If y=[(x+1)(2 x+1)(3 x+1) (n x+1)]^{4}, where n N and d y/d x at x=0 is 2 k, then the value of k is

Correct Answer is : n(n+1)n(n+1)n(n+1)

Correct Answer is : n(n+1)n(n+1)n(n+1)

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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 122 of 150

Marks: +1, -0

y=[(x+1)(2 x+1)(3 x+1) \ldots \ldots(n x+1)]^{4}

n \in N

\frac{d y}{d x}

x=0

2 k

k

Solution:

Concept:

Use the chain rule and the fact that the derivative of a product at

x=0

is the sum of the coefficients of

x

in each factor.

Explanation:

Let

u = (x+1)(2x+1)(3x+1)\cdots(nx+1)

Then

y = u^{4}

Differentiate:

\frac{dy}{dx} = 4u^{3}\frac{du}{dx}

x=0

, each factor becomes

1

, so

u(0)=1

Now find

\frac{du}{dx}

x=0

Each factor is of the form

(1 + mx)

, whose derivative at

x=0

m

\left.\frac{du}{dx}\right|_{x=0} = 1+2+3+\cdots+n = \frac{n(n+1)}{2}

Thus

\left.\frac{dy}{dx}\right|_{x=0} = 4(1)^{3}\cdot \frac{n(n+1)}{2} = 2n(n+1)

Given

\left.\frac{dy}{dx}\right|_{x=0} = 2k

2k = 2n(n+1)

, hence

k = n(n+1)

Answer:

k = n(n+1)

which matches Option B.

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