Question

Assertion :Let $f(x+y)=f\left(x\right)+f\left(y\right)-xy\forall x,yϵR$ & $\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}=5$, then area bounded by the curve $y=f\left(x\right)$, $x$-axis & the ordinates $x=0,x=10$ is $2{\int }_0^{5}f\left(x\right)dx$ Reason: Graph of $f\left(x\right)$ is symmetrical about the line $x=5$.

• 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 A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$f(x+y)=f\left(x\right)+f\left(y\right)-xy$ (A)
Putting $x=0=y$
$\therefore$ $f\left(0\right)=0$
Now $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)-hx-f\left(x\right)}{h}$ using (A)
$=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}-x$, $f^{\prime }\left(x\right)=5-x$
$\therefore$ $=5x-\frac{x^2}{2}+c$
Putting $x=0$ we get $c=0$
$\therefore$ $f\left(x\right)=5x-\frac{x^2}{2}=y$ (say)
For line of symmetry of f(x) we have $f^{\prime }\left(x\right)=0$
$\Rightarrow$ $5-\frac{2x}{2}=0$
$\Rightarrow$ $x=5$ so Reason (R) is true
(Note a function $f\left(x\right)$ is symmetrical about a line $x=a$ if $f(a+x)=f(a-x)$. In our case $x=a=5$ & $f(5+x)=f(5-x)$
$\therefore$ Required area $={\int }_0^{10}f\left(x\right)dx=2{\int }_0^{5}f\left(x\right)dx=\frac{250}{3}$ square units