Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}m\text{and}n\text{are positive integers satisfying}& \text{(P)}& 9\\ & 1+\cos 2\theta +\cos 4\theta +\cos 6\theta +\cos 8\theta +\cos 10\theta =\frac{\cos m\theta \cdot \sin n\theta }{\sin \theta },& & \\ & \text{then}(m+n)\text{is equal to}& & \\ \text{(B)}& \text{The minimum value of the expression}\frac{9x^2{\sin }^{2}x+4}{x\sin x}\text{for}x\in (0,\pi )\text{is}& \text{(Q)}& 10\\ \text{(C)}& \text{Let}f\left(x\right)=11-8\sin x-2{\cos }^{2}x.\text{If the maximum and minimum values of}f\left(x\right)& \text{(R)}& 11\\ & \text{are denoted by}M\text{and}m\text{respectively, then}\frac{M+8}{m}\text{has the value equal to}& & \\ \text{(D)}& \text{If}\tan 9\theta =\frac{3}{4}(\text{where}0<\theta <\frac{\pi }{18}),\text{then the value of}(\text{3 cosec 3}\theta -4\sec 3\theta )& \text{(S)}& 12\\ & \text{is equal to}& & \\ & & \text{(T)}& 13\end{array}$

Which of the following is a CORRECT combination ?

• A. $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(Q)}$
• B. $\text{(C)}\to \text{(T)},\text{(D)}\to \text{(Q)}$
• C. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(S)}$
• D. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(P)}$

Answer 1

The correct option is A $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(Q)}$

$\text{(C)}$
$f\left(x\right)=11-8\sin x-2{\cos }^{2}x$
$\Rightarrow f\left(x\right)=2({(\sin x-2)}^2+\frac{1}{2})$
$\therefore M=19$ and $m=3$
Hence, $\frac{M+8}{m}=\frac{19+8}{3}=9$
$\text{(C)}\to \text{(P)}$


$\text{(D)}$
$\tan 9\theta =\frac{3}{4}$
Then $\sin 9\theta =\frac{3}{5};\cos 9\theta =\frac{4}{5}$
Now, $\frac{3}{\sin 3\theta }-\frac{4}{\cos 3\theta }$
$=2\frac{(3\cos 3\theta -4\sin 3\theta )}{2\sin 3\theta \cos 3\theta }$
$=\frac{10[\frac{3}{5}\cos 3\theta -\frac{4}{5}\sin 3\theta ]}{\sin 6\theta }$
$=10\left[\frac{\sin 9\theta \cos 3\theta -\cos 9\theta \sin 3\theta }{\sin 6\theta }\right]$
$=10\frac{\sin (9\theta -3\theta )}{\sin 6\theta }=10$
$\text{(D)}\to \text{(Q)}$