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The value of k for which $k\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}},$ where $y=\sqrt{x}+\frac{1}{\sqrt{x}};$ is : |
$2x$ 2 $x$ $\sqrt{x}$ |
$2x$ |
The correct answer is Option (1) → $2x$ $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}$ so $2x\frac{dy}{dx}=\sqrt{x}-\frac{1}{\sqrt{x}}$ on comparison $k=2x$ |
