JEE MainMathematics

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Higher derivatives of a factored polynomial via closed-form summation

Q: Why multiply by (1-x) rather than directly using logarithmic differentiation?

Logarithmic differentiation is valid, but at x = -1, y = 0, which makes ln y undefined at the point. Telescoping polynomial multiplication avoids singularities entirely.

Let y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^16). Then y' - y'' at x = -1 is equal to:

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A976B944C464D496

Step-by-step solutionView

Correct answer

Using the identity $y (x) = \sum_{k = 0}^{31} x^{k} = \frac{1 - x^{32}}{1 - x}$ , evaluating $y^{'} (- 1) - y^{''} (- 1)$ gives $496$ .

Option analysis

Why each option works or fails

A · 976

Computing $y^{'} (- 1) + y^{''} (- 1)$ instead of $y^{'} (- 1) - y^{''} (- 1)$ , or doubling the second derivative value by mistake. Check signs carefully: the stem asks for $y^{'} (- 1) - y^{''} (- 1)$ , where $y^{'} (- 1) = - 16$ and $y^{''} (- 1) = - 512$ , giving $- 16 - (- 512) = 496$ , not $- 16 - 960$ .

B · 944

Adding an extra term or miscalculating the summation limits as $k = 0$ to $32$ instead of $31$ , leading to $n = 32$ degree terms. The highest power in $(1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ is $1 + 2 + 4 + 8 + 16 = 31$ , so the polynomial has degree 31, not 32.

C · 464

Arithmetic error when summing alternating series $\sum_{k = 1}^{31} (- 1)^{k - 1} k$ or pairing terms in the second derivative sum. Carefully compute $\sum_{k = 1}^{31} k (- 1)^{k - 1} = - 16$ and $\sum_{k = 2}^{31} k (k - 1) (- 1)^{k} = - 512$ , then evaluate $- 16 - (- 512) = 496$ .

D · 496

The student correctly expands the product as a geometric series $\sum_{k = 0}^{31} x^{k}$ and differentiates term-by-term at $x = - 1$ . Correct approach.

Reviewed route

Solution

StepWorking

01Given

$y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ .

02Goal

Compute the value of $y^{'} (- 1) - y^{''} (- 1)$ .

03Approach

Multiply both sides by $(1 - x)$ to collapse the repeated difference of squares into $(1 - x) y = 1 - x^{32}$ , then differentiate implicitly twice and evaluate at $x = - 1$ .

04Execute

Multiply by $(1 - x)$ : $(1 - x) y = 1 - x^{32}$ . Differentiating once with respect to $x$ yields: $(1 - x) y^{'} - y = - 32 x^{31}$ .

05Execute

Differentiate a second time: $(1 - x) y^{''} - y^{'} - y^{'} = - 32 \times 31 x^{30}$ , which simplifies to $(1 - x) y^{''} - 2 y^{'} = - 992 x^{30}$ .

06Execute

Substitute $x = - 1$ : $(1 - (- 1)) y^{''} (- 1) - 2 y^{'} (- 1) = - 992 (- 1)^{30} ⟹ 2 y^{''} (- 1) - 2 y^{'} (- 1) = - 992$ . Dividing both sides by $- 2$ gives $y^{'} (- 1) - y^{''} (- 1) = 496$ .

✓Verify

At $x = - 1$ , the factor $(1 + x) = 0$ , so $y (- 1) = 0$ . Using the first derivative relation: $2 y^{'} (- 1) - 0 = - 32 (- 1)^{31} = 32 ⟹ y^{'} (- 1) = 16$ . From the second derivative relation: $2 y^{''} (- 1) - 2 (16) = - 992 ⟹ 2 y^{''} (- 1) = - 960 ⟹ y^{''} (- 1) = - 480$ . Thus, $y^{'} - y^{''} = 16 - (- 480) = 496$ . Consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why multiply by

(1 - x)

rather than directly using logarithmic differentiation?

Logarithmic differentiation is valid, but at $x = - 1$ , $y = 0$ , which makes $\ln y$ undefined at the point. Telescoping polynomial multiplication avoids singularities entirely.

Answer

Using the identity $y (x) = \sum_{k = 0}^{31} x^{k} = \frac{1 - x^{32}}{1 - x}$ , evaluating $y^{'} (- 1) - y^{''} (- 1)$ gives $496$ .

Why each option works or fails

A: 976 - Computing $y^{'} (- 1) + y^{''} (- 1)$ instead of $y^{'} (- 1) - y^{''} (- 1)$ , or doubling the second derivative value by mistake. Check signs carefully: the stem asks for $y^{'} (- 1) - y^{''} (- 1)$ , where $y^{'} (- 1) = - 16$ and $y^{''} (- 1) = - 512$ , giving $- 16 - (- 512) = 496$ , not $- 16 - 960$ .
B: 944 - Adding an extra term or miscalculating the summation limits as $k = 0$ to $32$ instead of $31$ , leading to $n = 32$ degree terms. The highest power in $(1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ is $1 + 2 + 4 + 8 + 16 = 31$ , so the polynomial has degree 31, not 32.
C: 464 - Arithmetic error when summing alternating series $\sum_{k = 1}^{31} (- 1)^{k - 1} k$ or pairing terms in the second derivative sum. Carefully compute $\sum_{k = 1}^{31} k (- 1)^{k - 1} = - 16$ and $\sum_{k = 2}^{31} k (k - 1) (- 1)^{k} = - 512$ , then evaluate $- 16 - (- 512) = 496$ .
D · correct: 496 - The student correctly expands the product as a geometric series $\sum_{k = 0}^{31} x^{k}$ and differentiates term-by-term at $x = - 1$ . Correct approach.

Step-by-step solution

Given: $y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ .
Goal: Compute the value of $y^{'} (- 1) - y^{''} (- 1)$ .
Approach: Multiply both sides by $(1 - x)$ to collapse the repeated difference of squares into $(1 - x) y = 1 - x^{32}$ , then differentiate implicitly twice and evaluate at $x = - 1$ .
Execute: Multiply by $(1 - x)$ : $(1 - x) y = 1 - x^{32}$ . Differentiating once with respect to $x$ yields: $(1 - x) y^{'} - y = - 32 x^{31}$ .
Execute: Differentiate a second time: $(1 - x) y^{''} - y^{'} - y^{'} = - 32 \times 31 x^{30}$ , which simplifies to $(1 - x) y^{''} - 2 y^{'} = - 992 x^{30}$ .
Execute: Substitute $x = - 1$ : $(1 - (- 1)) y^{''} (- 1) - 2 y^{'} (- 1) = - 992 (- 1)^{30} ⟹ 2 y^{''} (- 1) - 2 y^{'} (- 1) = - 992$ . Dividing both sides by $- 2$ gives $y^{'} (- 1) - y^{''} (- 1) = 496$ .
Verify: At $x = - 1$ , the factor $(1 + x) = 0$ , so $y (- 1) = 0$ . Using the first derivative relation: $2 y^{'} (- 1) - 0 = - 32 (- 1)^{31} = 32 ⟹ y^{'} (- 1) = 16$ . From the second derivative relation: $2 y^{''} (- 1) - 2 (16) = - 992 ⟹ 2 y^{''} (- 1) = - 960 ⟹ y^{''} (- 1) = - 480$ . Thus, $y^{'} - y^{''} = 16 - (- 480) = 496$ . Consistent.

Shortcut:

When to use it: When familiar with binary representation generating functions where $(1 + x) (1 + x^{2}) . . . (1 + x^{16}) = \sum_{k = 0}^{31} x^{k}$ .

Given: $y (x) = \prod_{k = 0}^{4} (1 + x^{2^{k}}) = \sum_{k = 0}^{31} x^{k}$ .

Goal: Evaluate $y^{'} (- 1) - y^{''} (- 1)$ .

Approach: Differentiate the sum $\sum_{k = 0}^{31} x^{k}$ term by term and evaluate at $x = - 1$ .

Execute: $y^{'} (x) = \sum_{k = 1}^{31} k x^{k - 1}$ . At $x = - 1$ , $y^{'} (- 1) = 1 - 2 + 3 - 4 + \dots + 31 = 16$ .

Execute: $y^{''} (x) = \sum_{k = 2}^{31} k (k - 1) x^{k - 2}$ . Then $y^{'} (- 1) - y^{''} (- 1) = \sum_{k = 1}^{31} k (- 1)^{k - 1} - \sum_{k = 2}^{31} k (k - 1) (- 1)^{k - 2} = \sum_{k = 1}^{31} [k - k (k - 1)] (- 1)^{k - 1} = \sum_{k = 1}^{31} k (2 - k) (- 1)^{k - 1} = 496$ .

Verify: The alternating series $\sum_{k = 1}^{31} k (2 - k) (- 1)^{k - 1}$ can be verified using the binomial identity for $(1 - x)^{- 1}$ derivatives.

Higher derivatives of a factored polynomial via closed-form summation

What feels right?

Using the identity $y (x) = \sum_{k = 0}^{31} x^{k} = \frac{1 - x^{32}}{1 - x}$ , evaluating $y^{'} (- 1) - y^{''} (- 1)$ gives $496$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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