Question

Let $f(x+y)=f\left(x\right)\times f\left(y\right)$ for all $x$ and $y$ and $f\left(7\right)=5,f^{\prime }\left(0\right)=2$ then $f^{\prime }\left(7\right)$ will be

• A. $5$
• B. $2$
• C. $7$
• D. $10$

Answer 1

The correct option is D $10$

Given $f(x+y)=f\left(x\right).f\left(y\right),\forall x,y$

Now $f(0+0)=f\left(0\right).f\left(0\right)$

or, $f\left(0\right)\left[f\right(0)-1]=0$

or, $f\left(0\right)=0$ or $f\left(0\right)=1$...........(1).

But $f\left(0\right)$ can't be $0$ otherwise it will lead to the contradiction $2=0$ by putting the value of $f\left(0\right)$ in (2).

Also given, $f^{\prime }\left(0\right)=2$

or, $\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=2$

or, $\underset{h\to 0}{lim}\frac{f\left(0\right)f\left(h\right)-f\left(0\right)}{h}=2$

or, $f\left(0\right)$ $\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=2$.........(2).

So $f\left(0\right)=1$ gives from (2),

or, $\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=2$.........(3).

Now $f^{\prime }\left(7\right)$

$=$ $\underset{h\to 0}{lim}\frac{f(7+h)-f\left(7\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(7\right)f\left(h\right)-f\left(7\right)}{h}$

$=f\left(7\right)$ $\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$

$=5.2$ [Using (3) and since $f\left(7\right)=5$]

$=10$.