Q101 / 160 · TS EAMCET 2007 · Mathematics

If $f(x)=\left\{\begin{array}{lll}x-5, & \text { for } x \leq 1 \\ 4 x^2-9, & \text { for } 1 < x < 2 \\ 3 x+4, & \text { for } x \geq 2,\end{array}\right.$ then $f^{\prime}\left(2^{+}\right)$is equal to

  • A
    $0$
  • B
    $2$
  • Correct answer
    C
    $3$
  • D
    $4$

Solution

Given, $\begin{aligned} f(x) & = \begin{cases}x-5 & \text { for } x \leq 1 \\ 4 x^2-9 & \text { for } 1 < x < 2 \\ 3 x+4 & \text { for } x \geq 2\end{cases} \\ f^{\prime}\left(2^{+}\right) & =\lim _{x \rightarrow 2^{+}}\left(\frac{f(x)-f(2)}{x-2}\right) \\ & =\lim _{x \rightarrow 2^{+}} \frac{3 x+4-(6+4)}{x-2} \\ & =\lim _{x \rightarrow 2} \frac{3 x-6}{x-2} \\ & =3\end{aligned}$

Hence, the correct answer is option: 'C'.

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