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, $\frac{dy}{dx}=\frac{d}{dx}{\int }_{x^2}^{x^3}\frac{dt}{\sqrt{t^2+1}}$
$=\frac{1}{\sqrt{x^6+1}}\frac{d}{dx}(x^3)-\frac{1}{\sqrt{x^4+1}}\frac{d}{dx}(x^2)$
$=\frac{3x^2}{\sqrt{x^6+1}}-\frac{2x}{\sqrt{x^4+1}}$
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
$(y-0)=\frac{1}{\sqrt{2}}(x-1)$
$\Rightarrow$ $x=\sqrt{2}y+1$