Question

If $f\left(x\right)=x^2+3x,$ then the differential coefficient of $f\left(x\right)$ at $x=1,$ is

• A. $1$
• B. $2$
• C. $5$
• D. $4$

Answer 1

The correct option is C $5$

$R.H.D.=$
$f^{\prime }\left(1^{+}\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$
$=\underset{h\to 0}{lim}\frac{(1+h{)}^2+3(1+h)-4}{h}$
$=\underset{h\to 0}{lim}\frac{h^2+5h}{h}=\underset{h\to 0}{lim}(h+5)$
$\Rightarrow f^{\prime }\left(1^{+}\right)=5$

And,
$L.H.D.=$
$f^{\prime }\left(1^{-}\right)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{(1-h{)}^2+3(1-h)-4}{-h}$
$=\underset{h\to 0}{lim}\frac{h^2-5h}{-h}=\underset{h\to 0}{lim}(5-h)$
$\Rightarrow f^{\prime }\left(1^{-}\right)=5$

$\Rightarrow L.H.D.=R.H.D.=f^{\prime }\left(1\right)$
$\therefore$ $f^{\prime }\left(1\right)=5$