Limits, Continuity & Differentiability
Differentiation
nta_abhyas_2025
Grade 12
Question:
Let $f(x) = x^2 - 4x - 3, x > 2$ and $g(x)$ is the inverse of $f(x)$. Then the value of $\frac{1}{g'(2)}$, where $f(z) = 2$ is (here, $g'$ represents the first derivative of $g$)
Step-by-Step Solution
Key Concept: The derivative of an inverse function satisfies $g'(f(x)) = \frac{1}{f'(x)}$
Given $g(f(x)) = x$. Differentiating, $g'(f(x)) \cdot f'(x) = 1$, so $g'(f(x)) = \frac{1}{f'(x)}$. When $f(x) = 2$, we have $x^2 - 4x - 3 = 2$, giving $x^2 - 4x - 5 = 0$. This factors as $(x-5)(x+1) = 0$, so $x = 5$ or $x = -1$. Since $x > 2$, we have $x = 5$. Therefore $g'(2) = \frac{1}{f'(5)} = \frac{1}{2(5)-4} = \frac{1}{6}$.
Correct Answer: 7