A) \[y=\sqrt{3}x+1\]
B) \[x=\sqrt{3}y+1\]
C) \[x=\sqrt{2}y+1\]
D) \[y=\sqrt{3}(x+1)+1\]
Correct Answer: C
Solution :
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 \[(y-0)=\frac{1}{\sqrt{2}}(x-1)\] \[\Rightarrow \] \[x=\sqrt{2}y+1\]
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