Question

$\text{List I}$ has four entries and $\text{List II}$ has four entries. Each entry of $\text{List I}$ has a unique match in $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Let}f\text{be a real-valued differentiable}& \text{(P)}& 18\\ & \text{function on}R\text{such that}{f}^{\prime }\left(1\right)=6& & \\ & \text{and}{f}^{\prime }\left(2\right)=2.\text{Then}& & \\ & \underset{h\to 0}{lim}\frac{f(3\cos h+4\sin h-2)-f\left(1\right)}{f(3e^h-5\sec h+4)-f\left(2\right)}& & \\ & \text{is equal to}& & \\ \text{(B)}& \text{If}(0.5{)}^{{\log }_3{\log }_{1/5}(x^2-\frac{4}{5})}>1,\text{then the least}& \text{(Q)}& 4\\ & \text{positive integral value of}x\text{is}& & \\ \text{(C)}& \text{The value of}\underset{x\to 1}{lim}\frac{{\sin }^2(x^3+x^2+x-3)}{1-\cos (x^2-4x+3)}\text{is}& \text{(R)}& 5\\ \text{(D)}& \text{Number of points of non-differentiability of}& \text{(S)}& 2\\ & f\left(x\right)=min(|\sin x|,|\cos x|,\frac{1}{4})\text{in}(0,\pi )\text{is}& & \end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

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

$\text{(A)}$
$\underset{h\to 0}{lim}\frac{f(3\cos h+4\sin h-2)-f\left(1\right)}{f(3e^h-5\sec h+4)-f\left(2\right)}\left(\frac{0}{0}\right)\text{form}$
Using L'Hopitals' rule, we get
$\underset{h\to 0}{lim}\frac{(-3\sin h+4\cos h)f^{\prime }(3\cos h+4\sin h-2)}{(3e^h-5\sec h\tan h)f^{\prime }(3e^h-5\sec h+4)}$
$=\frac{4}{3}\times \frac{f^{\prime }\left(1\right)}{f^{\prime }\left(2\right)}$
$=\frac{4}{3}\times \frac{6}{2}=4$

$\text{(B)}$
Clearly, $x^2-\frac{4}{5}>0$
${\left(\frac{1}{2}\right)}^{{\log }_3{\log }_{1/5}(x^2-\frac{4}{5})}>1={\left(\frac{1}{2}\right)}^0$
$\Rightarrow {\log }_3{\log }_{1/5}(x^2-\frac{4}{5})<0={\log }_{3}1$
$\Rightarrow {\log }_{1/5}(x^2-\frac{4}{5})<1={\log }_{1/5}\left(\frac{1}{5}\right)$
$\Rightarrow x^2-\frac{4}{5}>\frac{1}{5}$
$\Rightarrow x^2>1$
$\Rightarrow |x|>1$