Question

$\begin{array}{cccc}& List-I& List-II& \\ P.& I{f}_{\alpha }=\frac{\pi }{7}then\frac{1}{cos\alpha }+\frac{2cos\alpha }{cos2\alpha }=& 1.& 2\\ Q.& undersetx\to \infty Lt\left[\right(x-1\left)\right(x-2\left)\right(x+3\left)\right(x+5\left)\right(x+10){]}^{\frac{1}{5}}-x=& 2.& 3\\ R.& {\int }_{-2}^2\frac{3x^{2}dx}{1+e^x}=& 3.& 4\\ S.& Letf\left(x\right)=x^3+x^2+2x-1.Theminimumintegralvalueofxifxsatisfiesf\left(f\right(x\left)\right)>f(2x+1)is& 4.& 8\end{array}$

• 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

$P:\frac{2sin\alpha }{sin2\alpha }$ $+\frac{2cos\alpha }{cos2\alpha }$ $=\frac{4sin3\alpha }{sin4\alpha }$ $=4$

$Q:\frac{-1-2+3+5+10}{5}$ $=3$

$R:2I$ $={\int }_{-2}^{2}3x^{2}dx$ $\Rightarrow 1$ $={\int }_0^{2}3x^{2}dx=8$

$S:f^{,}\left(x\right)$ $=3x^2+2x+2>0$

$f\left(x\right)>2x+1$

$\Rightarrow x^3+x^2+2x–1>2x+1$

$\Rightarrow x^3+x^2–2>0$

$\Rightarrow (x–1)(x^2+2x+2)>0$

$\Rightarrow x>1$