Question

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

• A. $\sqrt{2}y$ + 1 = x
• B. $\sqrt{3}$x + 1 = y
• C. $\sqrt{3}$x + 1 + $\sqrt{3}$= y
• D. None of these

Answer 1

Answer: (A)

$\sqrt{2}y$ + 1 = x

Answer 2

At x =1, y =0

$\frac{dy}{dx}$= $\frac{1}{\sqrt{1 + x^{6}}}$. 3x² – $\frac{1}{\sqrt{1 + x^{4}}}$. 2x

$\left( \frac{dy}{dx} \right)_{(1,0)}$= $\frac{3}{\sqrt{2}}$–$\frac{2}{\sqrt{2}}$= $\frac{1}{\sqrt{2}}$

equation is y = $\frac{1}{\sqrt{2}}$ (x –1) Ž $\sqrt{2}$y + 1 = x