Question

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

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Let}f\text{be a real-valued differentiable function on}R\text{such that}{f}^{\prime }\left(1\right)=6\text{and}{f}^{\prime }\left(2\right)=2.& \text{(P)}& 4\\ & \text{Then}\underset{h\to 0}{lim}\frac{f(3\cos h+4\sin h-2)-f\left(1\right)}{f(3e^h-5\sec h+4)-f\left(2\right)}\text{is equal to}& & \\ \text{(B)}& \text{For}a>0,\text{let}f:[-4a,4a]\to R\text{be an even function such that}f\left(x\right)=f(4a-x)\text{for all}& \text{(Q)}& 5\\ & x\in [2a,4a]\text{and}\underset{h\to 0}{lim}\frac{f(2a+h)-f\left(2a\right)}{h}=4.\text{Then}\underset{h\to 0}{lim}\frac{f(h-2a)-f(-2a)}{2h}\text{is equal to}& & \\ \text{(C)}& \text{Suppopse}f\text{is a differentiable function on}R.\text{Let}F\left(x\right)=f\left(e^x\right)\text{and}G\left(x\right)=e^{f\left(x\right)}.& \text{(R)}& 3\\ & \text{If}{f}^{\prime }\left(1\right)=e^3\text{and}f\left(0\right)=f^{\prime }\left(0\right)=3,\text{then}\frac{G^{\prime }\left(0\right)}{F^{\prime }\left(0\right)}\text{is equal to}& & \\ \text{(D)}& \text{Let}f\left(x\right)=max\{\cos x,x,2x-1\}\text{where}x\ge 0.\text{Then number of points of}& \text{(S)}& 2\\ & \text{non-differentiability of}f\left(x\right),\text{is equal to}& & \\ & & \text{(T)}& 1\end{array}$

Which of the following is a CORRECT combination?

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

Answer 1

The correct option is C $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(S)}$

$\left(\text{C}\right)$
We have $F\left(x\right)=f\left(e^x\right)$
$\Rightarrow F^{\prime }\left(x\right)=e^x{f}^{\prime }\left(e^x\right)$
And $G\left(x\right)=e^{f\left(x\right)}$
$\Rightarrow G^{\prime }\left(x\right)=e^{f\left(x\right)}{f}^{\prime }\left(x\right)\therefore \frac{G^{\prime }\left(0\right)}{F^{\prime }\left(0\right)}=\frac{f^{\prime }\left(0\right)e^{f\left(0\right)}}{e^0{f}^{\prime }\left(e^0\right)}=\frac{f^{\prime }\left(0\right)e^{f\left(0\right)}}{f^{\prime }\left(1\right)}=\frac{3e^3}{e^3}=3$
$\text{(C)}\to \text{(R)}$


$\left(\text{D}\right)$


$f\left(x\right)$ is non-differentiable at two points, $x=x_1$ and $x=1$
$\therefore$ Number of points of non-differentiability $=2$
$\text{(D)}\to \text{(S)}$