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Column – I Column – II
A) The area under the curve $y = \frac{\ln x^{2}}{x^{2}}$ from $x = 1 t o x = \infty$ is (p) 1
B) The $Δ A B C$ , let $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of A,B,C respectively. If $\bar{b} . (\bar{a} + \bar{c}) = \bar{b} . \bar{b} + \bar{a} . \bar{c}, | \bar{b} - \bar{a} | = 3, | \bar{c} - \bar{b} | = 4$ and the angle between the medians $\bar{A M} a n d \bar{B D}$ is $π - \cos^{- 1} (\frac{1}{k}), t h e n \frac{1}{9} [k] =$ (q) 2
C) If $c \leq 1$ and the system of equations $x + y - 1 = 0, 2 x - y - c = 0$ and $b x - 3 b y + c = 0$ is consistent then the maximum value of b is (r) 3
D) A one litre oil can shaped like a circular cylinder without top is to be designed with least material. If r, h be the radius, height of the cylinder respectively, then $\frac{h}{r} =$ (s) 4

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A-p;B-q;C-p;D-s

A-q;B-p;C-r;D-s

A-q;B-q;C-r;D-p

A-p;B-q;C-s;D-r

answer is C.

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Detailed Solution

P) $\int_{1}^{\infty} \frac{2 \ln x}{x^{2}} d x = 2 {\frac{- \ln x}{x} + \int \frac{1}{x^{2}} d x}_{1}^{\infty} = - 2 {(\frac{1 + \ln x}{x})}_{1}^{\infty}$

$= - 2 (0 - 1) = 2 (∵ \underset{x \to \infty}{L t} \frac{1 + \ln x}{x} = 0)$

Q) $\bar{b} . (\bar{a} + \bar{c}) = \bar{b} . \bar{b} + \bar{a} . \bar{c}$

$\Rightarrow (\bar{b} - \bar{c}) . (\bar{a} - \bar{b}) = 0 \Rightarrow A B ⊥ B C, A B = 3, B C = 4$

Let $B = (0, 0), C = (4, 0), A = (0, 3) \Rightarrow M = 2 i, \bar{B D} = 2 i + \frac{3 j}{2}, \bar{A M} = 2 i - 3 j$

$\cos θ = \frac{4 - \frac{9}{2}}{\sqrt{4 + \frac{9}{4}} \sqrt{4 + 9}} = \frac{- 1}{5 \sqrt{13}} θ = π - \cos^{- 1} (\frac{1}{5 \sqrt{13}}) \frac{1}{9} [k] = \frac{1}{9} [5 \sqrt{13}] = \frac{18}{9} = 2$

R) $|\begin{matrix} 1 & 1 & - 1 \\ 2 & - 1 & - c \\ b & - 3 b & c \end{matrix}| = 0 \Rightarrow \frac{5 b}{3 + 4 b} = c \leq 1 \Rightarrow \frac{- 3}{4} < b \leq 3$

Max value = 3

S) $V = π r^{2} h \to h = \frac{1}{π r^{2}}$

$S = π r^{2} + \frac{2}{r} \frac{d S}{d r} = 0 \Rightarrow h = r$

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