MHT CET · Maths · Differentiation
\(
y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots \ldots \ldots\left(1+x^{2 n}\right) \text {, }
\)
then the value of \(\frac{d y}{d x}\) at \(x=0\) is
- A 0
- B -1
- C 1
- D 2
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
\(
y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)
\)
Taking ' \(\log\) ' on both sides, we get
\(\log y=\log (1+x)+\log \left(1+x^2\right) +\log \left(1+x^4\right)+\) \(\ldots+\log \left(1+x^{2 \mathrm{n}}\right)\)
Differentiating w.r.t. \(x\), we get
\(
\frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\ldots+\frac{2 \mathrm{n} \times x^{2 \mathrm{n}-1}}{1+x^{2 \mathrm{n}}}
\)
At \(x=0\), (i) \(\Rightarrow y=1\)
\(\therefore \left(\right.\) ii) \(\left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0}=1+0+0+\ldots+0=1\)
y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)
\)
Taking ' \(\log\) ' on both sides, we get
\(\log y=\log (1+x)+\log \left(1+x^2\right) +\log \left(1+x^4\right)+\) \(\ldots+\log \left(1+x^{2 \mathrm{n}}\right)\)
Differentiating w.r.t. \(x\), we get
\(
\frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\ldots+\frac{2 \mathrm{n} \times x^{2 \mathrm{n}-1}}{1+x^{2 \mathrm{n}}}
\)
At \(x=0\), (i) \(\Rightarrow y=1\)
\(\therefore \left(\right.\) ii) \(\left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0}=1+0+0+\ldots+0=1\)
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