Question:
If
\[
f(x)=(1+x^3)(1+x^6)(1+x^{12})(1+x^{24})
\]
then \(f'(-1)=\)
If
\[
f(x)=(1+x^3)(1+x^6)(1+x^{12})(1+x^{24})
\]
then \(f'(-1)=\)
Show Hint
Whenever powers double repeatedly (\(x^3,x^6,x^{12}\)), search immediately for telescoping product identities.
Updated On: Jun 15, 2026
- \(24\)
- \(12\)
- \(48\)
- \(60\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Use logarithmic differentiation or algebraic identity.
Notice pattern:
\[
(1-x)(1+x)=1-x^2
\]
Repeated telescoping often simplifies expressions.
Step 1: Observe product pattern.
Using identity \[ (1-x)(1+x)(1+x^2)(1+x^4)\dots = 1-x^{2^n} \] Transforming expression: \[ f(x)=\frac{1-x^{48}}{1-x^3} \]
Step 2: Differentiate.
Using quotient rule. \[ f'(x) = \frac{-48x^{47}(1-x^3)+(1-x^{48})(3x^2)} {(1-x^3)^2} \]
Step 3: Substitute \(x=-1\).
After simplification: \[ f'(-1)=48 \] Thus \[ \boxed{48} \]
Step 1: Observe product pattern.
Using identity \[ (1-x)(1+x)(1+x^2)(1+x^4)\dots = 1-x^{2^n} \] Transforming expression: \[ f(x)=\frac{1-x^{48}}{1-x^3} \]
Step 2: Differentiate.
Using quotient rule. \[ f'(x) = \frac{-48x^{47}(1-x^3)+(1-x^{48})(3x^2)} {(1-x^3)^2} \]
Step 3: Substitute \(x=-1\).
After simplification: \[ f'(-1)=48 \] Thus \[ \boxed{48} \]
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