Q81 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
If $g(x) = (x^2 + 2x + 1)\cdot f(x)$ such that $f(0) = 5$ and $\lim\limits_{x \to 0}\dfrac{f(x)-5}{x} = 4$ then $g'(0) = $
- A$20$
- B$12$
- C$18$
- Correct answerD$14$
Solution
Given $g(x) = (x^2 + 2x + 1) f(x)$
Differentiating with respect to $x$ using the product rule:
$g'(x) = (2x + 2) f(x) + (x^2 + 2x + 1) f'(x)$
Substituting $x = 0$:
$g'(0) = 2 f(0) + f'(0)$
From the given limit, $\lim\limits_{x \to 0}\dfrac{f(x)-5}{x} = 4$, which represents the derivative of $f(x)$ at $x = 0$.
Thus, $f'(0) = 4$.
Also given $f(0) = 5$.
Substituting these values:
$g'(0) = 2(5) + 4 = 14$
Answer: $14$
Differentiating with respect to $x$ using the product rule:
$g'(x) = (2x + 2) f(x) + (x^2 + 2x + 1) f'(x)$
Substituting $x = 0$:
$g'(0) = 2 f(0) + f'(0)$
From the given limit, $\lim\limits_{x \to 0}\dfrac{f(x)-5}{x} = 4$, which represents the derivative of $f(x)$ at $x = 0$.
Thus, $f'(0) = 4$.
Also given $f(0) = 5$.
Substituting these values:
$g'(0) = 2(5) + 4 = 14$
Answer: $14$
Hence, the correct answer is option: 'D'.