coefficient of 2 in front of \(\tan^{-1}\) into a single term.

1. Identify the standard inverse trigonometric identity: \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\) for all \(x \in R\).


2. Substitute this identity into the given expression.


3. The expression simplifies to: \(\cot \left( \frac{\pi}{2} \right)\).


4. We know that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).


5. Thus, \(\cot \left( \frac{\pi}{2} \right) = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0\).
Quick Tip: Standard complementary pairs like \(\sin^{-1}x + \cos^{-1}x = \pi/2\) and \(\sec^{-1}x + \csc^{-1}x = \pi/2\) are frequently tested. Memorizing these leads to instant answers.

1. Let \(\theta = \cos^{-1} \frac{3}{5}\). This implies \(\cos \theta = \frac{3}{5}\).


2. We need to find \(\sin \theta\).


3. Use the fundamental trigonometric identity: \(\sin^2 \theta + \cos^2 \theta = 1\).


4. Rearrange to find \(\sin \theta\): \(\sin \theta = \sqrt{1 - \cos^2 \theta}\) (since the range of \(\cos^{-1}\) is \([0, \pi]\), \(\sin \theta\) is non-negative).


5. Substitute the value: \(\sin \theta = \sqrt{1 - \left( \frac{3}{5} \right)^2} = \sqrt{1 - \frac{9}{25}}\).


6. Calculate the value: \(\sin \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}\).
Quick Tip: Think in terms of right-angled triangles. If \(\cos \theta = 3/5\), then the adjacent side is 3 and the hypotenuse is 5. By the Pythagorean theorem (\(3, 4, 5\)), the opposite side must be 4, so \(\sin \theta = 4/5\).

1. Evaluate each term individually based on principal values.


2. For the first term: \(\tan^{-1}(1) = \frac{\pi}{4}\).


3. For the second and third terms, we can use the identity \(\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}\).


4. Letting \(x = -\frac{1}{2}\), we have \(\cos^{-1}(-1/2) + \sin^{-1}(-1/2) = \frac{\pi}{2}\).


5. Now, sum all components: \(\frac{\pi}{4} + \frac{\pi}{2}\).


6. Find a common denominator: \(\frac{\pi}{4} + \frac{2\pi}{4} = \frac{3\pi}{4}\).
Quick Tip: Instead of calculating \(\cos^{-1}(-1/2) = 2\pi/3\) and \(\sin^{-1}(-1/2) = -\pi/6\) separately, using the identity \(\cos^{-1}x + \sin^{-1}x = \pi/2\) makes the sum much easier.

1. Recall that \(\cos^{-1}(\cos \theta) = \theta\) only if \(\theta\) is in the principal value range \([0, \pi]\).


2. Here \(\theta = \frac{7\pi}{6}\), which is greater than \(\pi\). Thus, we must transform it.


3. Use the property \(\cos(2\pi - \theta) = \cos \theta\).


4. \(\cos \left( \frac{7\pi}{6} \right) = \cos \left( 2\pi - \frac{7\pi}{6} \right) = \cos \left( \frac{12\pi - 7\pi}{6} \right) = \cos \left( \frac{5\pi}{6} \right)\).


5. Now, \(\frac{5\pi}{6}\) is within the range \([0, \pi]\).


6. Therefore, \(\cos^{-1} \left( \cos \frac{5\pi}{6} \right) = \frac{5\pi}{6}\).
Quick Tip: For \(\cos^{-1}(\cos x)\), if \(x > \pi\), use \(2\pi - x\) or \(x - 2\pi\) to bring the value into the range \([0, \pi]\) while maintaining the cosine value.

1. Use the identity for \(\tan^{-1} x + \tan^{-1} y\).


2. Note that here \(x=2\) and \(y=3\), so \(xy = 2 \times 3 = 6\). Since \(xy > 1\), the formula is \(\pi + \tan^{-1} \left( \frac{x+y}{1-xy} \right)\).


3. Substitute values: \(\pi + \tan^{-1} \left( \frac{2+3}{1 - (2 \times 3)} \right) = \pi + \tan^{-1} \left( \frac{5}{1 - 6} \right)\).


4. Simplify: \(\pi + \tan^{-1} \left( \frac{5}{-5} \right) = \pi + \tan^{-1}(-1)\).


5. Since \(\tan^{-1}(-1) = -\frac{\pi}{4}\), the sum is \(\pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
Quick Tip: The condition \(xy > 1\) is critical for \(\tan^{-1}\) sums. If you ignore it and use the basic formula, you'd get \(-\pi/4\), which is wrong because both angles are positive, so their sum must be positive.

1. Let \(\theta = \cos^{-1} x\). This implies \(\cos \theta = x\).


2. We need to find \(\tan \theta\).


3. We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).


4. Using the identity \(\sin \theta = \sqrt{1 - \cos^2 \theta}\), we have \(\sin \theta = \sqrt{1 - x^2}\).


5. Therefore, \(\tan \theta = \frac{\sqrt{1 - x^2}}{x}\).
Quick Tip: Converting an inverse trigonometric function into another can be done by defining it as an angle in a right triangle where adjacent = \(x\) and hypotenuse = 1. The opposite side then is \(\sqrt{1-x^2}\).

1. Rewrite the second term by dividing the numerator and denominator by \(y\): \(\tan^{-1} \left( \frac{(x/y) - 1}{1 + (x/y)} \right)\).


2. Observe that this matches the form \(\tan^{-1} \left( \frac{a - b}{1 + ab} \right)\) where \(a = x/y\) and \(b = 1\).


3. By the standard identity, \(\tan^{-1} \left( \frac{a - b}{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b\).


4. So, \(\tan^{-1} \frac{x-y}{x+y} = \tan^{-1} \frac{x}{y} - \tan^{-1} 1\).


5. Substitute this back into the original expression: \(\tan^{-1} \frac{x}{y} - \left( \tan^{-1} \frac{x}{y} - \tan^{-1} 1 \right)\).


6. Simplify: \(\tan^{-1} \frac{x}{y} - \tan^{-1} \frac{x}{y} + \tan^{-1} 1 = \tan^{-1} 1 = \frac{\pi}{4}\).
Quick Tip: Look for algebraic manipulation to turn fractions inside inverse functions into standard addition/subtraction identities. Dividing by a common variable often reveals the pattern.

1. Let \(x = \tan \theta\). Then \(\theta = \tan^{-1} x\).


2. Substitute \(x\) into the expression: \(\cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right)\).


3. Recall the trigonometric identity for \(\cos 2\theta\): \(\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\).


4. The expression becomes \(\cos^{-1}(\cos 2\theta)\).


5. Since \(\cos^{-1}(\cos 2\theta) = 2\theta\), substitute back for \(\theta\): \(2\tan^{-1} x\).
Quick Tip: The expression \(\frac{1-x^2}{1+x^2}\) is a key signal to substitute \(x = \tan \theta\). This is one of the three forms for \(2\tan^{-1}x\), the others involving \(\sin^{-1} \frac{2x}{1+x^2}\) and \(\tan^{-1} \frac{2x}{1-x^2}\).

1. Use the half-angle identities: \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\).


2. The integral becomes \(\int \frac{2\sin^2 x}{2\cos^2 x} dx = \int \tan^2 x dx\).


3. Use the trigonometric identity \(\tan^2 x = \sec^2 x - 1\).


4. Now integrate: \(\int (\sec^2 x - 1) dx\).


5. The antiderivative of \(\sec^2 x\) is \(\tan x\), and the antiderivative of 1 is \(x\).


6. The final result is \(\tan x - x + k\).
Quick Tip: There is no direct formula for the integral of \(\tan^2 x\). Always convert it to \(\sec^2 x - 1\) to perform the integration easily.

1. This is an integral of the form \(\int \frac{1}{u} du\).


2. Let \(u = x - 1\). Then \(du = dx\).


3. The integral becomes \(\int \frac{1}{u} du\).


4. The standard antiderivative is \(\log|u| + k\).


5. Substitute back for \(u\): \(\log|x - 1| + k\).
Quick Tip: Any linear term in the denominator like \(ax+b\) results in \(\frac{1}{a} \log|ax+b|\). Here \(a=1\), so the result is straightforward.

1. Integrate by parts using \(\int u dv = uv - \int v du\).


2. Choose \(u = \log x\) and \(dv = dx\).


3. Then \(du = \frac{1}{x} dx\) and \(v = x\).


4. Substitute into the formula: \(x \log x - \int x \left( \frac{1}{x} \right) dx\).


5. Simplify the integral: \(x \log x - \int 1 dx\).


6. Integrate: \(x \log x - x + k\).
Quick Tip: Integration by parts is essential when you have a single function whose derivative is simpler than the function itself, such as \(\log x\) or \(\sin^{-1} x\). Always assume the other function is "1".

1. Use substitution: Let \(t = \sqrt{x}\). Then \(x = t^2\) and \(dx = 2t dt\).


2. Substitute into the integral: \(\int \cos(t) \cdot 2t dt = 2 \int t \cos t dt\).


3. Integrate by parts: \(\int u dv = uv - \int v du\). Let \(u = t\) and \(dv = \cos t dt\).


4. Then \(du = dt\) and \(v = \sin t\).


5. Applying the formula: \(2 [t \sin t - \int \sin t dt] = 2 [t \sin t - (-\cos t)] = 2 [t \sin t + \cos t]\).


6. Substitute back \(t = \sqrt{x}\): \(2(\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}) + k\).
Quick Tip: Substitution followed by integration by parts is a standard method for functions where the argument is a root, like \(\sqrt{x}\) or \(x^2\).

1. Recognize the standard integral form: \(\int e^x (f(x) + f'(x)) dx = e^x f(x) + k\).


2. Let \(f(x) = \tan^{-1} x\).


3. Find the derivative: \(f'(x) = \frac{1}{1+x^2}\).


4. Since the integrand is exactly \(e^x(f(x) + f'(x))\), the solution is immediate.


5. The integral evaluates to \(e^x \tan^{-1} x + k\).
Quick Tip: Whenever you see \(e^x\) multiplied by a sum, immediately check if one term is the derivative of the other. This saves considerable time compared to using parts.

1. This is a standard integral formula for a rational function.


2. Using partial fractions: \(\frac{1}{x^2 - a^2} = \frac{1}{(x-a)(x+a)} = \frac{1}{2a} \left[ \frac{1}{x-a} - \frac{1}{x+a} \right]\).


3. Integrate term by term: \(\frac{1}{2a} \int \left( \frac{1}{x-a} - \frac{1}{x+a} \right) dx\).


4. This gives: \(\frac{1}{2a} [\log|x-a| - \log|x+a|] + k\).


5. Use the property of logarithms \(\log A - \log B = \log(A/B)\): \(\frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + k\).
Quick Tip: Memorize the difference: \(x^2 - a^2\) has the \((x-a)\) factor in the numerator of the log, while \(a^2 - x^2\) has it in the denominator.

1. Use the double-angle identity for cosine: \(\cos \theta = 1 - 2\sin^2 \frac{\theta}{2}\).


2. Rearrange to find an expression for \(\sin^2 \frac{x}{2}\): \(\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}\).


3. Substitute this into the integral: \(\int \frac{1 - \cos x}{2} dx = \frac{1}{2} \int (1 - \cos x) dx\).


4. Integrate term by term: \(\frac{1}{2} [x - \sin x] + k\).


5. Distribute the constant: \(\frac{1}{2}x - \frac{1}{2}\sin x + k\).
Quick Tip: Integration of squared trig functions (\(\sin^2, \cos^2\)) almost always requires using linear multiple-angle identities before integrating.

1. Use substitution: Let \(u = \sin^{-1} x\).


2. Find the differential: \(du = \frac{1}{\sqrt{1-x^2}} dx\).


3. Substitute \(u\) and \(du\) into the original integral: \(\int \tan u du\).


4. The standard antiderivative of \(\tan u\) is \(\log|\sec u| + k\).


5. Substitute back \(u = \sin^{-1} x\): \(\log|\sec(\sin^{-1} x)| + k\).
Quick Tip: Recognize pairs of functions and their derivatives. \(\frac{1}{\sqrt{1-x^2}}\) is the perfect partner for \(\sin^{-1} x\) in a substitution problem.

1. Multiply the numerator and denominator by \(e^x\): \(\int \frac{e^x}{e^x(e^x + e^{-x})} dx = \int \frac{e^x}{e^{2x} + 1} dx\).


2. Let \(u = e^x\). Then \(du = e^x dx\).


3. Substitute into the integral: \(\int \frac{du}{u^2 + 1}\).


4. This is the standard form for \(\tan^{-1} u + k\).


5. Substitute back \(u = e^x\): \(\tan^{-1}(e^x) + k\).
Quick Tip: Integrals involving mixed powers of \(e^x\) and \(e^{-x}\) usually simplify by multiplying through by \(e^x\) to get a consistent power.

1. Find the antiderivative: \(\int \frac{1}{1+x^2} dx = \tan^{-1} x\).


2. Apply the limits of integration from 0 to 1: \([\tan^{-1} x]_0^1\).


3. Calculate: \(\tan^{-1}(1) - \tan^{-1}(0)\).


4. Substitute known values: \(\frac{\pi}{4} - 0 = \frac{\pi}{4}\).
Quick Tip: The definite integral of the derivative of an inverse trig function over \([0, 1]\) is a very common exam problem. Know your standard values like \(\tan^{-1}(1) = \pi/4\).

1. Use the chain rule for differentiation: \(\frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot f'(x)\).


2. Here, \(f(x) = x - a\).


3. Differentiate \(f(x)\) with respect to \(x\): \(f'(x) = 1\) (since \(a\) is a constant).


4. Therefore, \(\frac{d}{dx} e^{x-a} = e^{x-a} \cdot (1) = e^{x-a}\).
Quick Tip: The exponential function \(e^{x+c}\) always differentiates to itself because the derivative of \((x+c)\) is simply 1.

1. Apply the chain rule: \(\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}\).


2. Let \(u = x^2 + ax + 1\).


3. Find the derivative of \(u\): \(\frac{du}{dx} = 2x + a\).


4. Substitute back into the chain rule formula: \(\frac{1}{2\sqrt{x^2+ax+1}} \cdot (2x + a)\).


5. This simplifies to: \(\frac{2x+a}{2\sqrt{x^2+ax+1}}\).
Quick Tip: For any \(\sqrt{f(x)}\), the derivative is always \(\frac{f'(x)}{2\sqrt{f(x)}}\). This shortcut is highly efficient for multiple-choice questions.

1. Use the chain rule: \(\frac{d}{dx} \sin(f(x)) = \cos(f(x)) \cdot f'(x)\).


2. Let \(f(x) = x^2\).


3. The derivative of \(x^2\) is \(2x\).


4. Multiply: \(\cos(x^2) \cdot 2x\).


5. Rearrange: \(2x \cos x^2\).
Quick Tip: Be careful not to confuse \(\sin(x^2)\) with \((\sin x)^2 = \sin^2 x\). The former requires the derivative of the angle \(x^2\), which is \(2x\).

1. Apply the square root derivative rule: \(\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot u'\).


2. Let \(u = \cot x\).


3. Find the derivative of \(\cot x\): \(u' = -\csc^2 x\).


4. Combine the terms: \(\frac{1}{2\sqrt{\cot x}} \cdot (-\csc^2 x)\).


5. This gives: \(\frac{-\csc^2 x}{2\sqrt{\cot x}}\).
Quick Tip: Always remember that derivatives of trigonometric functions starting with 'c' (cos, cot, csc) are negative.

1. Recall the property of complementary inverse trigonometric functions: \(\tan^{-1} \theta + \cot^{-1} \theta = \frac{\pi}{2}\).


2. This property holds for any valid input \(\theta\). Here \(\theta = \sqrt{x}\).


3. Thus, the expression simplifies to the derivative of a constant: \(\frac{d}{dx} \left( \frac{\pi}{2} \right)\).


4. The derivative of any constant value is 0.
Quick Tip: Simplifying an expression before differentiating can often reduce a complex problem into the derivative of a constant.

1. Use the constant multiple rule for differentiation: \(\frac{d}{dx} [c \cdot f(x)] = c \cdot f'(x)\).


2. Here \(c = 2\) and \(f(x) = \tan^{-1} x\).


3. The derivative of \(\tan^{-1} x\) is \(\frac{1}{1+x^2}\).


4. Multiplying the two gives: \(2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}\).
Quick Tip: The derivative of \(\tan^{-1}x\) is one of the most frequently used formulas in calculus exams. Memorize it along with its integral form.

1. Use the chain rule: \(\frac{d}{dx} \cos(f(x)) = -\sin(f(x)) \cdot f'(x)\).


2. Let \(f(x) = \sqrt{x}\).


3. Differentiate \(f(x)\): \(f'(x) = \frac{1}{2\sqrt{x}}\).


4. Combine: \(-\sin\sqrt{x} \cdot \frac{1}{2\sqrt{x}}\).


5. The final derivative is \(\frac{-\sin\sqrt{x}}{2\sqrt{x}}\).
Quick Tip: When applying the chain rule, work from the outside in. First differentiate the trigonometric part, then multiply by the derivative of the inner function.

1. First, evaluate the limit inside the brackets: \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\).


2. Here \(n = 5\), so the limit is \(5a^4\).


3. Notice that \(5a^4\) is a constant value with respect to \(x\).


4. The problem asks for the derivative of this constant with respect to \(x\).


5. The derivative of any constant is always 0.
Quick Tip: Always simplify the interior of the derivative first. If the interior is a definite limit or a constant, the derivative will always be zero.

1. This is a standard derivative formula in calculus.


2. The derivative of \(\tan^{-1} x\) is \(\frac{1}{1+x^2}\).


3. Using the relationship \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\), differentiate both sides.


4. \(\frac{d}{dx}(\tan^{-1} x) + \frac{d}{dx}(\cot^{-1} x) = 0\).


5. Substituting the known derivative: \(\frac{1}{1+x^2} + \frac{d}{dx}(\cot^{-1} x) = 0\).


6. Solving gives \(\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}\).
Quick Tip: Derivatives of co-functions (\(\cos, \cot, \csc\)) are always the negative of the derivative of their corresponding primary functions (\(\sin, \tan, \sec\)).

1. Find the y-coordinate of the point on the curve: Substitute \(x = 3\) into \(y = x^2 + 4x + 1\).


2. \(y = (3)^2 + 4(3) + 1 = 9 + 12 + 1 = 22\). So the point is \((3, 22)\).


3. Find the slope of the tangent by differentiating: \(\frac{dy}{dx} = 2x + 4\).


4. Calculate the slope (\(m\)) at \(x = 3\): \(m = 2(3) + 4 = 10\).


5. Use the point-slope form of the line: \(y - y_1 = m(x - x_1)\).


6. \(y - 22 = 10(x - 3) \implies y - 22 = 10x - 30\).


7. Rearrange to standard form: \(10x - y = 30 - 22 \implies 10x - y = 8\).
Quick Tip: To find a tangent, you always need two things: a point and a slope. The derivative gives you the slope, and the original equation gives you the point.

1. Let \(u = \sqrt{x}\). Differentiating both sides with respect to \(x\), we get \(\frac{du}{dx} = \frac{1}{2\sqrt{x}}\), which implies \(2 du = \frac{1}{\sqrt{x}} dx\).


2. Change the limits of integration: when \(x = 0\), \(u = \sqrt{0} = 0\); when \(x = 1\), \(u = \sqrt{1} = 1\).


3. Substitute these into the integral: \(\int_0^1 e^u \cdot (2 du)\).


4. Factor out the constant: \(2 \int_0^1 e^u du\).


5. The integral of \(e^u\) is \(e^u\), so we have \(2 [e^u]_0^1\).


6. Apply the limits: \(2(e^1 - e^0) = 2(e - 1)\).
Quick Tip: Whenever the exponent of a function is a root like \(\sqrt{x}\), and its derivative \(1/\sqrt{x}\) is present in the integrand, the substitution \(u = \sqrt{x}\) is the most efficient path.

1. This is a product of an algebraic function and an exponential function, so we use integration by parts: \(\int u dv = uv - \int v du\).


2. Let \(u = x\) and \(dv = e^x dx\). This gives \(du = dx\) and \(v = e^x\).


3. Apply the formula: \([x e^x]_0^1 - \int_0^1 e^x dx\).


4. Evaluate the first term: \((1 \cdot e^1 - 0 \cdot e^0) = e\).


5. Evaluate the second term: \(\int_0^1 e^x dx = [e^x]_0^1 = (e^1 - e^0) = e - 1\).


6. Combine the results: \(e - (e - 1) = e - e + 1 = 1\).
Quick Tip: For integrals like \(\int x^n e^x dx\), the result is always 1 when the limits are from 0 to 1 and \(n=1\). This is a specific case of the Gamma function property.

1. Rewrite the integrand as a power of \(x\): \(\int_1^4 x^{-1/2} dx\).


2. Apply the power rule for integration \(\int x^n dx = \frac{x^{n+1}}{n+1}\): \([\frac{x^{1/2}}{1/2}]_1^4 = [2\sqrt{x}]_1^4\).


3. Substitute the upper limit: \(2\sqrt{4} = 2(2) = 4\).


4. Substitute the lower limit: \(2\sqrt{1} = 2(1) = 2\).


5. Subtract the lower value from the upper value: \(4 - 2 = 2\).
Quick Tip: Remember that the antiderivative of \(\frac{1}{\sqrt{x}}\) is simply \(2\sqrt{x}\). Knowing this basic form directly saves several calculation steps.

1. Use the standard integration formula: \(\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\).


2. Apply the limits from 0 to \(a\): \([\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)]_0^a\).


3. Evaluate at \(x = a\): \((\frac{a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(1)) = 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}\).


4. Evaluate at \(x = 0\): \((\frac{0}{2}\sqrt{a^2} + \frac{a^2}{2}\sin^{-1}(0)) = 0 + 0 = 0\).


5. The final definite integral is \(\frac{\pi a^2}{4} - 0 = \frac{\pi a^2}{4}\).
Quick Tip: Geometrically, this integral represents the area of a quarter-circle of radius \(a\) in the first quadrant. Since the area of a full circle is \(\pi a^2\), the result must be \(\frac{1}{4} \pi a^2\).

1. The function \(f(x) = |x|\) is an even function, which means \(f(-x) = f(x)\).


2. For a symmetric interval \([-a, a]\), \(\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx\) if \(f(x)\) is even.


3. Thus, \(\int_{-2}^2 |x| dx = 2 \int_0^2 |x| dx\).


4. In the interval \([0, 2]\), \(|x| = x\), so the integral is \(2 \int_0^2 x dx\).


5. Integrate: \(2 [\frac{x^2}{2}]_0^2 = [x^2]_0^2\).


6. Substitute the limits: \(2^2 - 0^2 = 4\).
Quick Tip: Visually, the integral of \(|x|\) from \(-a\) to \(a\) is the sum of the areas of two identical triangles, each with base \(a\) and height \(a\). The total area is \(2 \times (\frac{1}{2} a \cdot a) = a^2\). Here \(a=2\), so the area is 4.

1. The order of a differential equation is defined as the highest order derivative present in the equation.


2. Here, the derivatives are \(\frac{d^2s}{dt^2}\) (second order) and \(\frac{ds}{dt}\) (first order). The highest order is 2.


3. The degree of a differential equation is defined as the power of the highest order derivative after the equation is cleared of radicals and fractions.


4. In this equation, the highest order derivative \(\frac{d^2s}{dt^2}\) is raised to the power of 2.


5. Therefore, the order is 2 and the degree is 2.
Quick Tip: First find the highest order derivative to determine the order. Then, look only at its exponent to determine the degree. Ignore the powers of lower-order derivatives.

1. Put the differential equation in the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\).


2. Divide the entire equation by \((1+x^2)\): \(\frac{dy}{dx} + \frac{1}{1+x^2}y = \frac{e^{\tan^{-1} x}}{1+x^2}\).


3. Identify \(P(x) = \frac{1}{1+x^2}\).


4. The integrating factor (I.F.) is given by \(e^{\int P(x) dx}\).


5. Calculate the integral: \(\int \frac{1}{1+x^2} dx = \tan^{-1} x\).


6. Therefore, I.F. \(= e^{\tan^{-1} x}\).
Quick Tip: Always ensure the coefficient of \(\frac{dy}{dx}\) is 1 before identifying \(P(x)\). This is the step where most errors occur in I.F. calculations.

Given, \[ \frac{dy}{dx} = e^{x+y} \]

Using property of exponents: \[ e^{x+y} = e^x \cdot e^y \]

So the equation becomes: \[ \frac{dy}{dx} = e^x e^y \]

Step 1: Separate the variables
\[ e^{-y} dy = e^x dx \]

Step 2: Integrate both sides
\[ \int e^{-y} dy = \int e^x dx \]
\[ - e^{-y} = e^x + C \]

Step 3: Rearranging
\[ e^x + e^{-y} = -C \]

Let \(k = -C\) (arbitrary constant)
\[ \boxed{e^x + e^{-y} = k} \] Quick Tip: In variable separable differential equations, ensure all terms of one variable are moved to one side before integrating.

1. Separate the variables: \(\frac{1}{\cot y} dy = \frac{1}{x} dx\).


2. Simplify the trigonometric term: \(\tan y dy = \frac{1}{x} dx\).


3. Integrate both sides: \(\int \tan y dy = \int \frac{1}{x} dx\).


4. The integral results in: \(\log|\sec y| = \log|x| + \log|C|\).


5. Use log properties to combine: \(\log|\sec y| = \log|Cx|\).


6. This gives \(\sec y = Cx\), or rearranging, \(x \sec y = K\) where \(K = 1/C\).
Quick Tip: Constants of integration in log equations are best written as \(\log C\) to simplify the final algebraic form.

1. First, evaluate the expression inside the parentheses: \((2 * 3)\).


2. Using the definition \(a * b = a + 2b\), we substitute \(a = 2\) and \(b = 3\): \((2) + 2(3) = 2 + 6 = 8\).


3. Now, evaluate the result with the remaining term: \((8 * 4)\).


4. Substitute \(a = 8\) and \(b = 4\): \((8) + 2(4) = 8 + 8 = 16\).
Quick Tip: Always follow the order of operations by solving nested parentheses first in defined binary operations.

1. Calculate \(A^2\): \(A^2 = \begin{bmatrix} 1 & -1
-1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1
-1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & -1-1
-1-1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & -2
-2 & 2 \end{bmatrix}\).


2. Observe that \(A^2 = 2A\).


3. Now calculate \(A^3\): \(A^3 = A \cdot A^2 = A(2A) = 2(A^2)\).


4. Substitute \(A^2 = 2A\) into the expression: \(A^3 = 2(2A) = 4A\).
Quick Tip: For matrices with a repeating pattern, \(A^2\) often results in a scalar multiple of \(A\). Recognizing this pattern allows you to compute higher powers \(A^n\) using induction quickly.

1. Write the transpose \(A' = \begin{bmatrix} \cos \alpha & -\sin \alpha
\sin \alpha & \cos \alpha \end{bmatrix}\).


2. Perform the matrix addition \(A + A'\): \(\begin{bmatrix} \cos \alpha + \cos \alpha & \sin \alpha - \sin \alpha
-\sin \alpha + \sin \alpha & \cos \alpha + \cos \alpha \end{bmatrix} = \begin{bmatrix} 2 \cos \alpha & 0
0 & 2 \cos \alpha \end{bmatrix}\).


3. Equate this to the identity matrix \(I = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}\).


4. This gives the scalar equation: \(2 \cos \alpha = 1\).


5. Solve for \(\cos \alpha\): \(\cos \alpha = \frac{1}{2}\).


6. The principal value for \(\alpha\) such that \(\cos \alpha = \frac{1}{2}\) is \(\frac{\pi}{3}\).
Quick Tip: Transpose operations swap elements \(a_{12}\) and \(a_{21}\). When adding a matrix to its transpose, the diagonal elements always double.

1. For a \(2 \times 2\) matrix \(M = \begin{bmatrix} a & b
c & d \end{bmatrix}\), the adjoint is calculated by swapping the diagonal elements (\(a\) and \(d\)) and changing the signs of the off-diagonal elements (\(b\) and \(c\)).


2. Here, \(a = 3\), \(b = -5\), \(c = -1\), and \(d = 2\).


3. Swap \(a\) and \(d\): 2 moves to the \((1,1)\) position, and 3 moves to the \((2,2)\) position.


4. Change signs of \(b\) and \(c\): \(-5\) becomes \(+5\), and \(-1\) becomes \(+1\).


5. The resulting adjoint matrix is \(\begin{bmatrix} 2 & 5
1 & 3 \end{bmatrix}\).
Quick Tip: The "swap and flip sign" shortcut is only valid for \(2 \times 2\) matrices. For larger matrices, you must compute the transpose of the cofactor matrix.

1. Identify the orders of the matrices: \(A\) is \(1 \times 4\) and \(B\) is \(4 \times 1\).


2. Matrix multiplication is possible because the number of columns in \(A\) matches the number of rows in \(B\). The result will be a \(1 \times 1\) matrix.


3. Calculate the single element of the result: \((1 \times 1) + (2 \times 2) + (3 \times 3) + (4 \times 4)\).


4. Compute the sums: \(1 + 4 + 9 + 16 = 30\).


5. The result is the matrix \([30]\).
Quick Tip: The product of a row matrix and a column matrix is always a single scalar value enclosed in a matrix.

1. A matrix is defined as an arrangement of elements into rows and columns.


2. The dimensions of the matrix are given as \(m \times n\), where \(m\) is the number of rows and \(n\) is the number of columns.


3. By geometric definition, a square is a shape where the length and width are equal.


4. Similarly, a matrix is "square" if the number of rows is equal to the number of columns.


5. Therefore, the condition is \(m = n\).
Quick Tip: Operations like finding the determinant or the inverse of a matrix are only defined for square matrices.

1. Find the transpose \(A'\) by swapping rows and columns: \(A' = \begin{bmatrix} 3 & 1
-4 & -1 \end{bmatrix}\).


2. Perform the matrix addition \(A + A'\).


3. Add corresponding elements: \(\begin{bmatrix} 3+3 & -4+1
1+(-4) & -1+(-1) \end{bmatrix}\).


4. Calculate each term: \(\begin{bmatrix} 6 & -3
-3 & -2 \end{bmatrix}\).
Quick Tip: The sum of a matrix and its transpose always results in a symmetric matrix (where elements across the diagonal are identical).

1. In set theory, a function \(f\) maps elements of set \(A\) (domain) to set \(B\) (codomain).


2. The set of all actual output values is called the range, denoted by \(f(A)\).


3. An "onto" (surjective) function is one where the range equals the codomain: \(f(A) = B\).


4. An "into" function is defined as a function where the range is a proper subset of the codomain. This means at least one element in set \(B\) is not mapped to by any element in \(A\).


5. Symbolically, this is represented as \(f(A) \subset B\) (where \(f(A) \neq B\)).
Quick Tip: Remember: "Into" means the function fits completely inside the codomain without necessarily filling it. "Onto" means it covers the entire codomain.

1. Check Reflexive: For any \(a \in R\), is \(a < a\)? No, a number cannot be strictly less than itself. Not reflexive.


2. Check Symmetric: If \(a < b\), does it imply \(b < a\)? No, for example \(1 < 2\) is true, but \(2 < 1\) is false. Not symmetric.


3. Check Transitive: If \(a < b\) and \(b < c\), does it imply \(a < c\)? Yes, this is a fundamental property of inequality.


4. Since it is only transitive and not the other two, it cannot be an equivalence relation.
Quick Tip: Equality (=) is reflexive, symmetric, and transitive. Inequality (<, >) is only transitive. This distinction is fundamental in mathematical logic.

1. The magnitude of a vector \(\vec{v} = a\vec{i} + b\vec{j} + c\vec{k}\) is given by \(\sqrt{a^2 + b^2 + c^2}\).


2. Here, \(a = 3\), \(b = 4\), and \(c = 7\).


3. Calculate the sum of the squares: \(3^2 + 4^2 + 7^2 = 9 + 16 + 49\).


4. Summing the values: \(9 + 16 = 25\); \(25 + 49 = 74\).


5. The magnitude is \(\sqrt{74}\).
Quick Tip: Magnitudes represent the straight-line distance from the origin. Squaring each component treats the problem as a 3D extension of the Pythagorean theorem.

1. Evaluate the cross product inside the parentheses: \(\vec{j} \times \vec{k}\).


2. According to the cyclic property of basis unit vectors, \(\vec{j} \times \vec{k} = \vec{i}\).


3. Now, substitute this back into the expression: \(\vec{i} \cdot \vec{i}\).


4. The dot product of any unit vector with itself is 1 (since the angle between them is \(0^\circ\) and \(\cos 0^\circ = 1\)).


5. Therefore, the result is 1.
Quick Tip: The expression \(\vec{i} \cdot (\vec{j} \times \vec{k})\) is the volume of a unit cube with sides aligned with the axes. By definition, this volume is 1.

1. Identify the direction ratios of the two lines from the denominators: \(\vec{d_1} = (2, 7, -3)\) and \(\vec{d_2} = (-1, 2, 4)\).


2. Two lines are perpendicular if the sum of the products of their direction ratios is zero: \(a_1a_2 + b_1b_2 + c_1c_2 = 0\).


3. Calculate the sum: \((2)(-1) + (7)(2) + (-3)(4)\).


4. Multiply: \(-2 + 14 - 12\).


5. Total: \(14 - 14 = 0\).


6. Since the dot product of the direction vectors is 0, the angle between the lines is \(\frac{\pi}{2}\) (90 degrees).
Quick Tip: Before doing complex cosine calculations for angles, check if the dot product is zero. Perpendicularity is a very common feature in these exam questions.

1. Use the distance formula for a point \((x_1, y_1, z_1)\) and plane \(ax+by+cz-d=0\): \(D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\).


2. Rewrite the plane equation: \(x + 2y - 2z - 9 = 0\). Here \(a=1, b=2, c=-2, d=9\).


3. Substitute the point \((2, 3, -5)\): \(Numerator = |1(2) + 2(3) - 2(-5) - 9| = |2 + 6 + 10 - 9| = 9\).


4. Calculate the denominator: \(\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\).


5. Divide the results: \(D = \frac{9}{3} = 3\).
Quick Tip: The denominator of this distance formula is always the magnitude of the normal vector of the plane.

1. Two planes are perpendicular if their normal vectors are perpendicular.


2. The normal vectors are \(\vec{n_1} = (1, -4, \lambda)\) and \(\vec{n_2} = (2, 2, 3)\).


3. Perpendicularity condition: \(\vec{n_1} \cdot \vec{n_2} = 0\).


4. \((1)(2) + (-4)(2) + (\lambda)(3) = 0\).


5. \(2 - 8 + 3\lambda = 0\).


6. \(-6 + 3\lambda = 0 \implies 3\lambda = 6 \implies \lambda = 2\).
Quick Tip: Geometric conditions for planes (perpendicularity, parallelism) are always solved by treating their normal vectors as standard directional vectors.

1. A line is parallel to a plane if the line is perpendicular to the normal vector of the plane.


2. The direction ratios of the line are \((a_1, b_1, c_1)\).


3. The normal vector of the plane has direction ratios \((a_2, b_2, c_2)\).


4. For the line to be perpendicular to the normal, the dot product of these two vectors must be zero.


5. Therefore, \(a_1a_2 + b_1b_2 + c_1c_2 = 0\).
Quick Tip: Be careful: "Parallel" for line-plane means the vectors are "perpendicular". "Perpendicular" for line-plane means the vectors are "parallel".

1. The intercept form of a plane is \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).


2. Substitute the given intercepts: \(a = -2, b = 3, c = 4\).


3. The equation is \(\frac{x}{-2} + \frac{y}{3} + \frac{z}{4} = 1\).


4. Multiply all terms by the Least Common Multiple (LCM) of denominators, which is 12: \(12\left(\frac{x}{-2}\right) + 12\left(\frac{y}{3}\right) + 12\left(\frac{z}{4}\right) = 12(1)\).


5. Simplify: \(-6x + 4y + 3z = 12\).


6. Rearrange to standard form: \(6x - 4y - 3z + 12 = 0\).
Quick Tip: When converting intercept form to standard form, always multiply by the LCM of the intercepts to clear the fractions.

1. The direction cosines define unit vectors along the lines: \(\vec{u_1} = (l_1, m_1, n_1)\) and \(\vec{u_2} = (l_2, m_2, n_2)\).


2. The cosine of the angle \(\theta\) between two unit vectors is simply their dot product.


3. \(\cos \theta = \vec{u_1} \cdot \vec{u_2}\).


4. Expanding the dot product: \(\cos \theta = l_1l_2 + m_1m_2 + n_1n_2\).
Quick Tip: This formula is identical to the dot product of direction ratios, except here the denominators are not needed because the magnitudes of unit vectors are exactly 1.

1. First, solve the inner operation \((1 * 2)\).


2. Using \(a * b = a^3 + b^3\): \(1^3 + 2^3 = 1 + 8 = 9\).


3. Now, evaluate \(4 * 9\).


4. Substitute values: \(4^3 + 9^3 = 64 + 729\).


5. Summing the results: \(64 + 729 = 793\).
Quick Tip: Binary operations are often non-associative. Always calculate terms in the order indicated by the brackets.

1. The notation \(n(S)\) represents the cardinality (number of elements) of set \(S\).


2. The Cartesian product \(A \times B\) is the set of all possible ordered pairs where the first element is from set \(A\) and the second is from set \(B\).


3. The rule for the cardinality of a Cartesian product is \(n(A \times B) = n(A) \times n(B)\).


4. Substitute the given values: \(4 \times 2\).


5. The result is 8.
Quick Tip: If set \(A\) has \(m\) elements and set \(B\) has \(n\) elements, the total number of subsets (relations) from \(A\) to \(B\) is \(2^{mn}\), while the number of elements in the product is simply \(mn\).

1. Evaluate the determinant: \((a+ib)(a-ib) - (c+id)(-c+id)\).


2. Use the difference of squares identity \((x+y)(x-y) = x^2 - y^2\).


3. First part: \((a+ib)(a-ib) = a^2 - (ib)^2\). Since \(i^2 = -1\), this becomes \(a^2 - (-b^2) = a^2 + b^2\).


4. Second part: \((c+id)(-c+id) = (id+c)(id-c)\). This equals \((id)^2 - c^2 = -d^2 - c^2 = -(c^2 + d^2)\).


5. Subtract the second part from the first: \((a^2 + b^2) - [-(c^2 + d^2)] = a^2 + b^2 + c^2 + d^2\).
Quick Tip: The product of a complex number and its conjugate \((x+iy)(x-iy)\) is always \(x^2+y^2\). This is a recurring pattern in matrix determinants involving complex entries.

1. Identify the feasible region's corner points from the constraints \(x+y \le 50, x \ge 0, y \ge 0\).


2. The corner points are \((0, 0)\), \((50, 0)\), and \((0, 50)\).


3. Calculate the value of \(Z = 4x + y\) at each point:

- At \((0, 0)\): \(Z = 4(0) + 0 = 0\).

- At \((50, 0)\): \(Z = 4(50) + 0 = 200\).

- At \((0, 50)\): \(Z = 4(0) + 50 = 50\).


4. The maximum value among these is 200.


5. Comparing with the options: (A) 50, (B) 250, (C) 0. None of these matches 200.


6. Therefore, the answer is "none of these".
Quick Tip: For linear programming problems, the maximum or minimum value of the objective function always occurs at one of the vertices of the feasible region polygon.

1. Let the determinant be \(\Delta\).


2. Observe the relationship between the columns: \(12 + 11 = 23\), \(10 + 26 = 36\), and \(26 + 37 = 63\).


3. Apply the column transformation \(C_1 \to C_1 - (C_2 + C_3)\).


4. The first column becomes: \(23 - (12+11) = 0\), \(36 - (10+26) = 0\), and \(63 - (26+37) = 0\).


5. The determinant now has a column of zeros: \(\Delta = \begin{vmatrix} 0 & 12 & 11
0 & 10 & 26
0 & 26 & 37 \end{vmatrix}\).


6. By the properties of determinants, if any row or column consists entirely of zeros, the value of the determinant is 0.
Quick Tip: Always look for linear dependencies between rows or columns. Summing two columns to get a third is a very common trick in competitive exam determinants.

1. Let the given determinant be \(\Delta\).


2. Apply the row transformation \(R_1 \to R_1 + R_2 + R_3\).


3. The elements of the first row become:

\((a-b) + (b-c) + (c-a) = 0\)

\((b-c) + (c-a) + (a-b) = 0\)

\((c-a) + (a-b) + (b-c) = 0\)


4. The determinant now has an entire row of zeros in the first row.


5. According to determinant properties, if all elements in a row are zero, the determinant's value is 0.
Quick Tip: Cyclic expressions like \((a-b), (b-c), (c-a)\) often sum to zero. Performing \(R_1 + R_2 + R_3\) or \(C_1 + C_2 + C_3\) is the standard first step for such determinants.

1. Evaluate the \(2 \times 2\) determinant: \((x)(2x) - (4)(2) = 0\).


2. Simplify the equation: \(2x^2 - 8 = 0\).


3. Divide the entire equation by 2: \(x^2 - 4 = 0\).


4. This simplifies to \(x^2 = 4\).


5. Taking the square root of both sides gives \(x = \pm 2\).
Quick Tip: For a \(2 \times 2\) determinant \(\begin{vmatrix} a & b
c & d \end{vmatrix}\), the value is \(ad - bc\). Set this to zero and solve the resulting quadratic equation carefully.

1. Evaluate the determinant: \(\cos 15^\circ \cos 75^\circ - \sin 15^\circ \sin 75^\circ\).


2. Recognize the trigonometric identity: \(\cos(A+B) = \cos A \cos B - \sin A \sin B\).


3. Here, \(A = 15^\circ\) and \(B = 75^\circ\).


4. The expression simplifies to \(\cos(15^\circ + 75^\circ)\).


5. This equals \(\cos 90^\circ\).


6. The value of \(\cos 90^\circ\) is 0.
Quick Tip: Trigonometric determinants usually simplify into standard identities like \(\sin(A\pm B)\), \(\cos(A\pm B)\), or \(\cos^2 \theta + \sin^2 \theta\). Identifying these saves calculating specific values.

1. Use De Morgan's Law for sets: \(A' \cap B' = (A \cup B)'\).


2. Substitute this into the probability expression: \(1 - P((A \cup B)')\).


3. Recall the complement rule of probability: \(P(E) = 1 - P(E')\).


4. Here, let \(E = A \cup B\). Then \(1 - P(E') = P(E)\).


5. Therefore, \(1 - P(A' \cap B') = P(A \cup B)\).
Quick Tip: De Morgan's Laws are fundamental: "not A and not B" is equivalent to "not (A or B)". This logic applies directly to set-based probability operations.

1. By the definition of independent events, the occurrence of one does not affect the probability of the others.


2. The notation \(P(ABC)\) represents the joint probability of events \(A, B\), and \(C\) occurring together, i.e., \(P(A \cap B \cap C)\).


3. For independent events, the probability of intersection is equal to the product of their individual probabilities.


4. Therefore, \(P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)\).
Quick Tip: Independence translates to multiplication in probability. If events are mutually exclusive, their joint probability is 0 and their union probability is the sum.

1. The notation \(P(A/B)\) represents the conditional probability of \(A\) given that \(B\) has occurred.


2. Apply the formula for conditional probability: \(P(A/B) = \frac{P(A \cap B)}{P(B)}\).


3. Substitute the given values into the formula: \(P(A/B) = \frac{4/13}{9/13}\).


4. The denominators of 13 cancel each other out.


5. The resulting fraction is \(\frac{4}{9}\).
Quick Tip: \(P(A/B)\) effectively restricts the sample space to \(B\). Simply put the intersection value over the given value (the one after the slash).

1. "Odds against" is defined as the ratio of unfavorable outcomes to favorable outcomes.


2. Let the odds against event \(A\) be \(a:b\). Here \(a = 3\) and \(b = 7\).


3. The probability \(P(A)\) is calculated as the ratio of favorable outcomes (\(b\)) to total possible outcomes (\(a + b\)).


4. Substitute the values: \(P(A) = \frac{7}{3+7}\).


5. Therefore, \(P(A) = \frac{7}{10}\).
Quick Tip: If odds against are \(x:y\), probability is \(y/(x+y)\). If odds in favor are \(x:y\), probability is \(x/(x+y)\).

1. Use the logarithmic property to simplify the expression: \(\log x^n = n \log x\).


2. Now differentiate with respect to \(x\): \(\frac{d}{dx}(n \log x)\).


3. Since \(n\) is a constant, apply the rule \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\).


4. The derivative of \(\log x\) is \(\frac{1}{x}\).


5. Therefore, the derivative is \(n \cdot \frac{1}{x} = \frac{n}{x}\).
Quick Tip: Applying log properties (\(log a^b = b log a\)) before differentiating usually turns a chain rule problem into a simple constant multiple problem.

1. Find the first derivative \(\frac{d}{dx}(\sin 2x)\).


2. Using the chain rule: \(\cos(2x) \cdot \frac{d}{dx}(2x) = 2 \cos 2x\).


3. Find the second derivative by differentiating \(2 \cos 2x\).


4. Using the chain rule again: \(2 \cdot [-\sin(2x) \cdot \frac{d}{dx}(2x)]\).


5. This equals \(2 \cdot [-\sin(2x) \cdot 2] = -4 \sin 2x\).
Quick Tip: Differentiating \(\sin(ax)\) or \(\cos(ax)\) multiple times results in the coefficient \(a\) being raised to the power of the order of differentiation. Signs alternate every two derivatives.

1. Recognize the infinite structure: \(y = \sqrt{\sin x + y}\).


2. Square both sides to eliminate the outer radical: \(y^2 = \sin x + y\).


3. Differentiate implicitly with respect to \(x\): \(\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(y)\).


4. This gives \(2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}\).


5. Group terms with \(\frac{dy}{dx}\) on one side: \(2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x\).


6. Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(2y - 1) = \cos x\).


7. Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{\cos x}{2y-1}\).
Quick Tip: For any infinite radical series of the form \(y = \sqrt{f(x) + y}\), the derivative is always \(\frac{f'(x)}{2y-1}\).

1. Differentiate both sides of the equation \(x^n + y^n = a^n\) with respect to \(x\).


2. Apply the power rule and chain rule: \(n x^{n-1} + n y^{n-1} \frac{dy}{dx} = 0\) (since \(a^n\) is constant).


3. Move the \(x\) term to the other side: \(n y^{n-1} \frac{dy}{dx} = -n x^{n-1}\).


4. Divide both sides by \(n y^{n-1}\): \(\frac{dy}{dx} = \frac{-n x^{n-1}}{n y^{n-1}}\).


5. Simplify by cancelling \(n\): \(\frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}}\).
Quick Tip: For implicit functions of the form \(f(x) + g(y) = c\), the derivative is always \(\frac{dy}{dx} = -\frac{f'(x)}{g'(y)}\).

1. This is a parametric differentiation problem where \(x\) and \(y\) are functions of \(\theta\).


2. Find \(\frac{dx}{d\theta}\): \(a(0 - (-\sin \theta)) = a \sin \theta\).


3. Find \(\frac{dy}{d\theta}\): \(a(1 + \cos \theta)\).


4. Use the chain rule for parametric equations: \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 + \cos \theta)}{a \sin \theta}\).


5. Cancel \(a\) and apply trigonometric half-angle identities: \(1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}\) and \(\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\).


6. Substitute: \(\frac{dy}{dx} = \frac{2 \cos^2 (\theta/2)}{2 \sin (\theta/2) \cos (\theta/2)} = \frac{\cos (\theta/2)}{\sin (\theta/2)} = \cot \frac{\theta}{2}\).
Quick Tip: Parametric results often end in a simple trigonometric function. Mastering half-angle identities like \((1+\cos \theta)\) and \((1-\cos \theta)\) is crucial for these final simplifications.

1. For variable exponents, use logarithmic differentiation: Take \(\log\) on both sides: \(\log y = \log x^x\).


2. Use log properties: \(\log y = x \log x\).


3. Differentiate implicitly with respect to \(x\): \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log x)\).


4. Use the product rule on the right: \(x \cdot \frac{1}{x} + 1 \cdot \log x = 1 + \log x\).


5. Rearrange to find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y(1 + \log x)\).


6. Substitute \(y = x^x\) back into the result: \(\frac{dy}{dx} = x^x(1 + \log x)\).
Quick Tip: When the base and exponent both contain the variable \(x\), taking natural logs is the only way to differentiate. Never apply the power rule \(n x^{n-1}\) to functions like \(x^x\).

1. Recognize the trigonometric substitution pattern: Let \(x = \sin \theta\), then \(\theta = \sin^{-1} x\).


2. The expression inside the inverse function becomes: \(3 \sin \theta - 4 \sin^3 \theta\).


3. Recall the triple-angle identity: \(\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta\).


4. Thus, \(\sin^{-1}(3x - 4x^3) = \sin^{-1}(\sin 3\theta) = 3\theta\).


5. Substitute back for \(\theta\): \(3 \sin^{-1} x\).


6. Differentiate this result: \(\frac{d}{dx}(3 \sin^{-1} x) = 3 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{3}{\sqrt{1-x^2}}\).
Quick Tip: Substitution is much faster than the chain rule for inverse trig derivatives. Look for triple-angle (\(3x-4x^3\)) or double-angle (\(2x/(1+x^2)\)) forms.

1. Use the trigonometric product-to-sum identity: \(2 \sin A \cos B = \sin(A+B) + \sin(A-B)\).


2. Therefore, \(\sin 3x \cos 5x = \frac{1}{2}[\sin(3x+5x) + \sin(3x-5x)]\).


3. Simplify: \(\frac{1}{2}[\sin 8x + \sin(-2x)] = \frac{1}{2}[\sin 8x - \sin 2x]\).


4. Differentiate the simplified expression: \(\frac{d}{dx} \{ \frac{1}{2}(\sin 8x - \sin 2x) \}\).


5. Apply chain rule: \(\frac{1}{2} [8 \cos 8x - 2 \cos 2x]\).


6. Distribute the \(1/2\): \(4 \cos 8x - \cos 2x\).
Quick Tip: While you can use the product rule, converting products of trig functions to sums/differences simplifies the derivative and helps avoid messy algebra.

1. Apply the chain rule: \(\frac{d}{dx} \cos(f(x)) = -\sin(f(x)) \cdot f'(x)\).


2. Here, \(f(x) = x^3\).


3. Find the derivative of the inner function: \(f'(x) = 3x^2\).


4. Multiply: \(-\sin(x^3) \cdot 3x^2\).


5. Rearrange: \(-3x^2 \sin x^3\).
Quick Tip: Pay close attention to the exponent position. \(\cos x^3\) is a chain rule problem based on the angle, whereas \(\cos^3 x\) would be a chain rule problem based on the cosine power.

1. To find the inverse, start by setting \(y = f(x)\). So, \(y = 3x - 5\).


2. Solve the equation for \(x\) in terms of \(y\).


3. Add 5 to both sides: \(y + 5 = 3x\).


4. Divide both sides by 3: \(x = \frac{y+5}{3}\).


5. Now, replace \(x\) with \(f^{-1}(x)\) and \(y\) with \(x\): \(f^{-1}(x) = \frac{x+5}{3}\).


6. This can be rewritten as \(\frac{1}{3}(x+5)\).
Quick Tip: The steps to finding an inverse function are always the same: set \(y = f(x)\), swap \(x\) and \(y\), and solve for the new \(y\).

1. The integral of a sum is equal to the sum of the integrals: \(\int x dx + \int 2 dx\).


2. Apply the power rule \(\int x^n dx = \frac{x^{n+1}}{n+1}\) for the first term: \(\frac{x^{1+1}}{1+1} = \frac{x^2}{2}\).


3. Apply the rule \(\int k dx = kx\) for the second term: \(2x\).


4. Add the constant of integration \(k\).


5. The final result is \(\frac{x^2}{2} + 2x + k\).
Quick Tip: Don't forget the linear term \(2x\) when integrating \((x+2)\). Basic polynomial integration is the bedrock of calculus marks.

1. Use the double-angle trigonometric identity: \(\cos 2x = \cos^2 x - \sin^2 x\).


2. Factorize the numerator: \((\cos x - \sin x)(\cos x + \sin x)\).


3. Substitute into the integral: \(\int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} dx\).


4. Cancel the common factor \((\sin x + \cos x)\) from numerator and denominator: \(\int \frac{\cos x - \sin x}{\cos x + \sin x} dx\).


5. Let \(u = \sin x + \cos x\). Then differentiate to find \(du\): \(du = (\cos x - \sin x) dx\).


6. The integral simplifies to \(\int \frac{1}{u} du\).


7. This equals \(\log|u| + k\).


8. Substitute back \(u = \sin x + \cos x\): \(\log(\sin x + \cos x) + k\).
Quick Tip: Factoring numerator terms often reveals that they are the derivative of the denominator, leading to a quick logarithmic solution.

1. The given equation of the curve is \(x^2 + y^2 = 3\).


2. To find the slope, differentiate both sides with respect to \(x\): \(\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(3)\).


3. Using the power rule and chain rule: \(2x + 2y \frac{dy}{dx} = 0\).


4. Solve for \(\frac{dy}{dx}\): \(2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\).


5. Now, substitute the coordinates of the given point \((1, \sqrt{2})\) into the derivative expression.


6. \(m = \left. \frac{dy}{dx} \right|_{(1, \sqrt{2})} = -\frac{1}{\sqrt{2}}\).


7. Thus, the slope of the curve at the point \((1, \sqrt{2})\) is \(-\frac{1}{\sqrt{2}}\).
Quick Tip: The slope of a curve at a given point is always equal to the value of its first derivative, \(dy/dx\), at that specific point.

1. This is a parametric differentiation problem where both \(x\) and \(y\) are functions of the parameter \(t\).


2. Differentiate \(x\) with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(at^2) = a(2t) = 2at\).


3. Differentiate \(y\) with respect to \(t\): \(\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a(1) = 2a\).


4. Use the chain rule for parametric equations: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).


5. Substitute the calculated derivatives: \(\frac{dy}{dx} = \frac{2a}{2at}\).


6. Simplify the expression by cancelling the common factor \(2a\): \(\frac{dy}{dx} = \frac{1}{t}\).
Quick Tip: In parametric differentiation, always compute \(dx/dt\) and \(dy/dt\) separately before dividing them. Ensure that \(dx/dt\) is in the denominator.

1. We use the chain rule to differentiate the nested function \(y = \sec\{ \tan(\sqrt{x}) \}\).


2. Let \(u = \tan(\sqrt{x})\) and \(v = \sqrt{x}\). Then \(y = \sec u\), \(u = \tan v\), and \(v = \sqrt{x}\).


3. The derivative is \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}\).


4. Differentiate each layer:

\(\frac{dy}{du} = \sec u \tan u = \sec\{\tan(\sqrt{x})\} \tan\{\tan(\sqrt{x})\}\)

\(\frac{du}{dv} = \sec^2 v = \sec^2\sqrt{x}\)

\(\frac{dv}{dx} = \frac{1}{2\sqrt{x}}\)


5. Multiply the components together: \(\frac{dy}{dx} = \sec\{\tan(\sqrt{x})\} \tan\{\tan(\sqrt{x})\} \cdot \sec^2\sqrt{x} \cdot \frac{1}{2\sqrt{x}}\).


6. Final simplified form: \(\frac{dy}{dx} = \frac{\sec\{\tan(\sqrt{x})\}\tan\{\tan(\sqrt{x})\}\sec^2\sqrt{x}}{2\sqrt{x}}\).
Quick Tip: For chain rule problems with multiple functions, differentiate from the outermost function to the innermost, multiplying the derivatives as you go.

1. Recall the addition theorem of probability: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).


2. Substitute the given values into the formula: \(\frac{1}{2} = \frac{1}{4} + \frac{1}{3} - P(A \cap B)\).


3. Rearrange to solve for \(P(A \cap B)\): \(P(A \cap B) = \frac{1}{4} + \frac{1}{3} - \frac{1}{2}\).


4. Find a common denominator (12): \(P(A \cap B) = \frac{3}{12} + \frac{4}{12} - \frac{6}{12} = \frac{7 - 6}{12} = \frac{1}{12}\).


5. Now, check the condition for independence: \(P(A \cap B) = P(A) \cdot P(B)\).


6. Calculate the product of individual probabilities: \(P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}\).


7. Since \(P(A \cap B) = P(A) \cdot P(B) = \frac{1}{12}\), the events \(A\) and \(B\) are independent.
Quick Tip: To test for the independence of two events, always compare the probability of their intersection with the product of their individual probabilities.

1. Let event \(B\) be "the second die always exhibits 4".


2. The possible outcomes for event \(B\) are \(\{(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)\}\). Thus, \(n(B) = 6\).


3. Let event \(A\) be "the sum of the numbers is 8".


4. We need to find the conditional probability \(P(A|B)\), which is the probability of \(A\) given \(B\).


5. Identify the outcomes in the reduced sample space \(B\) that satisfy condition \(A\): only the pair \((4, 4)\) sums to 8.


6. Thus, \(n(A \cap B) = 1\).


7. Using the formula for conditional probability: \(P(A|B) = \frac{n(A \cap B)}{n(B)}\).


8. Substitute the values: \(P(A|B) = \frac{1}{6}\).
Quick Tip: In conditional probability, the "given" event becomes your new sample space. Look for outcomes within that restricted set that satisfy the target condition.

1. The objective function is \(Z = 3x + 2y\).


2. Identify the feasible region's corner points from the constraints \(3x + y \le 15, x \ge 0, y \ge 0\).


3. The intercepts for \(3x + y = 15\) are:

- When \(x = 0, y = 15 \implies (0, 15)\).

- When \(y = 0, 3x = 15 \implies x = 5 \implies (5, 0)\).


4. The intersection of \(x=0\) and \(y=0\) is the origin \((0, 0)\).


5. Evaluate \(Z\) at each vertex:

- At \((0, 0)\): \(Z = 3(0) + 2(0) = 0\).

- At \((5, 0)\): \(Z = 3(5) + 2(0) = 15\).

- At \((0, 15)\): \(Z = 3(0) + 2(15) = 30\).


6. The maximum value of \(Z\) is 30, achieved at point \((0, 15)\).
Quick Tip: For linear programming with linear constraints, the optimum value always lies on a corner of the feasible region. Calculate \(Z\) at all corners to find the maximum.

1. We know that \(\tan^{-1} 1 = \frac{\pi}{4}\).


2. Now, consider \(\tan^{-1} 2 + \tan^{-1} 3\). Since \(x=2\) and \(y=3\), we have \(xy = 6 > 1\).


3. For \(xy > 1\), the formula is: \(\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\).


4. Substitute \(x=2, y=3\): \(\pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right)\).


5. This simplifies to: \(\pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).


6. Now, sum all terms: \(\tan^{-1} 1 + (\tan^{-1} 2 + \tan^{-1} 3) = \frac{\pi}{4} + \frac{3\pi}{4}\).


7. The result is \(\frac{4\pi}{4} = \pi\). Hence proved.
Quick Tip: Be extremely careful with the \(\tan^{-1}\) sum identity. If the product of terms exceeds 1, you must add \(\pi\) to the basic formula to stay in the correct range.

1. First, simplify \(2 \tan^{-1} \frac{1}{3}\) using the formula \(2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)\).


2. \(2 \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{2/3}{1 - 1/9} \right) = \tan^{-1} \left( \frac{2/3}{8/9} \right) = \tan^{-1} \left( \frac{2}{3} \cdot \frac{9}{8} \right) = \tan^{-1} \frac{3}{4}\).


3. Now, add \(\tan^{-1} \frac{1}{7}\) using the formula \(\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)\).


4. Here \(xy = (3/4)(1/7) = 3/28 < 1\), so the basic formula applies.


5. \(\tan^{-1} \frac{3}{4} + \tan^{-1} \frac{1}{7} = \tan^{-1} \left( \frac{3/4 + 1/7}{1 - (3/4)(1/7)} \right) = \tan^{-1} \left( \frac{25/28}{1 - 3/28} \right)\).


6. This simplifies to \(\tan^{-1} \left( \frac{25/28}{25/28} \right) = \tan^{-1} 1\).


7. The final value is \(\frac{\pi}{4}\).
Quick Tip: When evaluating sums of multiple inverse tan terms, simplify coefficients of 2 first using double-angle formulas before performing addition.

1. Use the identities \(\sec^2 \theta = 1 + \tan^2 \theta\) and \(\csc^2 \phi = 1 + \cot^2 \phi\).


2. For the first term, let \(\theta = \tan^{-1} 2\). Then \(\tan \theta = 2\).


3. \(\sec^2(\tan^{-1} 2) = 1 + \tan^2(\tan^{-1} 2) = 1 + (2)^2 = 1 + 4 = 5\).


4. For the second term, let \(\phi = \cot^{-1} 3\). Then \(\cot \phi = 3\).


5. \(\csc^2(\cot^{-1} 3) = 1 + \cot^2(\cot^{-1} 3) = 1 + (3)^2 = 1 + 9 = 10\).


6. Summing the results: \(5 + 10 = 15\). Hence proved.
Quick Tip: Identities like \(\sec^2(\tan^{-1}x) = 1+x^2\) and \(\csc^2(\cot^{-1}x) = 1+x^2\) are extremely useful for simplifying compositions of trigonometric and inverse trigonometric functions.

1. Use the trigonometric identity \(1 + \cos x = 2 \cos^2 \frac{x}{2}\).


2. Substitute this into the integral: \(\int \frac{dx}{2 \cos^2 (x/2)}\).


3. Rewrite using the secant function: \(\frac{1}{2} \int \sec^2 \frac{x}{2} dx\).


4. We know that the integral of \(\sec^2 \theta\) is \(\tan \theta\). Apply this with the appropriate scale factor.


5. \(\frac{1}{2} \cdot \left[ \frac{\tan(x/2)}{1/2} \right] + C\).


6. The factors of \(1/2\) cancel out: \(\tan\frac{x}{2} + C\).
Quick Tip: Converting \((1 \pm \cos x)\) into half-angle forms (\(2\cos^2(x/2)\) or \(2\sin^2(x/2)\)) is a standard technique for eliminating denominators in trigonometric integrals.

1. Split the integrand: \(\int \sec^2 x \cdot \sec^2 x dx\).


2. Use the identity \(\sec^2 x = 1 + \tan^2 x\) for one factor: \(\int (1 + \tan^2 x) \sec^2 x dx\).


3. Perform substitution: let \(u = \tan x\). Then \(du = \sec^2 x dx\).


4. The integral becomes \(\int (1 + u^2) du\).


5. Integrate term by term: \(u + \frac{u^3}{3} + C\).


6. Substitute back \(u = \tan x\): \(\tan x + \frac{1}{3}\tan^3 x + C\).
Quick Tip: When the power of secant is even and positive, factor out \(\sec^2 x\) to use as \(du\) and convert the remaining secant terms into tangent terms.

1. Identify the relationship between the numerator and denominator: \(\frac{1}{\sqrt{1-x^2}}\) is the derivative of \(\sin^{-1} x\).


2. Perform substitution: let \(u = \sin^{-1} x\). Then \(du = \frac{1}{\sqrt{1-x^2}} dx\).


3. Substitute into the integral: \(\int u du\).


4. Apply the power rule for integration: \(\frac{u^2}{2} + C\).


5. Substitute back \(u = \sin^{-1} x\): \(\frac{(\sin^{-1} x)^2}{2} + C\).
Quick Tip: Always look for \(f(x)\) and \(f'(x)\) pairs within the integrand to simplify the problem via substitution.

1. Two planes are parallel if their normal vectors are proportional.


2. The normals are \((2, -1, 3)\) and \((6, -3, 9)\). Since \((6, -3, 9) = 3(2, -1, 3)\), the planes are parallel.


3. Normalize the second equation by dividing by 3: \(2x - y + 3z - 1 = 0\).


4. Now we have two planes in the form \(ax + by + cz + d_1 = 0\) and \(ax + by + cz + d_2 = 0\).


5. Here \(a=2, b=-1, c=3, d_1=4, d_2=-1\).


6. Use the distance formula: \(D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\).


7. Substitute values: \(D = \frac{|4 - (-1)|}{\sqrt{2^2 + (-1)^2 + 3^2}} = \frac{5}{\sqrt{4 + 1 + 9}} = \frac{5}{\sqrt{14}}\).
Quick Tip: Before using the distance formula for parallel planes, ensure that the coefficients of \(x, y, and z\) are identical in both equations.

1. Identify the direction ratios of the two lines from their denominators.


2. Line 1: \(\vec{a} = (-1, p, 2)\).


3. Line 2: \(\vec{b} = (p, 2, -3)\).


4. For two lines to be perpendicular, the dot product of their direction vectors must be zero: \(\vec{a} \cdot \vec{b} = 0\).


5. Apply the dot product formula: \((-1)(p) + (p)(2) + (2)(-3) = 0\).


6. Simplify the equation: \(-p + 2p - 6 = 0\).


7. Solve for \(p\): \(p - 6 = 0 \implies p = 6\).
Quick Tip: Perpendicularity condition \(a_1 a_2 + b_1 b_2 + c_1 c_2 = 0\) is the standard way to solve for unknown parameters in 3D geometry lines.

1. The notation \((f \circ f)(x)\) represents the composition of function \(f\) with itself, i.e., \(f(f(x))\).


2. Start with the definition \(f(x) = (3 - x^3)^{1/3}\).


3. Now substitute \(f(x)\) into the function again: \((f \circ f)(x) = [3 - (f(x))^3]^{1/3}\).


4. Substitute the expression for \(f(x)\): \([3 - \{(3 - x^3)^{1/3}\}^3]^{1/3}\).


5. Simplify the interior cube: \([3 - (3 - x^3)]^{1/3}\).


6. This simplifies to \([3 - 3 + x^3]^{1/3} = [x^3]^{1/3}\).


7. The cube root of \(x^3\) is \(x\). Thus, \((f \circ f)(x) = x\). Hence proved.
Quick Tip: When evaluating compositions like \(f(f(x))\), treat the first \(f(x)\) as a single block and substitute it carefully into the \(x\) position of the second function.

1. Let the determinant be \(\Delta\).


2. Observe the first row: \(102 = 6 \times 17\), \(18 = 6 \times 3\), and \(36 = 6 \times 6\).


3. We can factor out the constant 6 from the first row (\(R_1\)): \(\Delta = 6 \cdot \begin{vmatrix} 17 & 3 & 6
1 & 3 & 4
17 & 3 & 6 \end{vmatrix}\).


4. Now compare the rows of the new determinant. Row 1 (\(R_1\)) and Row 3 (\(R_3\)) are identical.


5. According to the properties of determinants, if any two rows or columns are identical, the value of the determinant is 0.


6. Therefore, \(\Delta = 6 \cdot 0 = 0\). Hence proved.
Quick Tip: Before expanding a determinant, always check for rows or columns that are multiples of each other. Proportional rows/columns immediately result in a zero determinant.

1. Expand the determinant along the first row (\(R_1\)).


2. \(\Delta = 0 \cdot \begin{vmatrix} 0 & b
a & 0 \end{vmatrix} - a \cdot \begin{vmatrix} c & b
c & 0 \end{vmatrix} + b \cdot \begin{vmatrix} c & 0
c & a \end{vmatrix}\).


3. Calculate the \(2 \times 2\) determinants:

- The first term is 0.

- The second term: \(-a(c \cdot 0 - c \cdot b) = -a(-bc) = abc\).

- The third term: \(b(c \cdot a - c \cdot 0) = b(ac) = abc\).


4. Sum the results: \(0 + abc + abc = 2abc\). Hence proved.
Quick Tip: Expansion along a row with a zero is the most efficient way to manually evaluate small determinants.

1. Evaluate the \(2 \times 2\) determinant: \((2x)(x) - (8)(4) = 0\).


2. This simplifies to \(2x^2 - 32 = 0\).


3. Move the constant to the other side: \(2x^2 = 32\).


4. Divide both sides by 2: \(x^2 = 16\).


5. Take the square root of both sides: \(x = \pm \sqrt{16} = \pm 4\).
Quick Tip: Always consider both the positive and negative roots when solving equations of the form \(x^2 = k\).

1. Start with the given equation \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\).


2. Take the cross product of the entire equation with \(\vec{a}\): \(\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{0}\).


3. Distribute the cross product: \((\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \vec{0}\).


4. Since \(\vec{a} \times \vec{a} = \vec{0}\), the equation becomes \(\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}\).


5. Rearranging gives: \(\vec{a} \times \vec{b} = -(\vec{a} \times \vec{c})\). By anti-commutativity, \(\vec{a} \times \vec{b} = \vec{c} \times \vec{a}\).


6. Similarly, take the cross product of the original equation with \(\vec{b}\): \(\vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \implies (\vec{b} \times \vec{a}) + \vec{0} + (\vec{b} \times \vec{c}) = \vec{0}\).


7. This gives \(\vec{b} \times \vec{c} = -(\vec{b} \times \vec{a}) = \vec{a} \times \vec{b}\).


8. Combining these, we have \(\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}\). Hence proved.
Quick Tip: When three vectors sum to zero, they form a triangle. The equality of their pairwise cross products corresponds to the fact that the area of the triangle can be calculated using any two sides.

1. Let the adjacent side vectors be \(\vec{a} = \vec{i} + 2\vec{j} + 3\vec{k}\) and \(\vec{b} = -3\vec{i} - 2\vec{j} + \vec{k}\).


2. The area of a parallelogram is given by the magnitude of the cross product of its adjacent sides: \(Area = |\vec{a} \times \vec{b}|\).


3. Calculate \(\vec{a} \times \vec{b}\) using a determinant: \(\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}
1 & 2 & 3
-3 & -2 & 1 \end{vmatrix}\).


4. Expand: \(\vec{i}(2+6) - \vec{j}(1+9) + \vec{k}(-2+6) = 8\vec{i} - 10\vec{j} + 4\vec{k}\).


5. Now find the magnitude: \(|\vec{a} \times \vec{b}| = \sqrt{8^2 + (-10)^2 + 4^2}\).


6. Calculate: \(\sqrt{64 + 100 + 16} = \sqrt{180}\).


7. Simplify the radical: \(\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}\).
Quick Tip: Area of a parallelogram = \(|\vec{a} \times \vec{b}|\). Area of a triangle = \(\frac{1}{2} |\vec{a} \times \vec{b}|\). Don't confuse the two formulas.

1. Let \(\vec{a} = 3\vec{i} + \vec{j} + 2\vec{k}\) and \(\vec{b} = 2\vec{i} - 2\vec{j} + 4\vec{k}\).


2. Use the cross product formula for sine: \(\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}\).


3. Magnitudes: \(|\vec{a}| = \sqrt{9+1+4} = \sqrt{14}\). \(|\vec{b}| = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6}\).


4. Cross Product: \(\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}
3 & 1 & 2
2 & -2 & 4 \end{vmatrix} = \vec{i}(4+4) - \vec{j}(12-4) + \vec{k}(-6-2) = 8\vec{i} - 8\vec{j} - 8\vec{k}\).


5. \(|\vec{a} \times \vec{b}| = \sqrt{64+64+64} = \sqrt{3 \times 64} = 8\sqrt{3}\).


6. Substitute into formula: \(\sin \theta = \frac{8\sqrt{3}}{\sqrt{14} \cdot \sqrt{24}} = \frac{8\sqrt{3}}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{4\sqrt{3}}{\sqrt{84}} = \frac{4\sqrt{3}}{2\sqrt{21}} = \frac{2\sqrt{3}}{\sqrt{3} \cdot \sqrt{7}} = \frac{2}{\sqrt{7}}\).
Quick Tip: You can also find \(\cos \theta\) first using dot product and then use \(\sin \theta = \sqrt{1-\cos^2\theta}\) if the cross product calculation seems too long.

1. Take natural logarithm on both sides: \(\sin y \log x = \sin x \log y\).


2. Differentiate both sides with respect to \(x\) using the product rule.


3. LHS: \(\cos y \frac{dy}{dx} \log x + \sin y \cdot \frac{1}{x}\).


4. RHS: \(\cos x \log y + \sin x \cdot \frac{1}{y} \frac{dy}{dx}\).


5. Equating them: \(\cos y \log x \frac{dy}{dx} + \frac{\sin y}{x} = \cos x \log y + \frac{\sin x}{y} \frac{dy}{dx}\).


6. Group terms with \(\frac{dy}{dx}\): \(\left( \cos y \log x - \frac{\sin x}{y} \right) \frac{dy}{dx} = \cos x \log y - \frac{\sin y}{x}\).


7. Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{\cos x \log y - (\sin y / x)}{\cos y \log x - (\sin x / y)}\).
Quick Tip: When variables appear in exponents on both sides, always use logarithmic differentiation followed by implicit differentiation.

1. Let \(A = \begin{bmatrix} 4 & 1
2 & 3 \end{bmatrix}\). First, find the determinant \(|A|\).


2. \(|A| = (4)(3) - (1)(2) = 12 - 2 = 10\). Since \(|A| \neq 0\), the inverse exists.


3. Find the adjoint of \(A\) (adj \(A\)). For a \(2 \times 2\) matrix, swap diagonal elements and flip signs of off-diagonal elements.


4. adj \(A = \begin{bmatrix} 3 & -1
-2 & 4 \end{bmatrix}\).


5. Use the formula \(A^{-1} = \frac{1}{|A|} adj A\).


6. \(A^{-1} = \frac{1}{10} \begin{bmatrix} 3 & -1
-2 & 4 \end{bmatrix}\).
Quick Tip: For any \(2 \times 2\) matrix, the inverse is \(\frac{1}{ad-bc} \begin{bmatrix} d & -b
-c & a \end{bmatrix}\). This is a crucial time-saving shortcut.

1. First, calculate \(4A\) by multiplying every element of \(A\) by 4.


2. \(4A = 4 \cdot \begin{bmatrix} 1 & 2
-3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 8
-12 & 16 \end{bmatrix}\).


3. Now, subtract matrix \(B\) from \(4A\).


4. \(4A - B = \begin{bmatrix} 4 & 8
-12 & 16 \end{bmatrix} - \begin{bmatrix} 0 & 7
2 & -4 \end{bmatrix}\).


5. Perform element-wise subtraction:

- \((1,1): 4 - 0 = 4\)

- \((1,2): 8 - 7 = 1\)

- \((2,1): -12 - 2 = -14\)

- \((2,2): 16 - (-4) = 16 + 4 = 20\).


6. The resulting matrix is \(\begin{bmatrix} 4 & 1
-14 & 20 \end{bmatrix}\).
Quick Tip: Scalar multiplication and subtraction are element-wise operations. Be careful with signs, especially when subtracting negative numbers.

1. The denominator is a quadratic expression. Complete the square: \(x^2 + 6x + 13 = (x+3)^2 - 9 + 13 = (x+3)^2 + 4\).


2. Rewrite the integral: \(\int \frac{dx}{(x+3)^2 + 2^2}\).


3. This is in the standard form \(\int \frac{dx}{u^2 + a^2}\), where \(u = x+3\) and \(a = 2\).


4. Use the formula \(\int \frac{dx}{u^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{u}{a} + C\).


5. Substitute values: \(\frac{1}{2} \tan^{-1}\left(\frac{x+3}{2}\right) + C\).
Quick Tip: Completing the square is the standard way to handle integrals of the form \(1/(quadratic)\). It converts them into standard inverse tan or log forms.

1. Perform substitution: Let \(u = a^2 + x^2\). Then \(du = 2x dx\), which means \(x dx = \frac{1}{2} du\).


2. Change the limits: When \(x = 0, u = a^2\). When \(x = a, u = a^2 + a^2 = 2a^2\).


3. Substitute into the integral: \(\int_{a^2}^{2a^2} \frac{(1/2) du}{\sqrt{u}} = \frac{1}{2} \int_{a^2}^{2a^2} u^{-1/2} du\).


4. Evaluate the antiderivative: \(\frac{1}{2} [2u^{1/2}]_{a^2}^{2a^2} = [\sqrt{u}]_{a^2}^{2a^2}\).


5. Apply limits: \(\sqrt{2a^2} - \sqrt{a^2} = a\sqrt{2} - a\).


6. Factor out \(a\): \(a(\sqrt{2} - 1)\).
Quick Tip: Substitution is much easier if you change the limits of integration immediately. This avoids the need to substitute back the original variables at the end.

1. Rewrite the equation using exponential notation: \(\frac{dy}{dx} = e^{ax + by}\).


2. Separate the variables: \(\frac{dy}{dx} = e^{ax} \cdot e^{by} \implies e^{-by} dy = e^{ax} dx\).


3. Integrate both sides: \(\int e^{-by} dy = \int e^{ax} dx\).


4. Calculate integrals: \(\frac{e^{-by}}{-b} = \frac{e^{ax}}{a} + K\).


5. Rearrange terms: \(\frac{e^{ax}}{a} + \frac{e^{-by}}{b} = -K\).


6. Let \(C = -K\) be the arbitrary constant: \(\frac{e^{ax}}{a} + \frac{e^{-by}}{b} = C\).
Quick Tip: For equations with \(\log(dy/dx)\), always convert to exponential form first to see if variables can be separated.

1. Use substitution: Let \(u = \log x\). Then \(du = \frac{1}{x} dx\).


2. Change limits: When \(x = 1, u = 0\). When \(x = 2, u = \log 2\).


3. Substitute: \(\int_0^{\log 2} u du\).


4. Integrate: \([\frac{u^2}{2}]_0^{\log 2}\).


5. Apply limits: \(\frac{(\log 2)^2}{2} - 0 = \frac{1}{2}(\log 2)^2\).
Quick Tip: Recognizing that \(1/x\) is the derivative of \(\log x\) immediately suggests the \(u\)-substitution method.

1. Let \(I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\).


2. Apply property \(\int_0^a f(x) = \int_0^a f(a-x)\): \(I = \int_0^{\pi/2} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\cos(\pi/2 - x)} + \sqrt{\sin(\pi/2 - x)}} dx\).


3. Simplify: \(I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\).


4. Add the two expressions for \(I\): \(2I = \int_0^{\pi/2} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\).


5. This becomes \(2I = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \pi/2\).


6. Thus, \(I = \pi/4\).
Quick Tip: For integrals of the form \(\int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} dx\), the answer is consistently \(\pi/4\).

1. This is a linear differential equation of form \(\frac{dy}{dx} + Py = Q\), where \(P = 2 \tan x\) and \(Q = \sin x\).


2. Calculate the Integrating Factor (I.F.) \(= e^{\int P dx} = e^{\int 2 \tan x dx} = e^{2 \log|\sec x|} = \sec^2 x\).


3. Multiply through by I.F. and integrate: \(y \cdot \sec^2 x = \int \sec^2 x \cdot \sin x dx\).


4. Simplify the integral: \(\int \frac{1}{\cos^2 x} \cdot \sin x dx = \int \sec x \tan x dx\).


5. The integral of \(\sec x \tan x\) is \(\sec x\).


6. Thus, \(y \sec^2 x = \sec x + C\).
Quick Tip: When solving linear differential equations, always identify \(P\) and \(Q\) correctly and calculate the I.F. as the very first step.