CUET · MATHS · PYQ PAPER 2023
The value of k for which \(k \frac{d y}{d x}=\sqrt{x}-\frac{1}{\sqrt{x}}\), where \(y=\sqrt{x}+\frac{1}{\sqrt{x}}\); is :
- A 2x
- B 2
- C x
- D \(\sqrt{x}\)
Answer & Solution
Correct Answer
(A) 2x
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x} = \frac{d}{d x}\left(x^{1/2}+x^{-1/2}\right) = \frac{1}{2}x^{-1/2}-\frac{1}{2}x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{x-1}{2x\sqrt{x}}\) \(k \left(\frac{x-1}{2x\sqrt{x}}\right) = \sqrt{x}-\frac{1}{\sqrt{x}} = \frac{x-1}{\sqrt{x}}\)…
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