Q41 / 90 · JEE Main 2014 (12 Apr Online) · Mathematics

If $f(x)=x^2-x+5, x>\frac{1}{2}$, and $\mathrm{g}(\mathrm{x})$ is its inverse function, then $\mathrm{g}^{\prime}(7)$ equals:

  • A
    $-\frac{1}{3}$
  • B
    $\frac{1}{13}$
  • Correct answer
    C
    $\frac{1}{3}$
  • D
    $-\frac{1}{13}$

Solution

$f(x)=y=x^2-x+5$ $$ \begin{aligned} &x^2-x+\frac{1}{4}-\frac{1}{4}+5=y \\ &\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=y \\ &\left(x-\frac{1}{2}\right)^2=y-\frac{19}{4} \\ &x-\frac{1}{2}=\pm \sqrt{y-\frac{19}{4}} \end{aligned} $$ $$ \begin{aligned} &x=\frac{1}{2} \pm \sqrt{y-\frac{19}{4}} \\ &\text { As } x>\frac{1}{2} \\ &x=\frac{1}{2}+\sqrt{y-\frac{19}{4}} \\ &g(x)=\frac{1}{2}+\sqrt{x-\frac{19}{4}} \\ &g^{\prime}(x)=\frac{1}{2 \sqrt{x-\frac{19}{4}}} \end{aligned} $$ $$ g^{\prime}(7)=\frac{1}{2 \sqrt{7-\frac{19}{4}}}=\frac{1}{2 \frac{\sqrt{28-19}}{2}}=\frac{1}{3} $$

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

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