Question

Assertion :Let $f$ be a real valued function satisfying $f\left(\frac{x}{y}\right)=f\left(x\right)-f\left(y\right)$ & $\underset{x\to 0}{lim}\frac{f(1+x)}{\left(x\right)}=4$. Then area bounded by the curve $y=f\left(x\right)$, the $y$-axis & the line $y=4$ is $4e$ square units. Reason: The function $f\left(x\right)$ is concave downward.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Given $f\left(\frac{x}{y}\right)=f\left(x\right)-f\left(y\right)$ ...(1)
Putting $x=y=1$, we get
$f\left(1\right)=0$
Now $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$ (fact)
$=\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})}{h}$ using (1)
$=\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})}{x\frac{h}{x}}=\frac{1}{x}\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})}{\frac{h}{x}}=\frac{4}{x}$ $[\because {lim}_{x\to 0}\frac{f(1+x)}{x}=4]$
$\therefore$$f\left(x\right)=4\log x+c$
For $x=1;f\left(1\right)=0+c$
$\therefore$ $c=0$ as $f\left(1\right)=0$
$\therefore$ $f\left(x\right)=4\log x=y$ (say)
$\therefore$ $x=e^{\frac{y}{4}}$
$\therefore$ Required area $={\int }_{-\infty }^{4}xdx={\int }_{-\infty }^4{e}^{\frac{y}{4}}dy$
$=4{\left(e^{\frac{y}{4}}\right)}_{-\infty }^4=4(e-0)4e$
Again $f\left(x\right)=4\log x$
$f^{\prime }\left(x\right)=\frac{4}{x}$ $\Rightarrow$ $f^{\prime \prime }\left(x\right)=-\frac{4}{x^2}<0\forall x\in R$
$\Rightarrow$ $f\left(x\right)$ is concave downward curve

Hence both assertion and reason are true but reason is not correct explanation of assertion.