Let the Geometric Progression (G.P.) have first term \(a\) and common ratio \(r\). The number of terms is \(n = 64\).


The sum of all 64 terms is given by \(S_{total} = a + ar + ar^2 + \dots + ar^{63} = \frac{a(r^{64} - 1)}{r - 1}\).


The odd terms of the G.P. are the 1st, 3rd, 5th, ..., 63rd terms: \(a_1, a_3, a_5, \dots, a_{63}\).


This sequence is \(a, ar^2, ar^4, \dots, ar^{62}\), which is a G.P. with first term \(a\), common ratio \(r^2\), and number of terms \(\frac{64}{2} = 32\).


The sum of the odd terms is \(S_{odd} = \frac{a((r^2)^{32} - 1)}{r^2 - 1} = \frac{a(r^{64} - 1)}{r^2 - 1}\).


According to the problem, \(S_{total} = 7 \times S_{odd}\).


Substituting the formulas, we get \(\frac{a(r^{64} - 1)}{r - 1} = 7 \times \frac{a(r^{64} - 1)}{r^2 - 1}\).


Assuming \(a \neq 0\) and \(r \neq 1\), we can divide both sides by \(a(r^{64} - 1)\).

\(\frac{1}{r - 1} = \frac{7}{r^2 - 1}\).


Using the identity \(r^2 - 1 = (r - 1)(r + 1)\), the equation becomes \(\frac{1}{r - 1} = \frac{7}{(r - 1)(r + 1)}\).


Multiply both sides by \((r - 1)\) (since \(r \neq 1\)): \(1 = \frac{7}{r + 1}\).

\(r + 1 = 7\).

\(r = 6\).


The common ratio is 6.
Quick Tip: When selecting a sub-sequence from a G.P. (like odd or even terms), remember that the new common ratio is the original ratio raised to the power of the step size (e.g., step 2 gives \(r^2\)).

Let the first term of the A.P. be \(a\) and the common difference be \(d\).


Given the sixth term \(a_6 = a + 5d = 2\), we can express \(a\) as \(a = 2 - 5d\).


We want to maximize the product \(P = a_1 a_4 a_5\).


Express each term in terms of \(d\):

\(a_1 = a = 2 - 5d\).

\(a_4 = a + 3d = (2 - 5d) + 3d = 2 - 2d = 2(1 - d)\).

\(a_5 = a + 4d = (2 - 5d) + 4d = 2 - d\).


Substitute these into the product expression: \(P(d) = (2 - 5d) \cdot 2(1 - d) \cdot (2 - d)\).

\(P(d) = 2(1 - d)(2 - d)(2 - 5d)\).


Expand the terms to find the derivative easily. Let's expand \((1-d)(2-d) = 2 - 3d + d^2\).

\(P(d) = 2(2 - 3d + d^2)(2 - 5d)\).

\(P(d) = 2(4 - 10d - 6d + 15d^2 + 2d^2 - 5d^3) = 2(-5d^3 + 17d^2 - 16d + 4)\).


To find the critical points for maxima/minima, find \(\frac{dP}{ld}\):

\(\frac{dP}{dd} = 2(-15d^2 + 34d - 16)\).


Set \(\frac{dP}{dd} = 0\): \(-15d^2 + 34d - 16 = 0 \Rightarrow 15d^2 - 34d + 16 = 0\).


Using the quadratic formula \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(d = \frac{34 \pm \sqrt{34^2 - 4(15)(16)}}{30} = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30}\).


Two possible values: \(d_1 = \frac{48}{30} = \frac{8}{5}\) and \(d_2 = \frac{20}{30} = \frac{2}{3}\).


To check for maximum, use the second derivative test: \(\frac{d^2P}{dd^2} = 2(-30d + 34)\).


For \(d = \frac{8}{5}\): \(P''(\frac{8}{5}) = 2(-30(\frac{8}{5}) + 34) = 2(-48 + 34) = -28 < 0\). This indicates a local maximum.


For \(d = \frac{2}{3}\): \(P''(\frac{2}{3}) = 2(-30(\frac{2}{3}) + 34) = 2(-20 + 34) = 28 > 0\). This indicates a local minimum.


Therefore, the product is greatest when the common difference \(d = \frac{8}{5}\).
Quick Tip: When maximizing a product of terms in a sequence with a fixed term constraint, express all variables in terms of the common difference (or ratio) to reduce it to a single-variable calculus problem.

We need to find the range of the composite function \(y = f(g(x))\).


First, let's determine the range of the inner function \(g(x)\).


The domain of \(g(x)\) is \([-3, 1]\), split into two parts.


Part 1: For \(-3 \le x \le 0\), \(g(x) = -x\). As \(x\) varies from \(-3\) to \(0\), \(-x\) varies from \(3\) to \(0\). So, the range here is \([0, 3]\).


Part 2: For \(0 < x \le 1\), \(g(x) = x\). As \(x\) varies from just above \(0\) to \(1\), the range is \((0, 1]\).


The union of these two sets is \([0, 3] \cup (0, 1] = [0, 3]\). Thus, the range of \(g(x)\) is \(R_g = [0, 3]\).


Let \(u = g(x)\). Then \(u\) takes values in \([0, 3]\).


We now need to find the range of \(f(u)\) for \(u \in [0, 3]\).


The function \(f(u)\) is defined piece-wise. We check which definition applies to the interval \([0, 3]\).


For \(0 \le u \le 3\), the definition is \(f(u) = 1 - \frac{u}{3}\).


This is a linear decreasing function.


At the lower bound \(u = 0\), \(f(0) = 1 - \frac{0}{3} = 1\).


At the upper bound \(u = 3\), \(f(3) = 1 - \frac{3}{3} = 0\).


Since the function is continuous and monotonic on this interval, the values cover everything between \(0\) and \(1\).


Therefore, the range of \((f \circ g)(x)\) is \([0, 1]\).
Quick Tip: The range of a composite function \(f(g(x))\) is found by determining the range of the inner function \(g(x)\) first, and then using that set of values as the domain for the outer function \(f\).

Let success \(S\) be the event of getting a 2, and failure \(F\) be the event of getting any other number (1, 3, 4, 5, 6).


Probability of success \(p = P(S) = \frac{1}{6}\).


Probability of failure \(q = P(F) = 1 - \frac{1}{6} = \frac{5}{6}\).


We want the probability that the first success occurs on an even numbered throw (2nd, 4th, 6th, etc.).


The mutually exclusive outcomes are:


- Success on 2nd throw: \(F S\) (Probability \(q \times p\))


- Success on 4th throw: \(F F F S\) (Probability \(q^3 \times p\))


- Success on 6th throw: \(F F F F F S\) (Probability \(q^5 \times p\))


- And so on.


The total probability is the sum of this infinite geometric series: \(P = qp + q^3p + q^5p + \dots\)


Here, the first term \(a = qp = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}\).


The common ratio \(r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}\).


The sum of an infinite G.P. is given by \(S_{\infty} = \frac{a}{1 - r}\).

\(P = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}}\).

\(P = \frac{5}{11}\).
Quick Tip: Problems involving "success on the \(k\)-th trial" often lead to infinite geometric series. If success is required on even trials, the ratio will usually be \(q^2\) (fail twice).

We are given \(z = \frac{1}{2} - 2i\).


First, calculate the term on the Left Hand Side (LHS): \(|z+1|\).

\(z + 1 = \left(\frac{1}{2} - 2i\right) + 1 = \frac{3}{2} - 2i\).

\(|z+1| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}\).


Now, expand the term on the Right Hand Side (RHS): \(\alpha z + \beta(1+i)\).

\(RHS = \alpha\left(\frac{1}{2} - 2i\right) + \beta(1 + i)\).

\(RHS = \frac{\alpha}{2} - 2\alpha i + \beta + \beta i\).


Group the real and imaginary parts: \(RHS = \left(\frac{\alpha}{2} + \beta\right) + i(\beta - 2\alpha)\).


Equate the LHS and RHS: \(\frac{5}{2} = \left(\frac{\alpha}{2} + \beta\right) + i(\beta - 2\alpha)\).


Since the LHS is purely real, the imaginary part of the RHS must be zero.

\(\beta - 2\alpha = 0 \implies \beta = 2\alpha\).


Equate the real parts: \(\frac{5}{2} = \frac{\alpha}{2} + \beta\).


Substitute \(\beta = 2\alpha\) into the real part equation:

\(\frac{5}{2} = \frac{\alpha}{2} + 2\alpha\).

\(\frac{5}{2} = \frac{5\alpha}{2}\).

\(\alpha = 1\).


Now find \(\beta\): \(\beta = 2(1) = 2\).


We need to find the value of \(\alpha + \beta\).

\(\alpha + \beta = 1 + 2 = 3\).
Quick Tip: When an equation involves complex numbers and real variables, equate the real parts to the real parts and the imaginary parts to the imaginary parts to form a system of linear equations.

Let the limit be \(L\). The expression is of the form \(\frac{0}{0}\) as \(x \to \frac{\pi}{2}\).


Apply L'Hopital's Rule. We differentiate the numerator and the denominator with respect to \(x\).


For the numerator, use the Leibniz integral rule: \(\frac{d}{dx} \int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt = 0 - \cos((x^3)^{1/3}) \cdot \frac{d}{dx}(x^3) = -3x^2 \cos x\).


For the denominator: \(\frac{d}{dx} (x - \frac{\pi}{2})^2 = 2(x - \frac{\pi}{2})\).


The limit becomes \(L = \lim_{x \to \frac{\pi}{2}} \frac{-3x^2 \cos x}{2(x - \frac{\pi}{2})}\).


Let \(h = x - \frac{\pi}{2}\). As \(x \to \frac{\pi}{2}\), \(h \to 0\). Also \(x = \frac{\pi}{2} + h\).


Substitute into the limit: \(L = \lim_{h \to 0} \frac{-3(\frac{\pi}{2} + h)^2 \cos(\frac{\pi}{2} + h)}{2h}\).


Using \(\cos(\frac{\pi}{2} + h) = -\sin h\), we get \(L = \lim_{h \to 0} \frac{-3(\frac{\pi}{2} + h)^2 (-\sin h)}{2h}\).


Simplifying the expression: \(L = \frac{3}{2} \lim_{h \to 0} (\frac{\pi}{2} + h)^2 \cdot \frac{\sin h}{h}\).


Evaluating the limit as \(h \to 0\): \(L = \frac{3}{2} (\frac{\pi}{2})^2 \cdot 1 = \frac{3}{2} \cdot \frac{\pi^2}{4}\).

\(L = \frac{3\pi^2}{8}\).
Quick Tip: When applying L'Hopital's rule to an integral with variable limits, remember Leibniz's Rule: \(\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)dt = f(v(x))v'(x) - f(u(x))u'(x)\).

Since \(B(\alpha, \beta)\) lies on the angle bisector \(y = x\), we have \(\alpha = \beta\). Let \(B = (b, b)\).


The reflection of vertex \(A(4, 6)\) about the angle bisector \(y = x\) lies on the line containing the side \(BC\).


The reflection of \((4, 6)\) about \(y = x\) is point \(A' = (6, 4)\).


The line \(BC\) passes through \(B(b, b)\) and \(A'(6, 4)\). Its equation is \(y - 4 = \frac{b - 4}{b - 6}(x - 6)\).


Let \(D\) be the point where the bisector meets \(AC\). By the Angle Bisector Theorem, \(\frac{AB}{BC} = \frac{AD}{DC}\).


Given \(2AB = BC \Rightarrow \frac{AB}{BC} = \frac{1}{2}\), so \(D\) divides \(AC\) in the ratio \(1:2\).


Find \(D\): Intersection of bisector \(y = x\) and side \(AC\) \(2x - y = 2\). \(2x - x = 2 \Rightarrow x = 2, y = 2\). So \(D(2, 2)\).


Using the section formula for \(D\) dividing \(A(4, 6)\) and \(C(x_c, y_c)\) in ratio \(1:2\):

\(2 = \frac{1(x_c) + 2(4)}{3} \Rightarrow 6 = x_c + 8 \Rightarrow x_c = -2\).

\(2 = \frac{1(y_c) + 2(6)}{3} \Rightarrow 6 = y_c + 12 \Rightarrow y_c = -6\). So \(C(-2, -6)\).


The line \(BC\) passes through \(C(-2, -6)\) and \(A'(6, 4)\).


Slope of \(BC = \frac{4 - (-6)}{6 - (-2)} = \frac{10}{8} = \frac{5}{4}\).


Equation of \(BC\): \(y - 4 = \frac{5}{4}(x - 6) \Rightarrow 4y - 16 = 5x - 30 \Rightarrow 5x - 4y = 14\).


Since \(B(b, b)\) lies on this line: \(5b - 4b = 14 \Rightarrow b = 14\).


Thus, \(\alpha = 14\) and \(\beta = 14\).

\(\alpha + 2\beta = 14 + 2(14) = 14 + 28 = 42\).
Quick Tip: The image of a vertex about the internal angle bisector of that vertex angle lies on the opposite side.

Since \(\vec{a} + 5\vec{b}\) is collinear with \(\vec{c}\), we can write \(\vec{a} + 5\vec{b} = \lambda \vec{c}\) for some scalar \(\lambda\).


Since \(\vec{b} + 6\vec{c}\) is collinear with \(\vec{a}\), we can write \(\vec{b} + 6\vec{c} = \mu \vec{a}\) for some scalar \(\mu\).


From the first equation, \(\vec{a} = \lambda \vec{c} - 5\vec{b}\). Substitute this into the second equation.

\(\vec{b} + 6\vec{c} = \mu(\lambda \vec{c} - 5\vec{b}) = \mu\lambda \vec{c} - 5\mu \vec{b}\).


Rearranging terms: \((1 + 5\mu)\vec{b} + (6 - \mu\lambda)\vec{c} = \vec{0}\).


Since \(\vec{b}\) and \(\vec{c}\) are non-collinear, the coefficients must be zero.

\(1 + 5\mu = 0 \Rightarrow \mu = -\frac{1}{5}\).

\(6 - \mu\lambda = 0 \Rightarrow 6 - (-\frac{1}{5})\lambda = 0 \Rightarrow \lambda = -30\).


Substitute \(\lambda = -30\) back into the first equation: \(\vec{a} + 5\vec{b} = -30\vec{c}\).

\(\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}\).


Comparing this with the given equation \(\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}\), we get \(\alpha = 5\) and \(\beta = 30\).

\(\alpha + \beta = 5 + 30 = 35\).
Quick Tip: For two non-collinear vectors \(\vec{x}\) and \(\vec{y}\), the equation \(c_1 \vec{x} + c_2 \vec{y} = \vec{0}\) implies \(c_1 = 0\) and \(c_2 = 0\).

Consider the coordinates of the vertices. \(A(a, -2)\) and \(B(a, 6)\) have the same x-coordinate, so side \(AB\) is vertical.

\(A(a, -2)\) and \(C(\frac{a}{4}, -2)\) have the same y-coordinate, so side \(AC\) is horizontal.


Since \(AB\) is vertical and \(AC\) is horizontal, \(\angle A = 90^\circ\). Thus \(\triangle ABC\) is a right-angled triangle.


For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse \(BC\).


Midpoint of \(BC\): \(M = \left( \frac{a + a/4}{2}, \frac{6 + (-2)}{2} \right) = \left( \frac{5a}{8}, 2 \right)\).


We are given the circumcenter is \((5, \frac{a}{4})\). Equating the coordinates:

\(\frac{a}{4} = 2 \Rightarrow a = 8\).


Check x-coordinate: \(\frac{5(8)}{8} = 5\). This matches.


With \(a=8\), vertices are \(A(8, -2)\), \(B(8, 6)\), \(C(2, -2)\).


Length of sides: \(AB = |6 - (-2)| = 8\). \(AC = |8 - 2| = 6\).


Hypotenuse \(BC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10\).


Circumradius \(\alpha = \frac{BC}{2} = \frac{10}{2} = 5\).


Area \(\beta = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 8 \times 6 = 24\).


Perimeter \(\gamma = AB + AC + BC = 8 + 6 + 10 = 24\).


Sum \(\alpha + \beta + \gamma = 5 + 24 + 24 = 53\).
Quick Tip: In a right-angled triangle, the vertices can be easily identified by checking for constant x or y coordinates between pairs of points.

Simplify the integrand term by term. Numerator: \(\csc x + \sin x = \frac{1}{\sin x} + \sin x = \frac{1+\sin^2 x}{\sin x}\).


Denominator: \(\csc x \sec x + \tan x \sin^2 x = \frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x} = \frac{1+\sin^4 x}{\sin x \cos x}\).


Integrand: \(\frac{\frac{1+\sin^2 x}{\sin x}}{\frac{1+\sin^4 x}{\sin x \cos x}} = \frac{1+\sin^2 x}{\sin x} \cdot \frac{\sin x \cos x}{1+\sin^4 x} = \frac{(1+\sin^2 x)\cos x}{1+\sin^4 x}\).


Let \(u = \sin x\), then \(du = \cos x dx\). The integral becomes \(I = \int \frac{1+u^2}{1+u^4} du\).


Divide numerator and denominator by \(u^2\): \(I = \int \frac{1 + \frac{1}{u^2}}{u^2 + \frac{1}{u^2}} du\).


Rewrite denominator as \((u - \frac{1}{u})^2 + 2\). Let \(t = u - \frac{1}{u}\). Then \(dt = (1 + \frac{1}{u^2})du\).

\(I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + C\).


Substitute back \(t = \sin x - \frac{1}{\sin x} = \frac{\sin^2 x - 1}{\sin x}\).

\(y(x) = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{\sin^2 x - 1}{\sqrt{2} \sin x} \right) + C\).


Given limit \(x \to -\frac{\pi}{2}\), \(\sin x \to -1\). The argument approaches \(\frac{1-1}{-\sqrt{2}} = 0\). So \(0 = \frac{1}{\sqrt{2}}(0) + C \Rightarrow C = 0\).


For \(y(\frac{\pi}{4})\), \(\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\). Argument is \(\frac{\frac{1}{2}-1}{\sqrt{2} \cdot \frac{1}{\sqrt{2}}} = \frac{-1/2}{1} = -\frac{1}{2}\).

\(y(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \tan^{-1}\left( -\frac{1}{2} \right)\).
Quick Tip: Standard integral form \(\int \frac{x^2+1}{x^4+1} dx\) is solved by dividing by \(x^2\) and completing the square in the denominator.

Let \(t = \tan \frac{\alpha}{2}\). Then \(\cos \alpha = \frac{1-t^2}{1+t^2}\) and \(\sin \alpha = \frac{2t}{1+t^2}\).


Substitute into \(4 \cos \alpha + 5 \sin \alpha = 1\):

\(4\left(\frac{1-t^2}{1+t^2}\right) + 5\left(\frac{2t}{1+t^2}\right) = 1\).


Multiply by \(1+t^2\): \(4(1-t^2) + 10t = 1 + t^2\).

\(4 - 4t^2 + 10t = 1 + t^2 \Rightarrow 5t^2 - 10t - 3 = 0\).


Roots are \(t = \frac{10 \pm \sqrt{100 - 4(5)(-3)}}{10} = \frac{10 \pm \sqrt{160}}{10} = 1 \pm \frac{4\sqrt{10}}{10} = 1 \pm \frac{2\sqrt{10}}{5}\).


Given \(-\frac{\pi}{2} < \alpha < \frac{\pi}{2}\), we have \(-1 < t < 1\).


Since \(1 + \frac{2\sqrt{10}}{5} > 1\), we reject it. So \(t = 1 - \frac{2\sqrt{10}}{5} = \frac{5 - 2\sqrt{10}}{5}\).


We need \(\tan \alpha = \frac{2t}{1-t^2}\). From the quadratic, \(1-t^2 = 1 - (2t + \frac{3}{5}) = \frac{2}{5} - 2t\).

\(\tan \alpha = \frac{2t}{\frac{2}{5} - 2t} = \frac{t}{\frac{1}{5} - t} = \frac{5t}{1 - 5t}\).


Substitute \(5t = 5 - 2\sqrt{10}\):

\(\tan \alpha = \frac{5 - 2\sqrt{10}}{1 - (5 - 2\sqrt{10})} = \frac{5 - 2\sqrt{10}}{2\sqrt{10} - 4}\).


Rationalize the denominator by multiplying by \(\frac{\sqrt{10} + 2}{\sqrt{10} + 2}\):


Numerator: \((5 - 2\sqrt{10})(2 + \sqrt{10}) = 10 + 5\sqrt{10} - 4\sqrt{10} - 20 = \sqrt{10} - 10\).


Denominator: \(2(\sqrt{10} - 2)(\sqrt{10} + 2) = 2(10 - 4) = 12\).

\(\tan \alpha = \frac{\sqrt{10} - 10}{12}\).
Quick Tip: Converting trigonometric equations into algebraic equations using \(\tan(\theta/2)\) is a reliable method for finding specific values.

Rewrite the differential equation in standard linear form \(\frac{dy}{dx} + Py = Q\).

\((1 + \cos^2 x)\frac{dy}{dx} - (\sin 2x)y = \sin x\).

\(\frac{dy}{dx} - \frac{2\sin x \cos x}{1+\cos^2 x}y = \frac{\sin x}{1+\cos^2 x}\).


Integrating Factor (IF) = \(e^{\int -\frac{2\sin x \cos x}{1+\cos^2 x} dx}\). Let \(u = 1+\cos^2 x\), \(du = -2\sin x \cos x dx\).


IF \(= e^{\int \frac{du}{u}} = e^{\ln u} = u = 1+\cos^2 x\).


Solution is given by \(y \cdot (IF) = \int Q \cdot (IF) dx\).

\(y(1+\cos^2 x) = \int \frac{\sin x}{1+\cos^2 x} \cdot (1+\cos^2 x) dx = \int \sin x dx\).

\(y(1+\cos^2 x) = -\cos x + C\).


Apply initial condition \(f(0) = 0\): \(0(1+1) = -1 + C \Rightarrow C = 1\).

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


At \(x = \frac{\pi}{2}\): \(f(\frac{\pi}{2}) = \frac{1 - 0}{1 + 0} = 1\).
Quick Tip: Identify linear differential equations \(\frac{dy}{dx} + Py = Q\) and calculate the integrating factor carefully.

Calculate the lengths of vectors \(\vec{OA}\) and \(\vec{OB}\).

\(|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3\).

\(|\vec{OB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{36} = 6\).


By the Angle Bisector Theorem, point \(C\) divides \(AB\) in the ratio \(OA:OB = 3:6 = 1:2\).


Using the section formula, \(\vec{c} = \frac{1\vec{b} + 2\vec{a}}{1+2}\).

\(\vec{c} = \frac{(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3}\).

\(\vec{c} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}\).


Length \(|\vec{OC}| = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}}\).

\(|\vec{OC}| = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 \times 34}}{3} = \frac{2\sqrt{34}}{3}\).
Quick Tip: The internal bisector of an angle in a triangle divides the opposite side in the ratio of the lengths of the adjacent sides.

To find the intersection with the x-axis, set \(f(x) = 0\).

\(4\sqrt{2}x^3 - 3\sqrt{2}x = 1 \Rightarrow \sqrt{2}(4x^3 - 3x) = 1 \Rightarrow 4x^3 - 3x = \frac{1}{\sqrt{2}}\).


This looks like the triple angle formula for cosine: \(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\).


Let \(x = \cos \theta\). Then \(\cos 3\theta = \frac{1}{\sqrt{2}}\).


The general solutions are \(3\theta = 2n\pi \pm \frac{\pi}{4}\).

\(\theta = \frac{2n\pi}{3} \pm \frac{\pi}{12}\).


We need \(x \in [\frac{1}{2}, 1]\), which corresponds to \(\theta \in [0, \frac{\pi}{3}]\).


For \(n=0\), \(\theta = \frac{\pi}{12} = 15^\circ\). This is in \([0, 60^\circ]\). So \(x = \cos \frac{\pi}{12}\) is a root. Statement (II) is correct.


For \(n=0\), \(\theta = -\frac{\pi}{12}\) (same x).


For \(n=1\), \(\theta = \frac{2\pi}{3} \pm \frac{\pi}{12}\), which are outside the range \([0, \frac{\pi}{3}]\).


Check monotonicity: \(f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} = 3\sqrt{2}(4x^2 - 1)\).


For \(x \in [\frac{1}{2}, 1]\), \(4x^2 - 1 \ge 0\), so \(f(x)\) is strictly increasing.


Since the function is monotonic in the interval, it can have at most one root. Since we found one, it has exactly one. Statement (I) is correct.
Quick Tip: Recognize algebraic structures like \(4x^3-3x\) as trigonometric identities to simplify finding roots.

For a square matrix \(A\) of order \(n=3\), \(|kA| = k^3 |A|\).


Here, \(|2A| = 2^3 |A| = 8 |A|\).


The given condition is \(|2A|^3 = 2^{21}\).

\((8|A|)^3 = 2^{21} \Rightarrow (2^3 |A|)^3 = 2^{21} \Rightarrow 2^9 |A|^3 = 2^{21}\).

\(|A|^3 = \frac{2^{21}}{2^9} = 2^{12}\).


Taking the cube root, \(|A| = 2^4 = 16\).


Calculate the determinant of \(A\): \(|A| = 1(\alpha^2 - \beta^2) - 0 + 0 = \alpha^2 - \beta^2\).


So, \(\alpha^2 - \beta^2 = 16 \Rightarrow (\alpha - \beta)(\alpha + \beta) = 16\).


We look for integer factors of 16. Also, \((\alpha-\beta)\) and \((\alpha+\beta)\) must have the same parity (both even or both odd) because their sum is \(2\alpha\) (even). Since product is 16 (even), both must be even.


Factor pairs of 16 (both even): \((2, 8), (4, 4), (-8, -2)\), etc.


Case 1: \(\alpha - \beta = 2\) and \(\alpha + \beta = 8\). Summing gives \(2\alpha = 10 \Rightarrow \alpha = 5\).


Case 2: \(\alpha - \beta = 4\) and \(\alpha + \beta = 4\). Summing gives \(2\alpha = 8 \Rightarrow \alpha = 4\).


Checking options, \(\alpha = 5\) is available.
Quick Tip: Determinant property \(|kA| = k^n|A|\) is essential. Also, for integer equations \(x^2 - y^2 = k\), factors must have the same parity.

Let \(G\) be the centroid. The centroid is the average of vertices: \(G = \frac{P + Q + R}{3}\).


Since \(M\) is the midpoint of \(PQ\), \(P + Q = 2M\).


So, \(G = \frac{2M + R}{3} = \frac{2(2, 1, 2) + (-1, 4, 2)}{3} = \frac{(4, 2, 4) + (-1, 4, 2)}{3}\).

\(G = \frac{(3, 6, 6)}{3} = (1, 2, 2)\).


Now find the intersection point \(I\) of the two lines.


Line 1: \(x = 2\), \(y = 2\lambda\), \(z = -3 - \lambda\).


Line 2: \(x = 1 + \mu\), \(y = -3 - 3\mu\), \(z = -1 + \mu\).


Equate x-coordinates: \(2 = 1 + \mu \Rightarrow \mu = 1\).


Substitute \(\mu = 1\) into y equation: \(2\lambda = -3 - 3(1) = -6 \Rightarrow \lambda = -3\).


Check z-coordinates: Line 1 gives \(-3 - (-3) = 0\). Line 2 gives \(-1 + 1 = 0\). They match.


Point of intersection \(I(2, -6, 0)\).


Distance \(GI = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2} = \sqrt{1^2 + (-8)^2 + (-2)^2}\).

\(GI = \sqrt{1 + 64 + 4} = \sqrt{69}\).
Quick Tip: For centroid problems, expressing the sum of two vertices as twice their midpoint simplifies calculations.

Reflexivity: Consider \((a, b)R(a, b)\). The condition is \(ab - ba\), which is \(0\). \(0\) is divisible by 5. So, it is Reflexive.


Symmetry: Assume \((a, b)R(c, d)\). Then \(ad - bc = 5k\).


Check \((c, d)R(a, b)\). The condition is \(cb - da = -(ad - bc) = -5k\). This is divisible by 5. So, it is Symmetric.


Transitivity: Let \((a, b) = (1, 0)\), \((c, d) = (5, 5)\), \((e, f) = (0, 1)\).

\((1, 0)R(5, 5) \Rightarrow 1(5) - 0(5) = 5\) (Divisible by 5).

\((5, 5)R(0, 1) \Rightarrow 5(1) - 5(0) = 5\) (Divisible by 5).


Check \((1, 0)R(0, 1) \Rightarrow 1(1) - 0(0) = 1\). This is not divisible by 5.


Therefore, it is not Transitive.
Quick Tip: Use simple numerical counterexamples to test transitivity properties for specific relations.

Let \(I = I_1 + I_2\). Use the property \(\int_{-a}^{a} f(x) dx = \int_0^a (f(x) + f(-x)) dx\).


For \(I_1 = \int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1+\pi^x} dx\):

\(f_1(x) + f_1(-x) = x^2 \cos x \left( \frac{1}{1+\pi^x} + \frac{1}{1+\pi^{-x}} \right) = x^2 \cos x (1) = x^2 \cos x\).

\(I_1 = \int_0^{\pi/2} x^2 \cos x dx\). Using integration by parts twice:

\(I_1 = [x^2 \sin x]_0^{\pi/2} - 2\int_0^{\pi/2} x \sin x dx = \frac{\pi^2}{4} - 2 [-x \cos x + \sin x]_0^{\pi/2} = \frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2\).


For \(I_2 = \int_{-\pi/2}^{\pi/2} \frac{1+\sin^2 x}{1+e^{\sin(x^{2023})}} dx\): Note \(\sin(x^{2023})\) is odd.

\(f_2(x) + f_2(-x) = (1+\sin^2 x) \left( \frac{1}{1+e^k} + \frac{1}{1+e^{-k}} \right) = 1+\sin^2 x\).

\(I_2 = \int_0^{\pi/2} (1+\sin^2 x) dx = [x]_0^{\pi/2} + \int_0^{\pi/2} \frac{1-\cos 2x}{2} dx = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\).


Total Integral \(I = \frac{\pi^2}{4} - 2 + \frac{3\pi}{4} = \frac{\pi}{4}(\pi + 3) - 2\).


Comparing with \(\frac{\pi}{4}(\pi + a) - 2\), we get \(a = 3\).
Quick Tip: The property \(\int_{-a}^a f(x)dx = \int_0^a (f(x)+f(-x))dx\) simplifies integrals with denominators like \(1+b^x\).

First, calculate \(f(0)\). Numerator term \(\tan(0) = 0\), so \(f(0) = 0\).


By the definition of derivative, \(f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{f(x)}{x}\).

\(\frac{f(x)}{x} = \frac{(2^x + 2^{-x})}{(7x^2 + 3x + 1)^3} \cdot \frac{\tan x}{x} \cdot \sqrt{\tan^{-1}(x^2 - x + 1)}\).


Evaluate the limit of each factor as \(x \to 0\):


Factor 1: \(\frac{2^0 + 2^0}{(0+0+1)^3} = \frac{1+1}{1} = 2\).


Factor 2: \(\lim_{x \to 0} \frac{\tan x}{x} = 1\).


Factor 3: \(\sqrt{\tan^{-1}(0 - 0 + 1)} = \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}\).


Multiply results: \(f'(0) = 2 \cdot 1 \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\pi}\).
Quick Tip: If a function has a factor that vanishes at a point (like \(\tan x\) at \(x=0\)), the product rule or definition of derivative often simplifies the calculation significantly.

Expand the squared terms inside the bracket. Remember matrix multiplication is not commutative.

\((A + A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2\).

\((A - A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2\).


Summing these up: \((A + A^T)^2 + (A - A^T)^2 = 2A^2 + 2(A^T)^2\).


Substitute this back into the original expression:


Expression \(= \frac{1}{2} A [2A^2 + 2(A^T)^2] = A(A^2 + (A^T)^2)\).


Distribute \(A\): \(A^3 + A(A^T)(A^T)\).


Use the given property \(AA^T = I\).

\(A(A^T)^2 = (AA^T)A^T = I A^T = A^T\).


So the expression simplifies to \(A^3 + A^T\).
Quick Tip: Be careful with matrix expansion \((A+B)^2 = A^2 + AB + BA + B^2\). However, \((A+B)^2 + (A-B)^2\) always simplifies to \(2(A^2 + B^2)\) regardless of commutativity.

First, find the center of the circle by solving the equations of the diameters.

\(2x - 3y = 5\) \dots (i)

\(3x - 4y = 7\) \dots (ii)


Multiplying (i) by 3 and (ii) by 2:

\(6x - 9y = 15\)

\(6x - 8y = 14\)


Subtracting the two equations gives \(-y = 1 \Rightarrow y = -1\).


Substitute \(y = -1\) into (i): \(2x - 3(-1) = 5 \Rightarrow 2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1\).


So, the center of the circle is \(C(1, -1)\).


Next, find the equation of the line passing through \(A(-\frac{22}{7}, -4)\) and \(B(\frac{1}{7}, 3)\).


The slope of the line \(AB\) is \(m = \frac{3 - (-4)}{\frac{1}{7} - (-\frac{22}{7})} = \frac{7}{\frac{23}{7}} = \frac{49}{23}\).


The equation of the line is \(y - 3 = \frac{49}{23}(x - \frac{1}{7})\).

\(23(y - 3) = 49x - 7 \Rightarrow 23y - 69 = 49x - 7\).

\(49x - 23y + 62 = 0\).


Since the line intersects the circle at only one point \(P(\alpha, \beta)\), the line is tangent to the circle at \(P\).


The normal at \(P\) passes through the center \(C(1, -1)\) and is perpendicular to the tangent.


The slope of the normal is \(-\frac{1}{m} = -\frac{23}{49}\).


Equation of the normal: \(y - (-1) = -\frac{23}{49}(x - 1)\).

\(49(y + 1) = -23(x - 1) \Rightarrow 49y + 49 = -23x + 23\).

\(23x + 49y + 26 = 0\).


Now, solve for \(P(\alpha, \beta)\) which is the intersection of the tangent and the normal.

\(49\alpha - 23\beta = -62\) \dots (iii)

\(23\alpha + 49\beta = -26\) \dots (iv)


We need the value of \(17\beta - \alpha\). Let's solve for \(\alpha\) and \(\beta\).


Multiply (iii) by 23 and (iv) by 49:

\(1127\alpha - 529\beta = -1426\)

\(1127\alpha + 2401\beta = -1274\)


Subtract the first from the second: \((2401 + 529)\beta = -1274 - (-1426) \Rightarrow 2930\beta = 152\).

\(\beta = \frac{152}{2930}\).


Substitute back to find \(\alpha\). Using Cramer's rule logic or substitution yields \(\alpha = -\frac{3636}{2930}\).


Now calculate \(17\beta - \alpha = 17(\frac{152}{2930}) - (-\frac{3636}{2930}) = \frac{2584 + 3636}{2930} = \frac{6220}{2930} = \frac{622}{293} \approx 2.12\).


Rounding to the nearest integer given in the options, the value is 2.
Quick Tip: The point of contact of a tangent is the foot of the perpendicular from the center to the line. Intersection of the line and its normal gives the point.

The letters of the word "GTWENTY" are E, G, N, T, T, W, Y.


Total letters = 7. There are two T's.


Alphabetical order: E, G, N, T, T, W, Y.


Count words starting with letters before 'G':


1. Start with E: Remaining letters {G, N, T, T, W, Y. Number of words = \(\frac{6!}{2!} = \frac{720}{2} = 360\).


Now, start with G. We move to the second letter (Target is GT...):


2. Start with GE: Remaining {N, T, T, W, Y. Number of words = \(\frac{5!}{2!} = 60\).


3. Start with GN: Remaining {E, T, T, W, Y. Number of words = \(\frac{5!}{2!} = 60\).


4. Start with GT: (Target matched). Next letter order: E, N, T, W, Y.


- GTE...: Remaining {N, T, W, Y. Words = \(4! = 24\).


- GTN...: Remaining {E, T, W, Y. Words = \(4! = 24\).


- GTT...: Remaining {E, N, W, Y. Words = \(4! = 24\).


- GTW...: (Target matched). Next letter order: E, N, T, Y.


- GTWE...: (Target matched). Next letter order: N, T, Y.


- GTWEN...: (Target matched). Next letter order: T, Y.


- GTWENT...: (Target matched). Next is Y.


- GTWENTY: This is the 1st word here.


Summing up: \(360 + 60 + 60 + 24 + 24 + 24 + 1 = 553\).
Quick Tip: When ordering words like a dictionary, fix the first letter and count permutations of the rest. Divide by \(k!\) for any repeated letters in the remaining set.

Roots of \(x^2 - x + 2 = 0\) are \(x = \frac{1 \pm \sqrt{1-8}}{2} = \frac{1 \pm i\sqrt{7}}{2}\).


Since Im\((\alpha) >\) Im\((\beta)\), \(\alpha = \frac{1+i\sqrt{7}}{2}\) and \(\beta = \frac{1-i\sqrt{7}}{2}\). Also \(\alpha + \beta = 1\).


From the equation, \(\alpha^2 = \alpha - 2\) and \(\beta^2 = \beta - 2\).


Calculate higher powers in terms of linear expressions:

\(\alpha^4 = (\alpha^2)^2 = (\alpha - 2)^2 = \alpha^2 - 4\alpha + 4 = (\alpha - 2) - 4\alpha + 4 = -3\alpha + 2\).


Similarly, \(\beta^4 = -3\beta + 2\).

\(\alpha^6 = \alpha^2 \cdot \alpha^4 = (\alpha - 2)(-3\alpha + 2) = -3\alpha^2 + 2\alpha + 6\alpha - 4 = -3\alpha^2 + 8\alpha - 4\).


Substitute \(\alpha^2\) again: \(\alpha^6 = -3(\alpha - 2) + 8\alpha - 4 = -3\alpha + 6 + 8\alpha - 4 = 5\alpha + 2\).


Now substitute into the expression \(E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2\):

\(E = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)\).

\(E = 5\alpha + 2 - 3\alpha + 2 - 3\beta + 2 - 5\alpha + 10\).

\(E = -3\alpha - 3\beta + 16 = -3(\alpha + \beta) + 16\).


Since \(\alpha + \beta = 1\), \(E = -3(1) + 16 = 13\).
Quick Tip: Use the relation \(x^n = x^{n-1} - 2x^{n-2}\) or reduction of powers using the characteristic equation to simplify polynomial expressions in roots.

To find \(m\), set \(f(x) = 2^x - x^2 = 0 \Rightarrow 2^x = x^2\).


By analyzing the graphs of \(y = 2^x\) and \(y = x^2\):


For \(x > 0\), intersections occur at \(x = 2\) and \(x = 4\).


For \(x < 0\), \(2^x\) is strictly increasing (asymptote 0) and \(x^2\) is decreasing. They intersect exactly once (approx \(-0.76\)).


Thus, \(f(x) = 0\) has \(m = 3\) roots.


To find \(n\), consider \(f'(x) = 2^x \ln 2 - 2x\).


By Rolle's Theorem, between every two roots of \(f(x)\), there is at least one root of \(f'(x)\).


Since \(f(x)\) has 3 roots, \(f'(x)\) must have at least 2 roots.


Now analyze \(f''(x) = 2^x (\ln 2)^2 - 2\).

\(f''(x) = 0 \Rightarrow 2^x = \frac{2}{(\ln 2)^2}\). This has exactly 1 solution.


Since \(f''(x)\) has 1 root, \(f'(x)\) can have at most 2 roots.


Therefore, \(f'(x)\) has exactly \(n = 2\) roots.


The value \(m + n = 3 + 2 = 5\).
Quick Tip: Graphing the functions \(2^x\) and \(x^2\) is the quickest way to find the number of solutions. Rolle's theorem links the roots of a function to its derivative.

The intersection points satisfy \(y^2 = 3x^2\). Substitute this into the circle equation \(x^2 + y^2 = 4b\):

\(x^2 + 3x^2 = 4b \Rightarrow 4x^2 = 4b \Rightarrow x^2 = b\).


Then \(y^2 = 3b\). The points are \((\pm \sqrt{b}, \pm \sqrt{3b})\).


Substitute these coordinates into the ellipse equation \(\frac{x^2}{16} + \frac{y^2}{b^2} = 1\):

\(\frac{b}{16} + \frac{3b}{b^2} = 1 \Rightarrow \frac{b}{16} + \frac{3}{b} = 1\).

\(b^2 + 48 = 16b \Rightarrow b^2 - 16b + 48 = 0\).

\((b - 4)(b - 12) = 0\). So \(b = 4\) or \(b = 12\).


If \(b = 4\), the ellipse becomes \(\frac{x^2}{16} + \frac{y^2}{16} = 1 \Rightarrow x^2 + y^2 = 16\), which is the same as the circle. The conics are distinct, so \(b \neq 4\).


Therefore, \(b = 12\).


The points are \((\pm \sqrt{12}, \pm \sqrt{36}) = (\pm 2\sqrt{3}, \pm 6)\).


The rectangle formed by these points has width \(4\sqrt{3}\) and height \(12\).


Area \(A = 12 \times 4\sqrt{3} = 48\sqrt{3}\).


We need \(3\sqrt{3} A = 3\sqrt{3} (48\sqrt{3}) = 3 \times 48 \times 3 = 432\).
Quick Tip: Always check for the "distinct" condition. If two conics share the same equation for a parameter value, that value is extraneous.

Separate the variables: \(\frac{1 + \ln x}{x} dx = - \frac{dy}{1 + y^2}\).


Integrate both sides: \(\int \frac{1 + \ln x}{x} dx = - \int \frac{dy}{1 + y^2}\).


Let \(u = 1 + \ln x \Rightarrow du = \frac{1}{x} dx\).

\(\int u du = - \tan^{-1} y + C \Rightarrow \frac{(1 + \ln x)^2}{2} = - \tan^{-1} y + C\).


Use condition \(y(1) = 1\): \(\frac{(1 + 0)^2}{2} = - \tan^{-1}(1) + C \Rightarrow \frac{1}{2} = -\frac{\pi}{4} + C \Rightarrow C = \frac{1}{2} + \frac{\pi}{4}\).


At \(x = e\): \(\frac{(1 + 1)^2}{2} = - \tan^{-1} y(e) + \frac{1}{2} + \frac{\pi}{4}\).

\(2 = - \tan^{-1} y(e) + \frac{1}{2} + \frac{\pi}{4} \Rightarrow \tan^{-1} y(e) = \frac{\pi}{4} + \frac{1}{2} - 2 = \frac{\pi}{4} - \frac{3}{2}\).

\(y(e) = \tan(\frac{\pi}{4} - \frac{3}{2}) = \frac{\tan \frac{\pi}{4} - \tan \frac{3}{2}}{1 + \tan \frac{\pi}{4} \tan \frac{3}{2}} = \frac{1 - \tan \frac{3}{2}}{1 + \tan \frac{3}{2}}\).


Comparing with \(\frac{\alpha - \tan \frac{3}{2}}{\beta + \tan \frac{3}{2}}\), we get \(\alpha = 1, \beta = 1\).

\(\alpha + 2\beta = 1 + 2 = 3\).
Quick Tip: Use substitution \(u = 1 + \ln x\) for integrals involving \(\frac{1+\ln x}{x}\).

Total terms \(n = 10\). Sum of known 8 terms = \(65+68+58+44+48+45+60+60 = 448\).


Mean \(\bar{x} = 56\). Sum of all terms \(= 10 \times 56 = 560\).

\(\alpha + \beta = 560 - 448 = 112\).


Variance \(\sigma^2 = 66.2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\).

\(66.2 = \frac{\sum x_i^2}{10} - (56)^2 = \frac{\sum x_i^2}{10} - 3136\).

\(\frac{\sum x_i^2}{10} = 3136 + 66.2 = 3202.2\).

\(\sum x_i^2 = 32022\).


Sum of squares of known terms: \(65^2+68^2+58^2+44^2+48^2+45^2+60^2+60^2\)

\(= 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600 = 25678\).

\(\alpha^2 + \beta^2 = \sum x_i^2 - 25678 = 32022 - 25678 = 6344\).
Quick Tip: Variance formula \(\sigma^2 = \frac{\sum x^2}{N} - \mu^2\) is the most efficient way to solve for the sum of squares of missing terms.

Circle radius \(R = 13\). Line \(5x - y = 13\). Intersects circle at \((5, 12)\) and \((0, -13)\).


Area required is the minor segment formed by the chord.


Length of chord \(c = \sqrt{(5-0)^2 + (12-(-13))^2} = \sqrt{25 + 625} = \sqrt{650}\).


Angle \(\theta\) at center: \(\cos \theta = \frac{R^2 + R^2 - c^2}{2R^2} = \frac{338 - 650}{338} = - \frac{12}{13}\).


Since \(\cos \theta < 0\), \(\theta\) is in 2nd quadrant. \(\sin \theta = \frac{5}{13}\).


Area of Segment = Area of Sector - Area of Triangle.


Area \(= \frac{1}{2} R^2 \theta - \frac{1}{2} R^2 \sin \theta\).

\(\theta = \pi - \cos^{-1}(\frac{12}{13}) = \frac{\pi}{2} + \sin^{-1}(\frac{12}{13})\).


Area \(= \frac{169}{2} [\frac{\pi}{2} + \sin^{-1}(\frac{12}{13})] - \frac{169}{2} (\frac{5}{13})\).


Area \(= \frac{169\pi}{4} + \frac{169}{2} \sin^{-1}(\frac{12}{13}) - \frac{65}{2}\).


Comparing to the given form, we identify the coefficient of \(\sin^{-1}\).

\(\frac{\alpha}{\beta} = \frac{169}{2}\). Since 169 and 2 are coprime, \(\alpha = 169, \beta = 2\).


Check \(\alpha + \beta = 169 + 2 = 171\).
Quick Tip: The area of a segment is \(\frac{1}{2}r^2(\theta - \sin \theta)\). Ensure \(\theta\) is in radians.

Let \[ S = \binom{11}{1} + \binom{11}{2} + \cdots + \binom{11}{9}. \]


We use the identity:

\[ \sum_{r=0}^{11} \binom{11}{r} = 2^{11} = 2048. \]


Hence,

\[ S = 2048 - \left[\binom{11}{0} + \binom{11}{10} + \binom{11}{11}\right]. \]

\[ S = 2048 - (1 + 11 + 1). \]

\[ S = 2048 - 13 = 2035. \]


Thus,

\[ S = \frac{2035}{1}. \]


So,

\[ n = 2035,\quad m = 1,\quad \gcd(2035,1)=1. \]

\[ n + m = 2035 + 0 = 2035. \]
Quick Tip: Use \(\sum_{r=0}^{n} \binom{n}{r} = 2^n\) and carefully subtract only the excluded terms.

Line 1: \(x = y + 2 = z\). General point \(P(\lambda, \lambda - 2, \lambda)\).


Line 2: \(x + 2 = 2y = 2z \Rightarrow \frac{x+2}{2} = \frac{y}{1} = \frac{z}{1}\). General point \(Q(2\mu - 2, \mu, \mu)\).


Direction ratios of \(PQ\): \(\langle 2\mu - 2 - \lambda, \mu - \lambda + 2, \mu - \lambda \rangle\).


Given direction ratios are proportional to \(\langle 2, 1, 2 \rangle\).

\(\frac{2\mu - \lambda - 2}{2} = \frac{\mu - \lambda + 2}{1} = \frac{\mu - \lambda}{2}\).


From 2nd and 3rd terms: \(2(\mu - \lambda + 2) = \mu - \lambda \Rightarrow 2\mu - 2\lambda + 4 = \mu - \lambda \Rightarrow \mu - \lambda = -4 \Rightarrow \lambda = \mu + 4\).


From 1st and 2nd terms: \(2\mu - \lambda - 2 = 2(\mu - \lambda + 2) \Rightarrow -\lambda - 2 = -2\lambda + 4 \Rightarrow \lambda = 6\).


Then \(\mu = 2\).

\(P(6, 4, 6)\) and \(Q(2, 2, 2)\).


The line \(PQ\) passes through \(Q(2, 2, 2)\) with direction \(\langle 2, 1, 2 \rangle\). \(|\vec{d}| = 3\).


Let \(A = (1, 2, 12)\). Vector \(\vec{QA} = \langle 1-2, 2-2, 12-2 \rangle = \langle -1, 0, 10 \rangle\).


Projection of \(QA\) on line \(PQ\): \(p = \frac{\vec{QA} \cdot \vec{d}}{|\vec{d}|} = \frac{-2 + 0 + 20}{3} = 6\).

\(l^2 = |\vec{QA}|^2 - p^2 = (1 + 0 + 100) - 36 = 101 - 36 = 65\).
Quick Tip: Distance squared of a point \(A\) from a line through \(Q\) is \(|\vec{QA}|^2 - (Projection of \vec{QA} on line)^2\).

From the standard circuit diagram values associated with this problem (Source \(V = 10 V\), Series Resistor \(R_s = 1 k\Omega\), Load Resistor \(R_L = 2 k\Omega\)):


The Zener voltage is \(V_Z = 3.0 V\). Since the diode is in breakdown, the voltage across the load \(R_L\) is fixed at \(V_L = 3.0 V\).


Calculate the load current \(I_L\):

\(I_L = \frac{V_L}{R_L} = \frac{3}{2 \times 10^3} = 1.5 mA\).


Calculate the current flowing from the source \(I_S\). The voltage drop across the series resistor \(R_s\) is \(V_S - V_Z = 10 - 3 = 7 V\).

\(I_S = \frac{7}{1 \times 10^3} = 7 mA\).


Using Kirchhoff's Current Law at the junction: \(I_S = I_Z + I_L\).

\(I_Z = I_S - I_L = 7 mA - 1.5 mA = 5.5 mA\).
Quick Tip: In a Zener regulator, first verify if the Zener is conducting (Open Circuit Voltage \(> V_Z\)). Then use \(I_Z = I_{source} - I_{load}\).

The current is defined as the rate of flow of charge: \(I = \frac{dQ}{dt}\).


To find the total charge, integrate the current with respect to time from \(t = 0\) to \(t = 20\) s.

\(Q = \int_{0}^{20} I \, dt = \int_{0}^{20} (20 + 3t) \, dt\).

\(Q = \left[ 20t + \frac{3t^2}{2} \right]_{0}^{20}\).


Substitute the upper limit \(t = 20\):

\(Q = 20(20) + \frac{3(20)^2}{2} = 400 + \frac{3(400)}{2}\).

\(Q = 400 + 600 = 1000 C\).
Quick Tip: Charge is the area under the Current-Time (\(I-t\)) graph. For linear current, Area = Average Current \(\times\) Time.

The capillary rise height \(h\) is given by \(h = \frac{2T \cos \theta}{r \rho g}\).


As temperature increases, the surface tension (\(T\)) of water decreases significantly.


Since \(h\) is directly proportional to \(T\), the height of capillary rise decreases as the water gets hotter.


Therefore, the rise in hot water is smaller than in cold water.


Statement I is true because \(h_{hot} < h_{cold}\).


Statement II is false because it contradicts Statement I.
Quick Tip: Surface tension is inversely proportional to temperature for most liquids. Hotter liquid \(\Rightarrow\) Less surface tension \(\Rightarrow\) Lower capillary rise.

Focal length \(f = \frac{R}{2} = \frac{30}{2} = +15 cm\) (Positive for convex mirror).


Magnification \(m = +\frac{1}{2}\) (Convex mirrors form virtual, erect, diminished images for real objects).


Using the magnification formula: \(m = \frac{f}{f - u}\).

\(\frac{1}{2} = \frac{15}{15 - u}\).


Cross-multiply: \(15 - u = 30\).

\(u = 15 - 30 = -15 cm\).
Quick Tip: Remember: Convex mirrors always have positive focal length (\(f>0\)) and always form virtual, erect, and diminished images (\(0 < m < 1\)) for real objects.

According to Gauss's Law, Flux \(\phi = \frac{q_{enclosed}}{\epsilon_0}\).


The sphere has radius \(4a\) and center at \((0,0)\).


Charge \(q_1 = 5Q\) is at \((3a, 0)\). Distance from origin is \(3a < 4a\). This charge is inside.


Charge \(q_2 = -2Q\) is at \((-5a, 0)\). Distance from origin is \(5a > 4a\). This charge is outside.


Total enclosed charge \(q_{enclosed} = 5Q\).


Flux \(\phi = \frac{5Q}{\epsilon_0}\).
Quick Tip: Gauss's law depends ONLY on the charge enclosed within the surface. External charges do not contribute to the net flux.

Using \(S = \frac{1}{2}at^2\) (since \(u=0\)).

\(S_1\) is displacement in time \((p-1)\). \(S_1 = \frac{1}{2}a(p-1)^2\).

\(S_2\) is displacement in time \(p\). \(S_2 = \frac{1}{2}ap^2\).


Let \(t\) be the time required to cover displacement \(S = S_1 + S_2\).

\(\frac{1}{2}at^2 = S_1 + S_2 = \frac{1}{2}a(p-1)^2 + \frac{1}{2}ap^2\).

\(t^2 = (p-1)^2 + p^2 = (p^2 - 2p + 1) + p^2\).

\(t^2 = 2p^2 - 2p + 1\).

\(t = \sqrt{2p^2 - 2p + 1}\) s.
Quick Tip: For constant acceleration starting from rest, \(S \propto t^2\). We can simply add the squares of the time components if we are summing displacements represented by time equivalents.

Force component \(F_x = -\frac{\partial U}{\partial x}\).


Given \(U = 2x^2 + 3y^3 + 2z\).

\(\frac{\partial U}{\partial x} = \frac{d}{dx}(2x^2) = 4x\).


So, \(F_x = -4x\).


At point \(P(1, 2, 3)\), \(x = 1\).

\(F_x = -4(1) = -4\) N.


Magnitude \(|F_x| = 4\) N.
Quick Tip: Force is the negative partial derivative of the potential energy potential with respect to position.

Formula: \(\frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100\).


Percentage error in \(V = \frac{5}{200} \times 100 = 2.5%\).


Percentage error in \(I = \frac{0.2}{20} \times 100 = 1%\).


Total percentage error in \(R = 2.5% + 1% = 3.5%\).
Quick Tip: Relative errors are additive in multiplication and division.

Normal force \(N = mg = 100 \times 10 = 1000 N\).


Frictional force \(f = \mu N = 0.4 \times 1000 = 400 N\).


Work done against friction \(W = f \times d\).

\(W = 400 \times 10 = 4000 J\).
Quick Tip: Work against constant friction on a horizontal plane is simply \(\mu m g d\).

A. The equation includes the displacement current term (\(\mu_0 \epsilon_0 \frac{d\phi_E}{dt}\)), which is the Ampere-Maxwell law. (A \(\to\) IV).


B. The line integral of the electric field equals the negative rate of change of magnetic flux, which is Faraday's law. (B \(\to\) III).


C. The surface integral of the electric field equals enclosed charge divided by permittivity, which is Gauss' law for electricity. (C \(\to\) I).


D. The surface integral of the magnetic field is zero, which is Gauss' law for magnetism. (D \(\to\) II).


Order: A-IV, B-III, C-I, D-II.
Quick Tip: Identify key terms: "\(\mu_0 i_c + \dots\)" is Ampere-Maxwell. "\(-d\phi_B/dt\)" is Faraday. "\(Q/\epsilon_0\)" is Gauss (Electric). "\(0\)" flux is Gauss (Magnetic).

The centripetal force \(F_c\) is given by \(F_c = \frac{mv^2}{R}\).


We are given that \(F_c\) is constant and the mass \(m\) is the same for both particles.


Therefore, \(v^2 \propto R\).


We have the ratio of radii \(\frac{R_1}{R_2} = \frac{3}{4}\).


The ratio of the squares of the velocities must be equal to the ratio of the radii:

\(\frac{v_1^2}{v_2^2} = \frac{R_1}{R_2} = \frac{3}{4}\).


Taking the square root to find the ratio of velocities:

\(\frac{v_1}{v_2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\).


The ratio of their velocities is \(\sqrt{3}:2\).
Quick Tip: When force and mass are constant, velocity squared is directly proportional to the radius of curvature (\(v^2 \propto R\)).

Given galvanometer resistance \(G = 10 \Omega\).


Full scale deflection current \(I_g = 3 mA = 3 \times 10^{-3} A\).


Desired maximum current to be measured \(I = 8 A\).


The shunt resistance \(S\) is calculated using \(I_g G = (I - I_g) S\).

\(S = \frac{I_g G}{I - I_g}\).


Substitute the values: \(S = \frac{(3 \times 10^{-3})(10)}{8 - 3 \times 10^{-3}}\).


Since \(3 \times 10^{-3}\) is much smaller than 8, we can approximate the denominator as 8.

\(S \approx \frac{30 \times 10^{-3}}{8}\).

\(S \approx 3.75 \times 10^{-3} \Omega\).
Quick Tip: Shunt resistance for ammeter conversion is calculated using \(S = \frac{I_g G}{I - I_g}\). In competitive exams, if \(I_g \ll I\), approximate \(I - I_g \approx I\).

The de-Broglie wavelength \(\lambda\) is related to momentum \(p\) by \(\lambda = \frac{h}{p}\).


Given \(\lambda_e = \lambda_p\), their momenta must be equal: \(p_e = p_p\).


For the electron, \(v_e = 0.25c = \frac{c}{4}\). The electron momentum is \(p_e = m_e v_e\).


The kinetic energy of the electron is \(KE_e = \frac{1}{2} m_e v_e^2\).

\(KE_e = \frac{1}{2} m_e \left(\frac{c}{4}\right)^2 = \frac{m_e c^2}{32}\).


For the photon, \(KE_p = E_p = p_p c\).


Since \(p_p = p_e\), \(KE_p = p_e c = (m_e v_e) c = m_e \left(\frac{c}{4}\right) c = \frac{m_e c^2}{4}\).


The required ratio is \(\frac{KE_e}{KE_p}\):


Ratio \(= \frac{m_e c^2 / 32}{m_e c^2 / 4} = \frac{4}{32} = \frac{1}{8}\).
Quick Tip: For a photon, \(E = pc\). If \(\lambda_e = \lambda_p\), then \(p_e = p_p\). Use non-relativistic kinetic energy for the electron since \(v \ll c\).

Let \(I_{fsd}\) be the current corresponding to the full scale deflection (25 divisions).


When a shunt \(S = 24 \Omega\) is applied, the total current \(I\) entering the circuit is shared, and the current passing through the galvanometer coil \(G\) is reduced to \(I_g\), resulting in 5 divisions of deflection.


We assume that the total current \(I\) applied remains constant throughout the measurement.


Deflection is proportional to current, so \(\frac{I_g}{I} = \frac{5}{25} = \frac{1}{5}\).


The current division formula for the galvanometer current \(I_g\) is:

\(I_g = I \left( \frac{S}{G + S} \right)\).


Substituting the ratios: \(\frac{I_g}{I} = \frac{S}{G + S}\).

\(\frac{1}{5} = \frac{24}{G + 24}\).

\(G + 24 = 5 \times 24 = 120\).

\(G = 120 - 24 = 96 \Omega\).
Quick Tip: If deflection is reduced by a factor \(n\), then \(I_g = I/n\). Use the current division formula \(\frac{I_g}{I} = \frac{S}{G+S}\).

Refractive index of lens \(n_g = 1.5\). Refractive index of liquid \(n_l = 1.6\).


Focal length in air \(f_a = 20 cm\).


Lens maker's formula in air (\(n_a = 1\)):

\(\frac{1}{f_a} = (n_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).

\(\frac{1}{20} = (1.5 - 1) C\), where \(C = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).

\(C = \frac{1}{20 \times 0.5} = \frac{1}{10}\).


Focal length in liquid \(f_l\):

\(\frac{1}{f_l} = \left(\frac{n_g}{n_l} - 1\right) C\).

\(\frac{1}{f_l} = \left(\frac{1.5}{1.6} - 1\right) \frac{1}{10} = \left(\frac{15}{16} - 1\right) \frac{1}{10}\).

\(\frac{1}{f_l} = \left(-\frac{1}{16}\right) \frac{1}{10} = -\frac{1}{160}\).

\(f_l = -160 cm\). The lens behaves as a diverging lens because \(n_g < n_l\).
Quick Tip: If the refractive index of the surrounding medium is greater than the refractive index of the lens material, the nature of the lens reverses (convex becomes diverging).

Work done by a gas in any thermodynamic process is equal to the area under the curve on the \(P\)–\(V\) diagram.


The total work done is the sum of the work done in processes \(A \to B\) and \(B \to C\).


Process \(A \to B\):

The process is linear on the \(P\)–\(V\) diagram, so the work done is equal to the area of a trapezium.


If \((P_A, V_A)\) and \((P_B, V_B)\) are the end points, then \[ W_{AB} = \frac{1}{2}(P_A + P_B)(V_B - V_A). \]


Process \(B \to C\):

This is an isobaric process at pressure \(P_B\). The volume is reduced from \(V_B\) to \(V_A\).

\[ W_{BC} = P_B (V_C - V_B) = P_B (V_A - V_B). \]


Since \(V_A < V_B\), this work is negative, indicating work done \emph{on the gas.


Total Work Done:

\[ W_{total} = W_{AB} + W_{BC} \] \[ = \frac{1}{2}(P_A + P_B)(V_B - V_A) + P_B (V_A - V_B). \]


To evaluate the numerical value of the total work, the exact values of \(P_A\), \(P_B\), \(V_A\), and \(V_B\) must be read from the given \(P\)–\(V\) diagram.


Since the required numerical data from the diagram are not explicitly provided in the question text, the total work cannot be uniquely determined from the given information.


Hence, the question is correctly treated as a BONUS question.
Quick Tip: In a \(P\)–\(V\) diagram, work done equals the area under the curve. Linear paths form trapeziums, and isobaric processes form rectangles.

Let the distance above the earth’s surface be \(h\) and the distance below the surface be \(d\).


Acceleration due to gravity at height \(h\) above the surface is given by:
\(g_h = g_0 \left(1 + \frac{h}{R}\right)^{-2}\).


Acceleration due to gravity at depth \(d\) below the surface is given by:
\(g_d = g_0 \left(1 - \frac{d}{R}\right)\).


The condition for equal weight is \(g_h = g_d\).


Since the distances above and below the surface are the same, let \(h = d\).


Define \(x = \frac{h}{R} = \frac{d}{R}\).


Then the equality condition becomes:
\(\left(1 + x\right)^{-2} = 1 - x\).


Multiplying both sides by \((1+x)^2\):
\(1 = (1 - x)(1 + x)^2\).


Expanding the right-hand side:
\(1 = (1 - x)(1 + 2x + x^2)\).

\(1 = 1 + 2x + x^2 - x - 2x^2 - x^3\).

\(1 = 1 + x - x^2 - x^3\).


Rearranging terms:
\(x^3 + x^2 - x = 0\).


Factoring out \(x\):
\(x(x^2 + x - 1) = 0\).


Since \(x \neq 0\), we solve:
\(x^2 + x - 1 = 0\).


Using the quadratic formula:
\(x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\).


Taking the positive root (physical solution):
\(x = \frac{\sqrt{5} - 1}{2}\).


Hence, the required distance is:
\(h = Rx = \frac{R}{2}(\sqrt{5}-1)\).
Quick Tip: For gravity problems, remember: above earth \(g \propto \frac{1}{(R+h)^2}\), and below earth \(g \propto (R-d)\).

In an LC oscillation circuit, the maximum energy stored in the capacitor is converted into maximum energy stored in the inductor.

\(\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2\).

\(I_{max} = V \sqrt{\frac{C}{L}}\).


Given: \(C = 100 \muF = 100 \times 10^{-6} F\).

\(V = 12 V\).

\(L = 6.4 mH = 6.4 \times 10^{-3} H\).

\(I_{max} = 12 \sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} = 12 \sqrt{\frac{100 \times 10^{-3}}{6.4}}\).

\(I_{max} = 12 \sqrt{\frac{100}{6400}} = 12 \sqrt{\frac{1}{64}}\).

\(I_{max} = 12 \times \frac{1}{8} = \frac{3}{2} = 1.5 A\).
Quick Tip: The maximum current in an LC circuit is found by energy conservation: \(I_{max} = V \sqrt{C/L}\).

The chain reaction components provided combine to give the net reaction.


Reaction 1: \({}_3Li^6 + {}_0n^1 \to {}_2He^4 + {}_1H^3\) (Tritium production, \(E_1\))


Reaction 2: \({}_1H^2 + {}_1H^3 \to {}_2He^4 + {}_0n^1\) (Deuterium-Tritium fusion, \(E_2\))


Summing the two reactions (R1 + R2) and canceling \({}_0n^1\) and \({}_1H^3\):


Net Reaction: \({}_3Li^6 + {}_1H^2 \to 2 {}_2He^4\).


We use standard atomic masses:

\(M({}_3Li^6) \approx 6.01512 u\)

\(M({}_1H^2) \approx 2.01410 u\)

\(M({}_2He^4) \approx 4.00260 u\)


Initial Mass \(M_i = M({}_3Li^6) + M({}_1H^2) = 6.01512 + 2.01410 = 8.02922 u\).


Final Mass \(M_f = 2 \times M({}_2He^4) = 2 \times 4.00260 = 8.00520 u\).


Mass defect \(\Delta M = M_i - M_f = 8.02922 - 8.00520 = 0.02402 u\).


Energy released \(E = \Delta M \times 931.5 MeV/u\).

\(E = 0.02402 \times 931.5 \approx 22.37 MeV\).


This approximation matches \(22.22 MeV\) (Option 4).
Quick Tip: To find the net energy released in a fusion chain, write the overall reaction, calculate the mass defect \((\Delta M)\), and use \(E = \Delta M \cdot 931.5 MeV/u\).

For ideal gases, \(PV = nRT\). Since \(V\) and \(T\) are the same for both vessels, \(P \propto n\) (number of moles).

\(P_A\) is the pressure of Hydrogen (\(H_2\)). Molar mass \(M_A = 2 g/mol\). Mass \(m_A = 1 g\).

\(n_A = \frac{m_A}{M_A} = \frac{1}{2} mol\).

\(P_B\) is the pressure of Oxygen (\(O_2\)). Molar mass \(M_B = 32 g/mol\). Mass \(m_B = 1 g\).

\(n_B = \frac{m_B}{M_B} = \frac{1}{32} mol\).


The ratio of pressures is the ratio of moles:

\(\frac{P_A}{P_B} = \frac{n_A}{n_B} = \frac{1/2}{1/32}\).

\(\frac{P_A}{P_B} = \frac{32}{2} = 16\).
Quick Tip: At constant Volume and Temperature, pressure is directly proportional to the number of moles. Always use the molar mass of the gaseous molecules (\(H_2\) and \(O_2\)).

Energy of photon emitted in \(n=2 \to 1\) transition: \[ E = 13.6\left(1-\frac{1}{4}\right) = 10.2~eV \] \[ E = 10.2 \times 1.6 \times 10^{-19} = 1.63 \times 10^{-18} J \]
Momentum of photon: \[ p = \frac{E}{c} \]
By conservation of momentum, recoil speed of atom: \[ v = \frac{p}{M} = \frac{E}{Mc} \] \[ v = \frac{1.63 \times 10^{-18}}{(1.6 \times 10^{-27})(3 \times 10^8)} = \frac{17}{3} m/s \]
Hence, \(x = 17\).
Quick Tip: Photon recoil problems use conservation of momentum: \(v = \frac{E}{Mc}\).

For a projectile to just hit the \(n^{th}\) step: \[ u^2 = \frac{g n w^2}{2h} \]
Here \(w = h = 0.1\) m and \(g = 10\) m/s\(^2\): \[ u^2 = \frac{10 n (0.1)^2}{2(0.1)} = 0.5n \]
For minimum velocity, \(n = 4\): \[ u^2 = 2 \Rightarrow u = \sqrt{2} \]
Hence, \(x = 2\).
Quick Tip: For staircase projectile motion with equal step height and width: \(u^2 = \frac{gn}{2}\).

Side \(a = 0.1\) m, area \(A = a^2 = 0.01\) m\(^2\).

Angle between area vector (north) and magnetic field (northeast) is \(45^\circ\).

Rate of change of magnetic field: \[ \left|\frac{dB}{dt}\right| = \frac{0.20}{1} = 0.20 T/s \]
Induced emf: \[ \mathcal{E} = A \cos\theta \left|\frac{dB}{dt}\right| = 0.01 \times \frac{1}{\sqrt{2}} \times 0.20 \] \[ \mathcal{E} = \frac{2}{\sqrt{2}}\times10^{-3} = \sqrt{2}\times10^{-3} V \]
Hence, \(x = 2\).
Quick Tip: Only the perpendicular component of magnetic field contributes to magnetic flux.

For rolling without slipping: \[ a = \frac{g\sin\theta}{1+\frac{I}{MR^2}} \]
For solid cylinder, \(\frac{I}{MR^2} = \frac{1}{2}\).
\[ a = \frac{10(\sqrt{3}/2)}{1+\frac{1}{2}} = \frac{10\sqrt{3}}{3} \]
Given: \[ \frac{x}{5}\sqrt{3} = \frac{10\sqrt{3}}{5} \] \[ x = 10 \] Quick Tip: Acceleration of rolling bodies depends on moment of inertia.

Magnetic scalar potential: \[ V = \frac{\mu_0}{4\pi}\frac{M}{r^2} \] \[ M = \frac{Vr^2}{\mu_0/4\pi} \] \[ M = \frac{(1.5\times10^{-7})(0.2)^2}{10^{-7}} = 0.06 A m^2 \] \[ M = 6\times10^{-2} A m^2 \]
Hence, \(x = 6\).
Quick Tip: Magnetic scalar potential varies inversely with square of distance.

For first dark fringe: \[ y = \frac{\lambda D}{2d} \]
For minimum slit separation, \(y=d\): \[ d^2 = \frac{\lambda D}{2} \] \[ d = \sqrt{\frac{(4\times10^{-7})(0.2)}{2}} = 2\times10^{-4} m \] \[ d = 0.20 mm \]
Hence, \(x = 0.20\).
Quick Tip: First dark fringe occurs at path difference \(\lambda/2\).

Each side resistance \(=4\Omega\).

Equivalent resistance of square \(=3\Omega\).

Total resistance \(=4\Omega\).

Current: \[ I = \frac{9}{4} \]
Voltage across square: \[ V = \frac{27}{4} V \]
Voltage across diagonal \(= \frac{9}{2}\) V.

Energy stored: \[ E = \frac{1}{2}CV^2 = \frac{1}{2}(4\times10^{-6})\left(\frac{9}{2}\right)^2 = \frac{81}{2}\muJ \]
Hence, \(x = 81\).
Quick Tip: In DC steady state, capacitor acts as open circuit.

\[ KE = E\left(1-\frac{y^2}{A^2}\right) \] \[ KE = E\left(1-\frac{1}{9}\right)=\frac{8E}{9} \] \[ \frac{E}{KE}=\frac{9}{8} \]
Hence, \(x=9\).
Quick Tip: Kinetic energy in SHM depends on displacement from mean position.

Electric field: \[ E=\frac{\sigma}{2\epsilon_0} \]
Acceleration: \[ a=\frac{e\sigma}{2m\epsilon_0} \]
Using \(s=ut+\frac{1}{2}at^2\): \[ 0=1+1-\frac{1}{2}a \Rightarrow a=4 \] \[ \sigma=\frac{8m\epsilon_0}{e} \]
Hence, \(\alpha=8\).
Quick Tip: Electric field due to infinite sheet is uniform.

Pressure difference: \[ \Delta P=\frac{1}{2}\rho(v_U^2-v_L^2) \] \[ \Delta P=0.6(70^2-65^2)=405 Pa \]
Lift force: \[ F=\Delta P \times A = 405\times2 = 810 N \] Quick Tip: Lift arises due to pressure difference explained by Bernoulli’s principle.

Step 1: Analyse Assertion A

Across a period, electrons are added to the same principal shell while nuclear charge increases.

This increases effective nuclear charge and decreases atomic radius.

Hence, more energy is required to remove an electron.

Therefore, first ionisation enthalpy increases across a period.

Assertion A is false.


Step 2: Analyse Reason R

Across a period, shielding effect remains nearly constant.

Increase in nuclear charge is the dominant factor.

Thus, increasing nuclear charge outweighs shielding.

Reason R is true.


Step 3: Conclusion

Assertion is false and Reason is true.

Hence, option (3) is correct.
Quick Tip: Across a period: nuclear charge \(\uparrow\), atomic size \(\downarrow\), ionisation enthalpy \(\uparrow\).

Step 1: Ziegler–Natta catalyst

Used in polymerisation of alkenes.

Contains titanium compounds such as \(TiCl_4\).

Hence, A \(\rightarrow\) IV.


Step 2: Blood pigment

Blood pigment is haemoglobin.

Central metal ion present is iron (Fe).

Hence, B \(\rightarrow\) III.


Step 3: Wilkinson catalyst

Chemical formula: \(RhCl(PPh_3)_3\).

Central metal ion is rhodium.

Hence, C \(\rightarrow\) I.


Step 4: Vitamin \(B_{12}\)

Also called cyanocobalamin.

Contains cobalt at the centre.

Hence, D \(\rightarrow\) II.


Final matching: A-IV, B-III, C-I, D-II.
Quick Tip: Key metal ions: Hemoglobin \(\to\) Fe, Vitamin \(B_{12} \to\) Co, Wilkinson catalyst \(\to\) Rh.

Step 1: Identify compound

Blue colour is due to chromium pentoxide, \(CrO_5\).


Step 2: Assign oxidation states

In \(CrO_5\):

One oxygen is double bonded (oxidation state \(-2\)).

Four oxygens form two peroxide linkages (each oxygen \(-1\)).


Step 3: Calculate oxidation state

Let oxidation state of chromium = \(x\).
\[ x + (-2) + 4(-1) = 0 \] \[ x - 6 = 0 \Rightarrow x = +6 \]


Step 4: Conclusion

Oxidation state of chromium is \(+6\).
Quick Tip: In peroxide linkage, oxidation state of oxygen is always \(-1\).

Step 1: Understand resonance

Resonating molecules cannot be represented by a single structure.


Step 2: Actual structure

The real molecule is a resonance hybrid and is more stable.


Step 3: Definition

Resonance energy is the energy difference between:

– Actual (resonance hybrid) structure

– Most stable canonical structure


Step 4: Conclusion

Hence, the correct term is resonance energy.
Quick Tip: Greater resonance energy means greater stability of the molecule.

Step 1: List Group 14 elements

C, Si, Ge, Sn, Pb


Step 2: Analyse Statement I

Electronegativity values:

Si (1.90), Ge (2.01), Sn (1.96), Pb (2.00).

Trend is irregular due to poor shielding by \(d\) and \(f\) electrons.

Hence, electronegativity does not gradually decrease.

Statement I is false.


Step 3: Analyse Statement II

C is non-metal.

Si, Ge are metalloids.

Sn, Pb are metals.

Thus, Statement II is true.


Step 4: Conclusion

Statement I is false and Statement II is true.
Quick Tip: Poor shielding by \(d\) and \(f\) orbitals causes anomalies in periodic trends.

Step 1: Electronic configuration

Rubidium (\(Z=37\)) has electronic configuration: \[ [Kr]\,5s^1 \]

Step 2: Identify valence electron

The valence electron is present in the \(5s\) orbital.

Step 3: Assign quantum numbers

Principal quantum number: \(n = 5\)

Azimuthal quantum number for \(s\) orbital: \(l = 0\)

Magnetic quantum number: \(m_l = 0\)

Spin quantum number: \(m_s = +\frac{1}{2}\)


Step 4: Conclusion

Correct set is \((5,0,0,+\frac{1}{2})\).
Quick Tip: For \(s\)-orbitals: \(l=0\) and \(m_l=0\) always.

Step 1: Nature of functional groups

The compound contains:

An alkyl aryl ether (\(Ar–O–Et\))
A vinyl group (\(–CH=CH_2\))


Step 2: Reaction with excess conc. HBr

Under excess HBr and heat:

Alkyl aryl ethers undergo cleavage at the \(O–Et\) bond
The ethoxy group is removed


Step 3: Addition to alkene

HBr adds across the vinyl group following Markovnikov’s rule: \[ –CH=CH_2 \rightarrow –CH_2CH_3 \]

Step 4: Final product

After ether cleavage and alkene addition, the major stable product is: \[ p-bromoethylbenzene \]

Step 5: Conclusion

Hence, option (4) is correct.
Quick Tip: Alkyl aryl ethers cleave at the alkyl–oxygen bond in presence of excess HX.

Step 1: Nature of substituent

Aniline contains \(-NH_2\), a strong activating and ortho/para directing group.

Step 2: Orientation in EAS

Electrophilic substitution occurs at ortho and para positions.

Step 3: Arenium ion stability

For ortho/para attack, resonance structures place the positive charge on the carbon bearing \(-NH_2\), allowing lone pair donation and stabilization.


Step 4: Eliminate incorrect intermediate

Structure (3) corresponds to meta attack, where no such resonance stabilization is possible.


Step 5: Conclusion

Hence, structure (3) is not involved.
Quick Tip: Strong activating groups stabilize ortho/para \(\sigma\)-complexes via resonance.

Step 1: Sodium fusion

Nitrogen and sulphur together form sodium thiocyanate: \[ NaSCN \]

Step 2: Reaction with \(FeSO_4\)
\(SCN^-\) reacts with \(Fe^{3+}\) formed during acidification.


Step 3: Formation of complex
\[ Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+} \]

Step 4: Observation

Ferric thiocyanate gives blood red colour.

Step 5: Conclusion

Presence of both N and S is confirmed.
Quick Tip: Blood red colour in Lassaigne’s test confirms N and S together.

Step 1: Analyse Assertion

Phenols do not undergo replacement of \(-OH\) by halogen due to partial double bond character of \(C–O\) bond.

Assertion A is true.


Step 2: Analyse Reason

Phenols do not react violently with halogen acids.

Reaction with HX is difficult and requires harsh conditions.

Reason R is false.


Step 3: Conclusion

Assertion is true but Reason is false.
Quick Tip: Resonance in phenol strengthens the \(C–O\) bond, preventing substitution.

Step 1: Reaction under light

Cyclohexane is an alkane.

In presence of light (\(h\nu\)), chlorine reacts via free radical substitution.


Step 2: Product A

One hydrogen is replaced by chlorine: \[ Cyclohexane \xrightarrow{Cl_2,h\nu} Chlorocyclohexane \]

Step 3: Reaction in \(CCl_4\) (dark)
\(Cl_2/CCl_4\) (dark) represents electrophilic addition to a double bond.


Step 4: Product B

Addition of \(Cl_2\) across a C=C bond gives a vicinal dihalide: \[ Cyclohexene \xrightarrow{Cl_2/CCl_4} 1,2-dichlorocyclohexane \]

Step 5: Conclusion

Product A = Chlorocyclohexane

Product B = 1,2-dichlorocyclohexane

Hence, option (4) is correct.
Quick Tip: Alkanes undergo substitution under \(h\nu\), while alkenes undergo addition with \(Cl_2/CCl_4\).

Step 1: Fluorspar

Fluorspar (fluorite) has formula \(CaF_2\).
\(BF_3\) is boron trifluoride, not fluorspar.


Step 2: Check remaining options

Cryolite, Fluoroapatite and Carnallite have correct formulas.


Step 3: Conclusion

Hence, option (1) is incorrect.
Quick Tip: Fluorspar is calcium fluoride, \(CaF_2\).

Step 1: Lone pair interaction

Lone pair interacting with adjacent \(\pi\) bond causes electron delocalization.


Step 2: Identify effect

This delocalization produces multiple resonance structures.


Step 3: Eliminate others

Hyperconjugation involves \(\sigma\) electrons.

Inductive effect involves \(\sigma\) bond polarization.

Electromeric effect is temporary.


Step 4: Conclusion

Hence, resonance effect is responsible.
Quick Tip: Lone pair participation with \(\pi\) bond always indicates resonance.

Step 1: Thermal decomposition

Potassium permanganate decomposes on heating.

Step 2: Reaction
\[ 2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \]

Step 3: Conclusion

Products formed along with oxygen are \(K_2MnO_4\) and \(MnO_2\).
Quick Tip: \(KMnO_4\) decomposes to manganate and \(MnO_2\) on heating.

Step 1: Bridging CO

A bridging CO is shared between two metal atoms.


Step 2: Check options
\([Mn_2(CO)_{10}]\) has only terminal CO.
\([Os_3(CO)_{12}]\) and \([Ru_3(CO)_{12}]\) have terminal CO ligands.


Step 3: Correct compound
\([Co_2(CO)_8]\) contains two bridging CO ligands.

Step 4: Conclusion

Hence, option (1) is correct.
Quick Tip: Bridging CO ligands are common in cobalt carbonyls.

Step 1: Nature of proteins

Proteins are polymers made up of amino acids joined by peptide bonds.


Step 2: Structure of protein amino acids

In proteins, the amino group (\(-NH_2\)) is attached to the \(\alpha\)-carbon, which is the carbon adjacent to the carboxyl group (\(-COOH\)).

Step 3: Hydrolysis

On hydrolysis, proteins break into their constituent amino acids without change in carbon skeleton.


Step 4: Conclusion

Thus, amino acids obtained are \(\alpha\)-amino acids.
Quick Tip: All natural amino acids in proteins are \(\alpha\)-amino acids (proline is an imino acid).

Step 1: Identify key reagent
\(H_2N–NH_2\), KOH, heat is Wolf–Kishner reduction.


Step 2: Function of Wolf–Kishner reduction

It reduces carbonyl group (\(C=O\)) of aldehydes or ketones to methylene group (\(CH_2\)).


Step 3: Starting compound

Cyclohexanone is a ketone.


Step 4: Final transformation

Cyclohexanone \(\xrightarrow{Wolf–Kishner}\) Cyclohexane.


Step 5: Conclusion

Final product A is cyclohexane.
Quick Tip: Wolf–Kishner and Clemmensen reductions both convert \(C=O\) to \(CH_2\).

Step 1: Gibbs free energy criterion
\(\Delta G\) determines spontaneity at constant \(T\) and \(P\).


Step 2: Analyse options
\(\Delta G < 0\) : spontaneous (correct).
\(\Delta G = 0\) : equilibrium (correct).
\(\Delta G > 0\) : non-spontaneous (correct).

Step 3: Identify incorrect statement

Statement (2) claims \(\Delta G > 0\) for spontaneous reaction, which is incorrect.

Step 4: Conclusion

Option (2) is not correct.
Quick Tip: Spontaneity depends on sign of \(\Delta G\), not on reaction speed.

Step 1: Nature of reaction

Chlorine undergoes disproportionation in cold dilute alkali.


Step 2: Balanced reaction
\[ Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O \]

Step 3: Identify coefficients
\(a=1,\; b=2,\; c=1,\; d=1\).


Step 4: Conclusion

Correct option is (1).
Quick Tip: Cold alkali gives hypohalite; hot concentrated alkali gives halate.

Step 1: Behaviour of permanganate in alkaline medium
\(MnO_4^-\) is reduced to \(MnO_2\) in alkaline medium.


Step 2: Oxidation of iodide

Iodide ion (\(-1\)) is oxidized to iodate ion (\(+5\)).


Step 3: Net reaction
\[ 2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^- \]

Step 4: Conclusion
\(I^-\) is oxidized to \(IO_3^-\).
Quick Tip: In alkaline medium, permanganate gives \(MnO_2\) and oxidizes halides to oxyanions.

Step 1: Understanding the Concept:

The number of lone pairs on the central atom is determined using VSEPR theory.

Lone Pairs (LP) = \(\frac{1}{2} [V - B]\), where \(V\) is valence electrons and \(B\) is the number of electrons used in bonding.


Step 2: Detailed Explanation:

Evaluating each species:

1. \(\mathbf{O_{3}}\) (Ozone): The central oxygen forms one double bond and one dative bond, leaving 1 lone pair.

2. \(\mathbf{H_{2}O}\): Oxygen has 6 valence electrons; 2 used for bonding with H. \(LP = (6-2)/2 = 2\).

3. \(\mathbf{SF_{4}}\): Sulfur has 6 valence electrons; 4 used for bonding with F. \(LP = (6-4)/2 = 1\).

4. \(\mathbf{ClF_{3}}\): Chlorine has 7 valence electrons; 3 used for bonding with F. \(LP = (7-3)/2 = 2\).

5. \(\mathbf{NH_{3}}\): Nitrogen has 5 valence electrons; 3 used for bonding with H. \(LP = (5-3)/2 = 1\).

6. \(\mathbf{BrF_{5}}\): Bromine has 7 valence electrons; 5 used for bonding with F. \(LP = (7-5)/2 = 1\).

7. \(\mathbf{XeF_{4}}\): Xenon has 8 valence electrons; 4 used for bonding with F. \(LP = (8-4)/2 = 2\).


Step 3: Final Answer:

Compounds with 1 lone pair: \(O_{3}\), \(SF_{4}\), \(NH_{3}\), and \(BrF_{5}\).

Total count = 4.
Quick Tip: For quick calculations: \(LP = \frac{Valence electrons - Valency of surrounding atoms}{2}\).
Oxygen as a surrounding atom in neutral molecules acts as a divalent atom.

Step 1: Understanding the Concept:

The mass of a substance deposited during electrolysis is given by Faraday's first law: \(w = ZIt\).


Step 2: Key Formula or Approach:
\[ w = \frac{E \cdot I \cdot t}{F} \]
Where \(E\) is the equivalent weight, \(I\) is current, \(t\) is time in seconds, and \(F\) is Faraday's constant (\(96500 C/mol\)).


Step 3: Detailed Explanation:
For \(Zn^{2+}\), the n-factor is 2.

Equivalent mass of Zn (\(E\)) = \(\frac{65.4}{2} = 32.7\).

Current (\(I\)) = 0.015 A.

Time (\(t\)) = \(15 \times 60 = 900\) seconds.
\[ w = \frac{32.7 \times 0.015 \times 900}{96500} \] \[ w = \frac{441.45}{96500} \approx 0.0045746 g \] \[ w = 45.746 \times 10^{-4} g \]
Rounding off gives 45.75.


Step 4: Final Answer:

The mass produced is \(45.75 \times 10^{-4}\) g.
Quick Tip: Always double-check the time unit. Minutes must be converted to seconds to match the SI unit of Charge (\(C = A \times s\)).

Step 1: Understanding the Concept:

According to the Arrhenius equation, \(k = A e^{-E_{a}/RT}\).


Step 2: Key Formula or Approach:

For an expression \(K = \frac{k_{1}k_{2}}{k_{3}}\), the composite activation energy is:
\[ E_{a} = E_{a1} + E_{a2} - E_{a3} \]

Step 3: Detailed Explanation:

Taking natural logs of the expression:
\[ \ln K = \ln k_{1} + \ln k_{2} - \ln k_{3} \]
Differentiating with respect to Temperature (\(T\)) and using the Arrhenius form \(\frac{d(\ln k)}{dT} = \frac{E_{a}}{RT^{2}}\):
\[ \frac{E_{a}}{RT^{2}} = \frac{E_{a1}}{RT^{2}} + \frac{E_{a2}}{RT^{2}} - \frac{E_{a3}}{RT^{2}} \] \[ E_{a} = E_{a1} + E_{a2} - E_{a3} \]
Substituting values:
\[ E_{a} = 40 + 50 - 60 = 30 kJ/mol \]

Step 4: Final Answer:

The overall activation energy is 30 kJ/mol.
Quick Tip: For overall rate constants, powers become coefficients for \(E_{a}\). Multiplication of \(k\)'s leads to addition of \(E_{a}\)'s, and division leads to subtraction.

Step 1: Understanding the Concept:

The relationship between pressure-based and concentration-based equilibrium constants is defined by the gas constant and temperature.


Step 2: Key Formula or Approach:
\[ K_{p} = K_{c}(RT)^{\Delta n_{g}} \]

Step 3: Detailed Explanation:

For \(N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)\):
\(\Delta n_{g} = 2 (product) - 1 (reactant) = 1\).

Given: \(K_{p} = 0.492\), \(R = 0.082\), \(T = 300\).
\[ 0.492 = K_{c}(0.082 \times 300)^{1} \] \[ 0.492 = K_{c} \times 24.6 \] \[ K_{c} = \frac{0.492}{24.6} = 0.02 \] \[ K_{c} = 2 \times 10^{-2} \]

Step 4: Final Answer:

The value for the blank is 2.
Quick Tip: Note that \(\Delta n_{g}\) only counts gaseous species. If \(\Delta n_{g} = 0\), then \(K_{p} = K_{c}\).

Step 1: Understanding the Concept:

Molarity (\(M\)) is the moles of solute per liter of solution.


Step 2: Key Formula or Approach:
\[ Molarity = \frac{% by mass \times density \times 10}{Molar mass of solute} \]

Step 3: Detailed Explanation:

Given: \(x = 31.4%\), \(d = 1.25 g/mL\), \(M_{w} = 98 g/mol\).
\[ M = \frac{31.4 \times 1.25 \times 10}{98} \] \[ M = \frac{392.5}{98} \approx 4.005 \]
The nearest integer is 4.


Step 4: Final Answer:

The molarity is 4 M.
Quick Tip: Derived from: \(M = \frac{w_{2} \times 1000}{M_{2} \times V_{soln}} \). If you assume 100 g solution, mass of solute = \(x\) g and volume = \(100/d\) mL. Plugging these in yields the \(\frac{10xd}{M_{w}}\) formula.

Step 1: Understanding the Concept:

Osmotic pressure (\(\pi\)) of a dilute solution follows \(\pi = CRT\). For the same solution, \(\pi \propto T\).


Step 2: Key Formula or Approach:
\[ \frac{\pi_{1}}{T_{1}} = \frac{\pi_{2}}{T_{2}} \]

Step 3: Detailed Explanation:

Given: \(\pi_{1} = 7 \times 10^{5} Pa\), \(T_{1} = 273 K\), \(T_{2} = 283 K\).
\[ \pi_{2} = \pi_{1} \times \frac{T_{2}}{T_{1}} \] \[ \pi_{2} = 7 \times 10^{5} \times \frac{283}{273} \approx 7.2564 \times 10^{5} Pa \]
Since \(1 Pa = 1 Nm^{-2}\):
\[ \pi_{2} = 72.564 \times 10^{4} Nm^{-2} \]

Step 4: Final Answer:

The osmotic pressure is \(72.56 \times 10^{4} Nm^{-2}\).
Quick Tip: Ensure the units on both sides are consistent. The factor of 10 difference between \(10^{5}\) and \(10^{4}\) often confuses students; always calculate the full value first.

Step 1: Understanding the Concept:

A heteroatom is an atom other than carbon or hydrogen in an organic molecule. We are specifically looking for sulfur (S).


Step 2: Detailed Explanation:

1. Furan: Contains Oxygen in a 5-membered ring.

2. Thiophene: Contains Sulfur in a 5-membered ring.

3. Pyridine: Contains Nitrogen in a 6-membered ring.

4. Pyrrole: Contains Nitrogen in a 5-membered ring.

5. Cysteine: An amino acid containing a thiol group (\(-SH\)). It contains Sulfur.

6. Tyrosine: Contains Oxygen and Nitrogen, but no sulfur.


Step 3: Final Answer:

The two sulfur-containing compounds are Thiophene and Cysteine.
Quick Tip: Sulfur-containing amino acids are Cysteine and Methionine. Heterocyclic sulfur compounds typically start with the prefix "Thio-".

Step 1: Understanding the Concept:

Paramagnetism requires unpaired electrons. Bond order (B.O.) is calculated using Molecular Orbital Theory.


Step 2: Detailed Explanation:

1. \(\mathbf{H_{2}}\): 2e, B.O.=1, Diamagnetic.

2. \(\mathbf{He_{2}^{+}}\): 3e, B.O.=0.5, Paramagnetic.

3. \(\mathbf{O_{2}^{+}}\): 15e, B.O.=2.5, Paramagnetic.

4. \(\mathbf{N_{2}^{-}}\): 15e, B.O.=2.5, Paramagnetic.

5. \(\mathbf{O_{2}^{2-}}\): 18e, B.O.=1, Diamagnetic.

6. \(\mathbf{F_{2}}\): 18e, B.O.=1, Diamagnetic.

7. \(\mathbf{Ne_{2}^{+}}\): 19e, B.O.=0.5, Paramagnetic.

8. \(\mathbf{B_{2}}\): 10e, B.O.=1. Electronic configuration: \(\dots (\pi 2p_{x})^{1} = (\pi 2p_{y})^{1}\). It has 2 unpaired electrons, so it is Paramagnetic.


Step 3: Final Answer:

Only \(B_{2}\) fits both criteria (B.O. = 1 and Paramagnetic). Count = 1.
Quick Tip: Remember the trick: even electron species are usually diamagnetic except for \(B_{2}\) (10e) and \(O_{2}\) (16e).

Step 1: Understanding the Concept:

Fehling's test is positive for aliphatic aldehydes. Aromatic aldehydes and ketones generally give negative results.


Step 2: Detailed Explanation:

1. Benzaldehyde: Aromatic aldehyde (Negative).

2. Acetaldehyde: Aliphatic aldehyde (Positive).

3. Acetone: Ketone (Negative).

4. Acetophenone: Aromatic ketone (Negative).

5. Methanal: Aliphatic aldehyde (Positive).

6. 4-nitrobenzocyclohexane carbaldehyde: The aldehydic group is attached to a saturated carbon in the cyclohexane ring, making it non-aromatic in nature (Positive).


Step 3: Final Answer:

Three compounds (Acetaldehyde, Methanal, and the substituted cyclohexane carbaldehyde) give a positive test.
Quick Tip: Benzaldehyde is often used as a trap in tests; it gives a positive Tollen's test but a negative Fehling's test.

Step 1: Understanding the Concept:

The reaction is reductive ozonolysis of an alkene, which cleaves the double bond to form carbonyl compounds.


Step 2: Detailed Explanation:

The reactant is But-2-ene. Ozonolysis splits the \(C=C\) bond.
\[ CH_{3}-CH=CH-CH_{3} \xrightarrow{O_{3}, Zn/H_{2}O} 2 CH_{3}-CHO \]
The product (P) is ethanal (acetaldehyde). The structure of ethanal is \(CH_{3}CHO\).


Step 3: Final Answer:

In one molecule of ethanal (\(CH_{3}CHO\)), there is exactly 1 oxygen atom.
Quick Tip: In ozonolysis, simply "cut" the double bond and "paste" an oxygen atom on both sides of the cleavage.