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 value of constant $b$ is

• A. $5$
• B. $3$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

Answer: (A)

$5$

Given it is differentiable everywhere,
Therefore,

$f^{\prime }\left(x\right)=\{\begin{array}{cc}2x+3& ,x\le 1\\ b& ,x>1\end{array}$

At $x=1,f^{\prime }\left(1\right)=2\left(1\right)+3=5$

When $x>1$

As $f$ is differentiable at every point, both left-hand and right-hand derivatives should be equal.

$\therefore f^{\prime }\left(1\right)=b$

$\Rightarrow 5=b$

Hence, $b=5$.