Question

$f\left(x\right)=\{\begin{array}{cc}{x}^2+3x+a& \text{for}x\le 1\\ bx+2& \text{for}x>1\end{array}$
is everywhere differentiable. Then $f^{\prime }\left(1\right)=?$

• A. $3$
• B. $5$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$5$

Since $f\left(x\right)$ is everywhere differentiable , it should be continuous and differentiable at $x=1$

By using continuity , we get $a+4=b+2$

$\Rightarrow b=a+2$

By using differentiability, we get $b=5$

So we get $a=3,b=5$

Therefore $f^1\left(x\right)=b=5$

$\Rightarrow f^1\left(1\right)=5$