Question

If $f(x)$ be differentiable function and curve $y=f(x)$ passes through $(1,1)$ and satisfies the relation $2 f(x+y)+f(x-y)+3 y^2=3 f(x)+2 x y$, then $\displaystyle\lim _{x \rightarrow 1} \frac{f(x)-1}{x-1}$ is equal to

• A. $3$
• B. $0$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

Put $x=1,y=h$ in the relation $2\left(\frac{f \left(\right. 1 + h \left.\right) - f \left(\right. 1 \left.\right)}{h}\right)-\frac{\left(\right. f \left(\right. 1 - h \left.\right) - f \left(\right. 1 \left.\right) \left.\right)}{- h}+3h=2$
take limit $h \rightarrow 0$
$2f^{'}\left(\right.1\left.\right)-f^{'}\left(\right.1\left.\right)=2\Rightarrow f^{'}\left(\right.1\left.\right)=2$