Question

If the function $f:R\to R$ defined by
$f\left(x\right)=\{\begin{array}{c}ax,x<2\\ a{x}^2-bx+3,x\ge 2\end{array}$ is differentiable, then the value of $f^{\prime }(-3)+f^{\prime }\left(3\right)$ is equal to:

• A. $3$
• B. $4$
• C. $\frac{15}{2}$
• D. $0$

Answer 1

Answer: (A)

$3$

$f\left(x\right)=\{\begin{array}{ccc}ax& ;& x<2\\ a{x}^2-bx+3& ;& x\ge 2\end{array}$

Since $f\left(x\right)$ is differentiable function

Then $f\left(x\right)$ must be continuous.

Therefore, $f(2+h)=f(2-h)=f\left(2\right)$

$\Rightarrow 4a-2b+3=0$

$\Rightarrow 2a-2b+3=0$ ...(1)

Also, $f^{\prime }(2+h)=f^{\prime }(2-h)$

Thus $2a\left(2\right)-b=a$

$\Rightarrow 3a=b$ ...(2)

Solving equation (1) & (2), we get

$a=\frac{3}{4}$ & $b=\frac{9}{4}$

Now $f^{\prime }(-3)=a$

and $f^{\prime }\left(3\right)=2a\left(3\right)-b=6a-b$

Therefore, $f^{\prime }(-3)+f^{\prime }\left(3\right)=7a-b=\frac{21}{4}-\frac{9}{4}=\frac{12}{4}=3$