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Match List-I with List-II and select the correct answer using the code given below the list.
List-I List-II
(I) Let $p, q$ be real numbers such that $p^{2} + q > q^{2} + p$ . Then value $p^{2} + q$ is greater than (P) 3
(II) If equation $2 t^{3} - 9 t^{2} + 30 - a = 0$ has three real and distinct roots, then possible integral values of a is/are (Q) 5
(III) Let $f : [3, \infty) \to [1, \infty)$ be defined by $f (x) = π^{x (x - 3)}$ . Number of solutions of $f (x) = f^{- 1} (x)$ is always greater than (R) $- \frac{1}{2}$
(IV) Sum of series $\sum_{r = 1}^{\infty} {sin}^{- 1} [\frac{2 r + 1}{r (r + 1) (\sqrt{r^{2} + 2 r} + \sqrt{r^{2} - 1})}]$ is always less than (S) $- 1$
(T) 8
The correct option is:

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(I)-(P, Q, R) (II)-(Q, S) (III)-(P) (IV)-(S, T)

(I)- (P, Q, R) (II)-(P, R) (III)-(T) (IV)-(Q, T)

(I)- (P, Q, R, S) (II)-(P, Q, R) (III)-(T) (IV)-(P, Q, R, T)

(I)-(R, S) (II)-(Q, T) (III)-(R, S) (IV)-(P, Q, T)

answer is D.

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Detailed Solution

P. $p^{2} + q + q^{2} + p < 2 (p^{2} + q)$
   $0 \leq p^{2} + p + \frac{1}{4} + q^{2} + q + \frac{1}{4} < 2 (p^{2} + q) + \frac{1}{2}$
   $p^{2} + q > - \frac{1}{4}$
Q. $f (t) = 2 t^{3} - 9 t^{2} + 30 - a$
   $f^{'} (t) = 6 t (t - 3) = 0 \Rightarrow t = 0.3$
   $f (0) . f (3) < 0 \Rightarrow 3 < a < 30$
R. If $x \in [3, \infty)$ , then $y = π^{x (x - 3)}$ is increasing function
⇒ Clearly one-one and onto function
Equation $f (x) = f^{- 1} (x)$ has the same solution as $f (x) = x$
Now curve $y = x$ and $y = π^{x (x - 3)}$ intersect at one point for $x \geq 3$
Number of solution $= 1$
S. $T_{r} = {sin}^{- 1} (\frac{\sqrt{r^{2} + 2 r} - \sqrt{r^{2} - 1}}{r (r + 1)})$

$T_{r} = {sin}^{- 1} (\frac{1}{r} \sqrt{1 - \frac{1}{{(r + 1)}^{2}}} - \frac{1}{r + 1} \sqrt{1 - \frac{1}{r^{2}}})$
$T_{r} = {sin}^{- 1} (\frac{1}{r}) - {sin}^{- 1} (\frac{1}{r + 1}) S_{n} = {cos}^{- 1} (\frac{1}{n + 1}) \Rightarrow S_{\infty} = \frac{π}{2}$

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