Question

The equation of the tangent to the curve $ y=\int_{{{x}^{2}}}^{{{x}^{3}}}{\frac{dt}{\sqrt{{{t}^{2}}+1}}} $ at $ x=1 $ is

• A. $ y=\sqrt{3}x+1 $
• B. $ x=\sqrt{3}y+1 $
• C. $ x=\sqrt{2}y+1 $
• D. $ y=\sqrt{3}(x+1)+1 $

Answer 1

Answer: (C)

Given curve is $ y=\int_{{{x}^{2}}}^{{{x}^{3}}}{\frac{dt}{\sqrt{{{t}^{2}}+1}}} $
At $ x=1,\,\,\,\,y=0 $
Now, $ \frac{dy}{dx}=\frac{d}{dx}\,\,\int_{{{x}^{2}}}^{{{x}^{3}}}{\frac{dt}{\sqrt{{{t}^{2}}+1}}} $
$ =\frac{1}{\sqrt{{{x}^{6}}+1}}\,\frac{d}{dx}({{x}^{3}})-\frac{1}{\sqrt{{{x}^{4}}+1}}\,\frac{d}{dx}({{x}^{2}}) $
$ =\frac{3{{x}^{2}}}{\sqrt{{{x}^{6}}+1}}-\frac{2x}{\sqrt{{{x}^{4}}+1}} $
At $ x=1, $
$ {{\left( \frac{dy}{dx} \right)}_{x=1}}=\frac{3}{\sqrt{2}}-\frac{2}{\sqrt{2}}=\frac{1}{\sqrt{2}} $
Hence, required equation of tangent is
$ (y-0)=\frac{1}{\sqrt{2}}(x-1) $
$ \Rightarrow $ $ x=\sqrt{2}y+1 $