Step 1: Understanding the Concept:

The problem asks for the indefinite integral of a trigonometric expression.

Using fundamental trigonometric identities, we can simplify the integrand into a form that is easier to integrate.


Step 2: Key Formula or Approach:

We use the following identities:

1. \(1 - \cos x = 2\sin^2(x/2)\)

2. \(\csc x = \frac{1}{\sin x}\)

3. \(\sin x = 2\sin(x/2)\cos(x/2)\)


Step 3: Detailed Explanation:

Let the given integral be \(I\):
\[ I = \int (1 - \cos x)\csc^2 x \, dx \]

Since \(\csc^2 x = \frac{1}{\sin^2 x}\), we can rewrite the integral as:
\[ I = \int \frac{1 - \cos x}{\sin^2 x} \, dx \]

Substituting the half-angle identities for \(1 - \cos x\) and \(\sin x\):
\[ I = \int \frac{2\sin^2(x/2)}{(2\sin(x/2)\cos(x/2))^2} \, dx \]
\[ I = \int \frac{2\sin^2(x/2)}{4\sin^2(x/2)\cos^2(x/2)} \, dx \]

Cancelling the common term \(2\sin^2(x/2)\) in the numerator and denominator:
\[ I = \int \frac{1}{2\cos^2(x/2)} \, dx \]
\[ I = \frac{1}{2} \int \sec^2(x/2) \, dx \]

Using the standard integral \(\int \sec^2(u) du = \tan(u) + C\):
\[ I = \frac{1}{2} \left[ \frac{\tan(x/2)}{1/2} \right] + C \]
\[ I = \tan(x/2) + C \]


Step 4: Final Answer:

The value of the integral is \(\tan(x/2) + C\).
Quick Tip: Always look for half-angle simplifications when you see \((1 \pm \cos x)\). It often reduces the expression to a single squared trigonometric function.

Step 1: Understanding the Concept:

For a function to be monotonically increasing, its first derivative \(f'(x)\) must be greater than or equal to zero (\(f'(x) \ge 0\)) in the given interval.


Step 2: Key Formula or Approach:

The derivative of a polynomial \(ax^n\) is \(nax^{n-1}\).

We find the critical points by solving \(f'(x) = 0\) and use the sign-scheme (wavy curve) method.


Step 3: Detailed Explanation:

Given: \(f(x) = 2x^3 - 9x^2 + 12x + 29\)

Differentiating with respect to \(x\):
\[ f'(x) = 6x^2 - 18x + 12 \]

Factoring out the constant 6:
\[ f'(x) = 6(x^2 - 3x + 2) \]

Factorizing the quadratic expression:
\[ f'(x) = 6(x - 1)(x - 2) \]

To find where the function is increasing, set \(f'(x) > 0\):
\[ 6(x - 1)(x - 2) > 0 \]

Applying the sign scheme:

- Roots are \(x = 1\) and \(x = 2\).

- For \(x \in (-\infty, 1)\), \(f'(x) > 0\) (Positive).

- For \(x \in (1, 2)\), \(f'(x) < 0\) (Negative).

- For \(x \in (2, \infty)\), \(f'(x) > 0\) (Positive).

The function is monotonically increasing where \(f'(x) > 0\), which is \((-\infty, 1) \cup (2, \infty)\).


Step 4: Final Answer:

The interval is \((-\infty, 1) \cup (2, \infty)\).
Quick Tip: For any cubic function, the derivative is a parabola. If the leading coefficient is positive, the function always increases outside the roots of its derivative.

Step 1: Understanding the Concept:

"Ordinates and abscissa are equal" means \(x = y\). We need to find this point on the curve and then find the slope of the tangent at that point using differentiation.


Step 2: Key Formula or Approach:

1. Find intersection point with \(x = y\).

2. Slope \(m = \frac{dy}{dx}\).

3. Tangent equation: \(y - y_1 = m(x - x_1)\).


Step 3: Detailed Explanation:

First, find the point on the curve where \(x = y\):
\[ x = \sqrt{9 - 2x^2} \]

Squaring both sides: \(x^2 = 9 - 2x^2 \implies 3x^2 = 9 \implies x^2 = 3\).

Since \(y = \sqrt{...}\) implies \(y \ge 0\), and \(x = y\), we have \(x = \sqrt{3}\). So the point is \((\sqrt{3}, \sqrt{3})\).

Now, differentiate \(y = (9 - 2x^2)^{1/2}\):
\[ \frac{dy}{dx} = \frac{1}{2}(9 - 2x^2)^{-1/2} \cdot (-4x) = \frac{-2x}{\sqrt{9 - 2x^2}} \]

Substitute \(x = \sqrt{3}\) to find the slope \(m\):
\[ m = \frac{-2\sqrt{3}}{\sqrt{9 - 2(3)}} = \frac{-2\sqrt{3}}{\sqrt{3}} = -2 \]

Equation of tangent at \((\sqrt{3}, \sqrt{3})\) with slope \(m = -2\):
\[ y - \sqrt{3} = -2(x - \sqrt{3}) \]
\[ y - \sqrt{3} = -2x + 2\sqrt{3} \]
\[ 2x + y = 3\sqrt{3} \]


Step 4: Final Answer:

The equation of the tangent is \(2x + y = 3\sqrt{3}\).
Quick Tip: Notice that the curve is an ellipse. For any symmetric curve, the point where \(x=y\) is often a point of symmetry, making calculations easier.

Step 1: Understanding the Concept:

To find the minimum/maximum value of a rational function \(y = \frac{ax^2 + bx + c}{px^2 + qx + r}\), we can express it as a quadratic in \(x\) and use the discriminant condition \(D \ge 0\) for real \(x\).


Step 2: Key Formula or Approach:

Let \(y = \frac{x^2 - x + 1}{x^2 + x + 1}\).

For \(x \in \mathbb{R}\), the discriminant of the resulting quadratic in \(x\) must be non-negative.


Step 3: Detailed Explanation:
\[ y(x^2 + x + 1) = x^2 - x + 1 \]
\[ yx^2 + yx + y - x^2 + x - 1 = 0 \]
\[ (y - 1)x^2 + (y + 1)x + (y - 1) = 0 \]

Since \(x\) is real, \(D \ge 0\):
\[ (y + 1)^2 - 4(y - 1)(y - 1) \ge 0 \]
\[ (y + 1)^2 - 4(y - 1)^2 \ge 0 \]

Factoring as \(A^2 - B^2 = (A-B)(A+B)\):
\[ [(y + 1) - 2(y - 1)][(y + 1) + 2(y - 1)] \ge 0 \]
\[ (y + 1 - 2y + 2)(y + 1 + 2y - 2) \ge 0 \]
\[ (3 - y)(3y - 1) \ge 0 \]
\[ (y - 3)(3y - 1) \le 0 \]

This inequality holds for \(y \in [1/3, 3]\).

Thus, the minimum value is \(1/3\) and the maximum value is \(3\).


Step 4: Final Answer:

The minimum value is \(1/3\).
Quick Tip: For expressions like \(\frac{x^2 - x + 1}{x^2 + x + 1}\), the minimum and maximum are always reciprocal pairs (\(1/3\) and \(3\)).

Step 1: Understanding the Concept:

This question is identical to Question 4. We are finding the global minimum of the rational expression.


Step 2: Detailed Explanation:

Following the method from the previous problem:

Let \(y = \frac{x^2 - x + 1}{x^2 + x + 1}\).

Cross-multiplying and forming the quadratic in \(x\):
\[ (y-1)x^2 + (y+1)x + (y-1) = 0 \]

Discriminant condition for real roots \(D \ge 0\):
\[ (y+1)^2 - 4(y-1)^2 \ge 0 \]

Which simplifies to:
\[ (3y-1)(y-3) \le 0 \implies \frac{1}{3} \le y \le 3 \]

The smallest value of \(y\) is \(1/3\).


Step 3: Final Answer:

The minimum value is \(1/3\).
Quick Tip: In exams, if you see the same function repeatedly, remember its range. For \(x^2 \pm x + 1\), the range of the ratio is usually \([1/3, 3]\).

Step 1: Understanding the Concept:

Integration by parts or substitution can be used for integrals involving \(\log x\). Replacing \(\log x\) with a variable \(t\) converts it into a standard integral form.


Step 2: Key Formula or Approach:

Let \(\log x = t \implies x = e^t\) and \(dx = e^t dt\).

Use \(\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin bx - b \cos bx)\).


Step 3: Detailed Explanation:

Substitute \(\log x = t \implies dx = e^t dt\):
\[ I = \int \sin(t) e^t \, dt \]

This is a standard integral of the form \(\int e^{at} \sin(bt) dt\) with \(a=1, b=1\):
\[ I = \frac{e^t}{1^2 + 1^2} (1 \cdot \sin t - 1 \cdot \cos t) + C \]
\[ I = \frac{e^t}{2} (\sin t - \cos t) + C \]

Substituting back \(t = \log x\) and \(e^t = x\):
\[ I = \frac{x}{2} (\sin(\log x) - \cos(\log x)) + C \]


Step 4: Final Answer:

The solution is \((x/2)[\sin(\log x) - \cos(\log x)] + C\).
Quick Tip: The integral of \(\int e^x \sin x dx\) is cyclic. Memorize the result as \(\frac{e^x}{2}(\sin x - \cos x)\) to save time in such substitutions.

Step 1: Understanding the Concept:

We find the intersection of a line and a plane by expressing the line in parametric form and substituting it into the plane equation. The distance from the origin is then found using the distance formula \(\sqrt{x^2+y^2+z^2}\).


Step 2: Key Formula or Approach:

Point on line: \((2\lambda+1, 3\lambda-1, 4\lambda+2)\).

Substitute into plane equation \(x+2y+3z=15\).


Step 3: Detailed Explanation:

The coordinates of any point \(P\) on the line are:
\(x = 2\lambda + 1\), \(y = 3\lambda - 1\), \(z = 4\lambda + 2\).

Since \(P\) lies on the plane \(x + 2y + 3z = 15\):
\[ (2\lambda + 1) + 2(3\lambda - 1) + 3(4\lambda + 2) = 15 \]
\[ 2\lambda + 1 + 6\lambda - 2 + 12\lambda + 6 = 15 \]
\[ 20\lambda + 5 = 15 \implies 20\lambda = 10 \implies \lambda = 1/2 \]

Now, find the coordinates of \(P\):
\(x = 2(1/2) + 1 = 2\)
\(y = 3(1/2) - 1 = 1/2\)
\(z = 4(1/2) + 2 = 4\)

Distance \(OP\) from origin \((0,0,0)\):
\[ OP = \sqrt{2^2 + (1/2)^2 + 4^2} = \sqrt{4 + 0.25 + 16} \]
\[ OP = \sqrt{20.25} = 4.5 = 9/2 \]


Step 4: Final Answer:

The distance of point \(P\) from the origin is \(9/2\).
Quick Tip: Parametric form of a line is your best friend for intersection problems. Always set the line ratios equal to a parameter like \(\lambda\).

Step 1: Understanding the Concept:

This is a trigonometric equation with two variables. We look for conditions under which the maximum value or specific values are attained.


Step 2: Detailed Explanation:

Given: \(\cos x + \cos y - \cos(x+y) = 3/2\).

Let's test the condition \(x = y\):
\[ 2\cos x - \cos 2x = 3/2 \]
\[ 2\cos x - (2\cos^2 x - 1) = 3/2 \]
\[ -2\cos^2 x + 2\cos x + 1 = 3/2 \]
\[ -2\cos^2 x + 2\cos x - 1/2 = 0 \]
\[ 4\cos^2 x - 4\cos x + 1 = 0 \]

This is a perfect square:
\[ (2\cos x - 1)^2 = 0 \implies \cos x = 1/2 \]

This yields a valid solution for \(x\) and \(y\). Specifically, \(x = y = \pi/3\).

Since the expression is symmetric in \(x\) and \(y\), the relationship \(x = y\) is the defining condition.


Step 3: Final Answer:

The condition is \(x = y\).
Quick Tip: In competitive exams, if an equation is symmetric in \(x\) and \(y\), often the solution lies in \(x = y\). Checking specific values like \(x=y=60^\circ\) helps confirm the options quickly.

Step 1: Understanding the Concept:

The circumcentre is the point equidistant from all three vertices. It is the intersection of the perpendicular bisectors of the sides.


Step 2: Key Formula or Approach:

Let circumcentre be \(O(h,k)\). Then \(OA^2 = OB^2 = OC^2\).


Step 3: Detailed Explanation:

Vertices are \(A(-2,3), B(6,-1), C(4,3)\).

Notice \(A\) and \(C\) have the same \(y\)-coordinate (\(y=3\)). Side \(AC\) is horizontal.

The perpendicular bisector of \(AC\) must be vertical and pass through its midpoint.

Midpoint of \(AC = (\frac{-2+4}{2}, \frac{3+3}{2}) = (1,3)\).

So, the bisector of \(AC\) is the line \(x = 1\). The circumcentre must have \(x\)-coordinate \(h=1\).

Now, set distance from \(O(1,k)\) to \(A(-2,3)\) equal to distance to \(B(6,-1)\):
\[ (1 - (-2))^2 + (k - 3)^2 = (1 - 6)^2 + (k - (-1))^2 \]
\[ 3^2 + (k - 3)^2 = (-5)^2 + (k + 1)^2 \]
\[ 9 + k^2 - 6k + 9 = 25 + k^2 + 2k + 1 \]
\[ 18 - 6k = 26 + 2k \]
\[ -8 = 8k \implies k = -1 \]

The coordinates are \((1, -1)\).


Step 4: Final Answer:

The circumcentre is \((1, -1)\).
Quick Tip: Check if any vertices have the same \(x\) or \(y\) coordinates first. It simplifies finding perpendicular bisectors to vertical or horizontal lines, saving significant time.

Step 1: Understanding the Concept:

We use inverse trigonometric identities, particularly the relationship between \(\cos^{-1} x\) and \(\sin^{-1} x\).


Step 2: Key Formula or Approach:

1. \(\cos^{-1} \sqrt{1-p} = \sin^{-1} \sqrt{p}\) for \(p \in [0, 1]\).

2. \(\sin^{-1} x + \cos^{-1} x = \pi/2\).


Step 3: Detailed Explanation:

Let \(\cos^{-1} \sqrt{p} = \theta\). Then \(\cos \theta = \sqrt{p}\).

Then \(\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - p}\).

This implies \(\theta = \sin^{-1} \sqrt{1-p}\) or conversely \(\cos^{-1} \sqrt{1-p} = \sin^{-1} \sqrt{p}\).

Substituting this into the equation:
\[ \cos^{-1} \sqrt{p} + \sin^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-a} = 3\pi/4 \]

Using the identity \(\cos^{-1} x + \sin^{-1} x = \pi/2\):
\[ \pi/2 + \cos^{-1} \sqrt{1-a} = 3\pi/4 \]
\[ \cos^{-1} \sqrt{1-a} = 3\pi/4 - \pi/2 = \pi/4 \]

Taking cosine on both sides:
\[ \sqrt{1-a} = \cos(\pi/4) = 1/\sqrt{2} \]

Squaring both sides:
\[ 1 - a = 1/2 \implies a = 1/2 \]


Step 4: Final Answer:

The value of \(a\) is \(1/2\).
Quick Tip: Whenever you see inverse trig terms like \(\sqrt{p}\) and \(\sqrt{1-p}\), think of the identity \(\sin^{-1} x + \cos^{-1} x = \pi/2\). It almost always leads to a simplification.

Step 1: Understanding the Concept:

In a triangle, use the Law of Cosines to find the third side and then determine the largest angle, which always lies opposite the longest side.


Step 2: Key Formula or Approach:

Cosine Rule: \(a^2 = b^2 + c^2 - 2bc \cos A\).


Step 3: Detailed Explanation:

Calculate side \(a\):
\[ a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos 30^\circ \]
\[ a^2 = 3 + 1 - 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 4 - 3 = 1 \implies a = 1 \]

Since \(a=1\) and \(c=1\), the triangle is isosceles with \(a=c\).

Angles opposite equal sides are equal, so \(\angle C = \angle A = 30^\circ\).

Remaining angle \(\angle B = 180^\circ - (30^\circ + 30^\circ) = 120^\circ\).

The angles are \(30^\circ, 30^\circ, 120^\circ\). The largest is \(120^\circ\).


Step 4: Final Answer:

The largest angle is \(120^\circ\).
Quick Tip: If the third side comes out equal to one of the given sides, the triangle is isosceles. This instantly gives you the second angle, making the sum calculation very fast.

Step 1: Understanding the Concept:

The area of a parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\) is given by \(|\vec{a} \times \vec{b}|\). We use vector cross product properties to find the area of the new parallelogram.


Step 2: Key Formula or Approach:

1. \(Area_1 = |\vec{a} \times \vec{b}| = 16\).

2. Cross product property: \((p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)(\vec{a} \times \vec{b})\).


Step 3: Detailed Explanation:
The new adjacent sides are \(\vec{u} = 3\vec{a} + 2\vec{b}\) and \(\vec{v} = \vec{a} + 3\vec{b}\).

Area of new parallelogram \(A_2 = |\vec{u} \times \vec{v}|\).
\[ A_2 = |(3\vec{a} + 2\vec{b}) \times (\vec{a} + 3\vec{b})| \]

Expanding using distributive property:
\[ A_2 = |3(\vec{a} \times \vec{a}) + 9(\vec{a} \times \vec{b}) + 2(\vec{b} \times \vec{a}) + 6(\vec{b} \times \vec{b})| \]

Since \(\vec{v} \times \vec{v} = 0\) and \(\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})\):
\[ A_2 = |0 + 9(\vec{a} \times \vec{b}) - 2(\vec{a} \times \vec{b}) + 0| \]
\[ A_2 = |7(\vec{a} \times \vec{b})| = 7 |\vec{a} \times \vec{b}| \]

Substituting the original area:
\[ A_2 = 7 \times 16 = 112 \]


Step 4: Final Answer:

The area is \(112\) sq. units.
Quick Tip: For area transformation problems, calculate the determinant of the coefficients: \(|3(3) - 2(1)| = 7\). Multiply the original area by this determinant value.

Step 1: Understanding the Concept:

This is a first-order linear differential equation of the form \(dy/dx + P(x)y = Q(x)\). We solve it using an Integrating Factor (IF).


Step 2: Key Formula or Approach:

1. \(IF = e^{\int P(x) dx}\).

2. Solution: \(y \cdot IF = \int (Q(x) \cdot IF) dx + C\).


Step 3: Detailed Explanation:

Comparing the given equation with the standard form: \(P(x) = 1/x\) and \(Q(x) = \sin x\).
\[ IF = e^{\int \frac{1}{x} dx} = e^{\ln x} = x \]

The solution is:
\[ y \cdot x = \int (x \sin x) \, dx + C \]

Using integration by parts \(\int u dv = uv - \int v du\):

Let \(u = x, dv = \sin x dx \implies du = dx, v = -\cos x\).
\[ xy = -x \cos x - \int (-\cos x) dx + C \]
\[ xy = -x \cos x + \sin x + C \]

Rearranging the terms to match the answer key structure:
\[ xy + x \cos x - \sin x = C \]


Step 4: Final Answer:

The solution is \(xy + x \cos x - \sin x = C\).
Quick Tip: Notice that the left side \(dy/dx + y/x\) can be written as \(\frac{1}{x} \frac{d}{dx}(xy)\). Multiplying the whole equation by \(x\) at the start is often faster than calculating the IF separately.

Step 1: Understanding the Concept:

First, rationalize the complex number \(x\). Then, derive a quadratic equation satisfied by \(x\) to simplify the high-degree polynomial calculation.


Step 2: Detailed Explanation:

Rationalize \(x\):
\[ x = \frac{5}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{5(1 + 2i)}{1 - (2i)^2} = \frac{5(1 + 2i)}{5} = 1 + 2i \]

So, \(x - 1 = 2i\).

Squaring both sides:
\[ (x - 1)^2 = (2i)^2 \implies x^2 - 2x + 1 = -4 \implies x^2 - 2x + 5 = 0 \]

Now evaluate the polynomial \(P(x) = x^3 + x^2 - x + 22\).

Divide \(P(x)\) by \(x^2 - 2x + 5\):
\[ x^3 + x^2 - x + 22 = x(x^2 - 2x + 5) + 3x^2 - 6x + 22 \]
\[ = x(x^2 - 2x + 5) + 3(x^2 - 2x + 5) + 7 \]

Since \(x^2 - 2x + 5 = 0\):
\[ P(x) = x(0) + 3(0) + 7 = 7 \]


Step 3: Final Answer:

The value of the polynomial is \(7\).
Quick Tip: Instead of plugging in complex values directly, always find the quadratic equation the complex number satisfies. The polynomial value is simply the remainder of the division.

Step 1: Understanding the Concept:

This question repeats the logic from Question 3. We find the coordinate where \(x=y\), determine the slope of the curve at that point, and find the tangent line.


Step 2: Detailed Explanation:

Point of intersection: \(x = \sqrt{9 - 2x^2} \implies 3x^2 = 9 \implies x = \sqrt{3}\).

Since \(x=y\), the point is \((\sqrt{3}, \sqrt{3})\).

Differentiating \(y^2 = 9 - 2x^2\) implicitly:
\[ 2y \frac{dy}{dx} = -4x \implies \frac{dy}{dx} = - \frac{2x}{y} \]

At \((\sqrt{3}, \sqrt{3})\): \(m = -2(\sqrt{3}) / \sqrt{3} = -2\).

Tangent equation: \(y - \sqrt{3} = -2(x - \sqrt{3})\).
\[ 2x + y = 3\sqrt{3} \]


Step 3: Final Answer:

The equation is \(2x + y = 3\sqrt{3}\).
Quick Tip: Implicit differentiation is often faster than differentiating a square root function directly.