Why multiply by rather than directly using logarithmic differentiation?

Logarithmic differentiation is valid, but at , , which makes undefined at the point. Telescoping polynomial multiplication avoids singularities entirely.

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Limits, Continuity and Differentiability: Mathematics | JEE Main

Let

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

. Then

y' - y''

x = -1

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

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

None. 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: $(1-x)y' - y = -32x^{31}$ .

05execute

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

06execute

Substitute $x = -1$ : $(1 - (-1))y''(-1) - 2y'(-1) = -992(-1)^{30} \implies 2y''(-1) - 2y'(-1) = -992$ . 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: $2y'(-1) - 0 = -32(-1)^{31} = 32 \implies y'(-1) = 16$ . From the second derivative relation: $2y''(-1) - 2(16) = -992 \implies 2y''(-1) = -960 \implies y''(-1) = -480$ . Thus, $y' - y'' = 16 - (-480) = 496$ . Consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 - None. 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: $(1-x)y' - y = -32x^{31}$ .
execute: Differentiate a second time: $(1-x)y'' - y' - y' = -32 \times 31 x^{30}$ , which simplifies to $(1-x)y'' - 2y' = -992 x^{30}$ .
execute: Substitute $x = -1$ : $(1 - (-1))y''(-1) - 2y'(-1) = -992(-1)^{30} \implies 2y''(-1) - 2y'(-1) = -992$ . 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: $2y'(-1) - 0 = -32(-1)^{31} = 32 \implies y'(-1) = 16$ . From the second derivative relation: $2y''(-1) - 2(16) = -992 \implies 2y''(-1) = -960 \implies y''(-1) = -480$ . 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.

Limits, Continuity and Differentiability: Mathematics | JEE Main

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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