Q83 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
If $y = [(x+1)(2x+1)(3x+1)\ldots\ldots(nx+1)]^4$, where $n \in N$ and $\dfrac{dy}{dx}$ at $x = 0$ is $2k$, then the value of $k$ is
- A$\dfrac{n(n+1)}{2}$
- Correct answerB$n(n+1)$
- C$2n(n+1)$
- D$4n(n+1)$
Solution
Given $y = [(x+1)(2x+1)(3x+1)\ldots(nx+1)]^4$
Taking natural logarithm on both sides:
$\ln y = 4 [\ln(x+1) + \ln(2x+1) + \ln(3x+1) + \ldots + \ln(nx+1)]$
Differentiating with respect to $x$:
$\dfrac{1}{y} \dfrac{dy}{dx} = 4 \left[ \dfrac{1}{x+1} + \dfrac{2}{2x+1} + \dfrac{3}{3x+1} + \ldots + \dfrac{n}{nx+1} \right]$
At $x = 0$, $y = [(0+1)(0+1)(0+1)\ldots(0+1)]^4 = 1$
Substituting $x = 0$ and $y = 1$ in the derivative expression:
$\dfrac{1}{1} \left( \dfrac{dy}{dx} \right)_{x=0} = 4 [1 + 2 + 3 + \ldots + n]$
$\left( \dfrac{dy}{dx} \right)_{x=0} = 4 \left[ \dfrac{n(n+1)}{2} \right] = 2n(n+1)$
Given that $\left( \dfrac{dy}{dx} \right)_{x=0} = 2k$
$2k = 2n(n+1)$
$k = n(n+1)$
Answer: $n(n+1)$
Taking natural logarithm on both sides:
$\ln y = 4 [\ln(x+1) + \ln(2x+1) + \ln(3x+1) + \ldots + \ln(nx+1)]$
Differentiating with respect to $x$:
$\dfrac{1}{y} \dfrac{dy}{dx} = 4 \left[ \dfrac{1}{x+1} + \dfrac{2}{2x+1} + \dfrac{3}{3x+1} + \ldots + \dfrac{n}{nx+1} \right]$
At $x = 0$, $y = [(0+1)(0+1)(0+1)\ldots(0+1)]^4 = 1$
Substituting $x = 0$ and $y = 1$ in the derivative expression:
$\dfrac{1}{1} \left( \dfrac{dy}{dx} \right)_{x=0} = 4 [1 + 2 + 3 + \ldots + n]$
$\left( \dfrac{dy}{dx} \right)_{x=0} = 4 \left[ \dfrac{n(n+1)}{2} \right] = 2n(n+1)$
Given that $\left( \dfrac{dy}{dx} \right)_{x=0} = 2k$
$2k = 2n(n+1)$
$k = n(n+1)$
Answer: $n(n+1)$
Hence, the correct answer is option: 'B'.