Question

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

Consider a differentiable function $f$ satisfying the relation $f(x-y+1)=\frac{f\left(x\right)}{f(y-1)}$ for all $x,y\in R$ and $f^{\prime }\left(0\right)=2,$ $f\left(0\right)=1.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}\int f\left(x\right)dx=\frac{2f\left(x\right)}{p}+C\text{where}C\text{is constant of integration,}& \text{(P)}& 1\\ & \text{then the value of}p\text{is}& & \\ \text{(B)}& \text{The value of}\frac{d^{10}}{d{x}^{10}}\left(f\left(\frac{x}{2}\right)\right)\text{at}x=0\text{is}& \text{(Q)}& 2\\ \text{(C)}& \text{The number of solutions of the equation}f\left(x\right)=x^2\text{is}& \text{(R)}& 3\\ \text{(D)}& \text{The value of}\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(x^2\right)}{\sin x}\text{is}& \text{(S)}& 4\end{array}$

Which of the following is a CORRECT combination?

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

Answer 1

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

$f(x-y+1)=\frac{f\left(x\right)}{f(y-1)}$
Put $x=0$
$f(1-y)=\frac{1}{f(y-1)}$
Let $y-1=u$
Then, $f(-u)=\frac{1}{f\left(u\right)}$
or $f\left(u\right)f(-u)=1$
Only possible $f\left(x\right)$ is $f\left(x\right)=e^{kx}$
Also, $f\left(0\right)=1$ and $f^{\prime }\left(0\right)=2$
$f^{\prime }\left(x\right)=k{e}^{kx}$
$\Rightarrow f^{\prime }\left(0\right)=k=2$
Hence, $f\left(x\right)=e^{2x}$

$\left(C\right)$
$e^{2x}=x^2\Rightarrow (e^x{)}^2-x^2=0\Rightarrow (e^x+x)(e^x-x)=0$
As $e^x>x,$ so $e^x-x>0$
and there is one solution for $e^x=-x$



$\therefore$ Number of solutions $=1$

$\left(D\right)$
$\underset{x\to 0}{lim}\frac{e^{2x}-e^{2x^2}}{x}=\underset{x\to 0}{lim}{e}^{2x^2}\cdot \left(\underset{x\to 0}{lim}\frac{e^{2x-2x^2}-1}{2x-2x^2}\right)\cdot \left(\underset{x\to 0}{lim}\frac{2x-2x^2}{x}\right)=1\times 1\times 2=2$