Question

If $f\left(x\right)=\left(x-a\right)g\left(x\right)$ and $g\left(x\right)$ is continuous $x=a$ then $f^{{}^{\prime }}\left(1\right)=$

• A. $g\left(1\right)$
• B. $g^{{}^{\prime }}\left(1\right)$
• C. $-g^{{}^{\prime }}\left(1\right)$
• D. $g^{{}^{\prime }}\left(-1\right)$

Answer 1

The correct option is A $g\left(1\right)$

From given identity ,

$f\left(a\right)=0$

$f(a+h)=(a+h-a)g(a+h)=hg(a+h)$

$f(a-h)=(a-h-a)g(a-h)=-hg(a-h)$

$f^{\prime }\left(a\right)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}=\underset{h\to 0}{lim}\frac{hg(a+h)-f\left(a\right)}{h}$ .......(RHD,Similiarly LHD can be worked out )

Since $g\left(x\right)$ is continuous at $x=a$ , we can write above equation as

$f^{\prime }\left(a\right)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}=\underset{h\to 0}{lim}\frac{hg\left(a\right)-f\left(a\right)}{h}=\underset{h\to 0}{lim}\frac{hg\left(a\right)}{h}=g\left(a\right)$

So finally we arrive at $f^{\prime }\left(a\right)=g\left(a\right)$

So $f^{\prime }\left(1\right)=g\left(1\right)$