MCQ
If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ then $\frac{\text{dy}}{\text{dx}}$ at x = 1 is
- A$1$
- B$\frac{1}{2}$
- C$\frac{1}{\sqrt{\text{2}}}$
- D$0$
Solution:
$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to x, we get
$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$
$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$
Putting x = 1, we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$
Thus,
$\frac{\text{dy}}{\text{dx}}$ at x = 1 is 0.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
The complex number z which satisfies the condition $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$ lies on: