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Match the sta...

Question

Match the statements in Column-I with the values in Column-II.
$\begin{matrix} Column I & Column II (A) & A line from the origin meets the lines & (p) & - 4 \frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1} and \frac{x - \frac{8}{2}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1} at P and q respectively. If length P Q = d, then d^{2} is (B) & The values of x satisfying & (q) & 0 {tan}^{- 1} (x + 3) - {tan}^{- 1} (x - 3) = {sin}^{- 1} (\frac{3}{5}) are (C) & Non-zero vectors \to a, \to b and \to c satisfy & (r) & 4 \to a . \to b = 0, (\to b - \to a) . (\to b + \to c) = 0 and 2 | \to b + \to c | = | \to b - \to a | If \to a = μ \to b + 4 \to c, then the possible values of μ are (D) & Let f be the function on [- π, π] given by & (s) & 5 f (0) = 9 and f (x) = sin (\frac{9 x}{2}) / sin (\frac{x}{2}) for x \neq 0. The value of \frac{2}{π} π \int - π f (x) d x is (t) & 6 \end{matrix}$

A-R; B-P; C-Q; D-S

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A-S; B-Q; C-P; D-R

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A-Q; B-P; R-R; D-T

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A-T; B-P; C-Q; D-R

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Solution

The correct option is D A-T; B-P; C-Q; D-R
(A) Equation of line passing through origin is $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$
$∴ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 & - 1 1 & - 2 & 1 a & b & c \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow a (- 1) - b (3) + c (- 5) = 0$
$- a - 3 b - 5 c = 0$
$a + 3 b + 5 c = 0$ ... (1)
Also, $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{8}{3} & - 3 & 1 2 & - 1 & 1 a & b & c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 0$
$a (- 2) - b (\frac{2}{3}) + c (\frac{10}{3}) = 0$
$2 a + \frac{2 b}{3} - \frac{10 c}{3} = 0$
$3 a + b - 5 c = 0$ ... (2)
Now, $a + 3 b + 5 c = 0$
$3 a + b - 5 c = 0$
$\frac{a}{- 20} = \frac{b}{20} = \frac{c}{- 8}$
$\frac{a}{5} = \frac{b}{- 5} = \frac{c}{4}$
Equation of line is
$\frac{x}{5} = \frac{y}{- 5} = \frac{z}{+ 4}$ ... (2)
$\frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1}$ ... (3)
Now,
$\frac{x - \frac{8}{3}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1}$ ... (4)
Point on (2) is $(5 λ, - 5 λ, + 4 λ)$
Point on (3) is $(2 + k_{1}, 1 - 2 k_{1}, - 1 + k_{1})$
Point on (4) is $(\frac{8}{3} + 2 k_{2}, - 3 - k_{2}, 1 + k_{2})$
On solving,
$2 + k_{1} + 1 - 2 k_{1} = 0$
$- k_{1} + 3 = 0$
$k_{1} = 3$
$P \equiv (5, - 5, 2)$
Again for $Q$ ,
$\frac{8}{3} + 2 k_{2} - 3 - k_{2} = 0$
$k_{2} - \frac{1}{3} = 0$
$k_{2} = \frac{1}{3}$
$Q \equiv (\frac{10}{3}, \frac{- 10}{3}, \frac{4}{3})$
$P Q = \sqrt{{(\frac{5}{3})}^{2} + {(\frac{5}{3})}^{2} + {(\frac{2}{3})}^{2}}$
$= \frac{\sqrt{54}}{3}$
$P Q^{2} = d^{2} = \frac{54}{9} = 6$
(B) ${tan}^{- 1} (\frac{x + 3 - x + 3}{1 + (x^{2} - 9)}) = {tan}^{- 1} (\frac{3}{4})$
$\frac{6}{x^{2} - 8} = \frac{3}{4}$
$3 x^{2} - 24 = 24$
$3 x^{2} = 48$
$x = \pm 4$
(C) $(\to b - \to a) . ⎛ ⎜ ⎝ \to b + \frac{\to a - μ \to b}{4} ⎞ ⎟ ⎠ = 0$
$(\to b - \to a) . (4 \to b + \to a - μ \to b) = 0$
$(4 - μ) b^{2} - a^{2} = 0$ ... (1)
Also, $2 ∣ ∣ ∣ ∣ \to b + \frac{\to a - μ \to b}{4} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ \to b - \to a ∣ ∣ ∣$
$2 ∣ ∣ ∣ ∣ \frac{(4 - μ) \to b + \to a}{4} ∣ ∣ ∣ ∣$
$\frac{(4 - μ)^{2} b^{2}}{4} + \frac{a^{2}}{4} = b^{2} + a^{2}$
$\frac{3 a^{2}}{4} = \frac{(4 - μ)^{2} - 4}{4} . b^{2}$
$3 a^{2} = ((4 - μ)^{2} - 4) b^{2}$ ... (2)
From equations (1) and (2),
$3 (4 - μ) = (4 - μ)^{2} - 4$
$(4 - μ)^{2} - 3 (4 - μ) - 4 = 0$
$\Rightarrow μ = 0, 5$
$μ = 5$ is not admissibl.
(D) $f (0) = 9$ ,
$f (x) = \frac{sin (\frac{9 x}{2})}{sin \frac{x}{2}}$
$= (3 - 4 {sin}^{2} \frac{x}{2}) (3 - 4 {sin}^{2} \frac{3 x}{2})$
$= 9 - 12 {sin}^{2} \frac{x}{2} - 12 {sin}^{2} \frac{3 x}{2} + 16 {sin}^{2} \frac{x}{2} . {sin}^{2} \frac{3 x}{2}$
$= 9 - 6 (1 - cos x) - 6 (1 - cos 3 x) + 4 (1 - cos x) (1 - cos 3 x)$
$= 1 + 6 cos x + 6 cos 3 x - 4 cos x - 4 cos 3 x + 4 cos x cos 3 x$
$I = \frac{2}{π} π \int - π \frac{sin \frac{9 x}{2}}{sin \frac{x}{2}}$
$= \frac{4}{π} π \int 0 1 + 2 cos x + 2 cos 3 x + 2 (cos 4 x + cos x)$
$= \frac{4}{π} \times π$
$= 4$

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List II

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\begin{matrix} Column I & Column II a. For θ \in R, if max {5 sin θ + 3 sin (θ - α)} = 7, then the set of possible values of α is & p. 2 n π + \frac{3 π}{4}, n \in Z b. x \neq \frac{n π}{2} and (cos x)^{{sin}^{2} x - 3 sin x + 2} = 1 & q. 2 n π \pm \frac{π}{3}, n \in Z c. \sqrt{sin x} + 2^{1 / 4} cos x = 0 & r. 2 n π + {cos}^{- 1} (\frac{1}{3}), n \in Z d. {log}_{5} tan x = ({log}_{5} 4) ({log}_{4} (3 sin x)) & s. no solution \end{matrix}

Column I

Column II

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