Question

Let $y (x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)$. Then $y^{\prime}-y^{\prime \prime}$ at $x=-1$ is equal to :

• A. 976
• B. 944
• C. 496
• D. 464

Answer 1

The Correct Option is C

The correct answer is (C) : 496
\(y=\frac{1−x^{32}}{1-x​}\)
\(⇒ y−xy=1−x^{32} \)
\(y'−xy'−y=−32x^{31 }\)
\(y''−xy''−y'−y'=−(32)(31)x^{30}\)
at x=−1
\(⇒ y'−y''=496\)

Answer 2

1. Expand \(y(x)\) as: \[ y(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16}). \] 2. Differentiate \(y(x)\) using the product rule: \[ y'(x) = \frac{d}{dx}\left[(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16})\right]. \] 3. Use the derivative of each factor: \[ \frac{d}{dx}(1 + x) = 1, \quad \frac{d}{dx}(1 + x^2) = 2x, \quad \frac{d}{dx}(1 + x^4) = 4x^3, \ldots \] 4. Combine terms to find: \[ y'(x) = y(x) \left[\frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4x^3}{1 + x^4} + \frac{8x^7}{1 + x^8} + \frac{16x^{15}}{1 + x^{16}}\right]. \] 5. Compute \(y' - y''\) at \(x = -1\). Substitute \(x = -1\) into each term and simplify carefully. 6. After evaluation: \[ y' - y'' = 496. \] Thus, \(y' - y''\) at \(x = -1\) is \(496\). Apply the product rule systematically for functions involving multiple terms. Substitute the value of \(x\) at the end to avoid errors.