Question

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List – I and List – II are given as options (A), (B), (C) and (D), out of which one is correct.

$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ P.& If\alpha =\frac{\pi }{7}then\frac{1}{cos\alpha }+\frac{2cos\alpha }{cos2\alpha }& 1.& 2\\ Q.& \underset{x\to \infty }{Lt}\left[\right(x-1\left)\right(x-2\left)\right(x+3\left)\right(x+5\left)\right(x+10){]}^{\frac{1}{5}}-x=& 2.& 3\\ R.& {\int }_{-2}^2\frac{3x^{2}dx}{1+e^x}=& 3.& 4\\ S.& \text{Let f(x)}=x^3+x^2+2x-1.\text{The minimum integral value of x}& 4.& 8\\ & \text{if x satisfies f(f(x)) > f(2x + 1) is}& & \end{array}$

• A. P-2 Q-3 R-4 S-1
• B. P-3 Q-2 R-1 S-4
• C. P-2 Q-3 R-1 S-4
• D. P-3 Q-2 R-4 S-1

Answer 1

The correct option is D

P-3 Q-2 R-4 S-1

$P:\frac{2sin\alpha }{sin2\alpha }+\frac{2cos\alpha }{cos2\alpha }=\frac{4sin3\alpha }{sin4\alpha }=4Q:1\frac{-1-2+3+5+10}{5}=3R:2l={\int }_{-2}^{2}3x^{2}dx\Rightarrow l={\int }_0^{2}3x^{2}dx=8$
$S:f^{\prime }\left(x\right)=3x^2+2x+2>0$
$f\left(x\right)>2x+1$
$\Rightarrow x^3+x^2+2x-1>2x+1\Rightarrow x^3+x^2-2>0\Rightarrow (x-1)(x^2+2x+2)>0\Rightarrow x>1$