The equation of the tangent to the curvey=∫x2x3 dt1+t2 at x=1 is

dydx=11+t2t=x3ddxx3−11+t2t=x2ddxx2=3x21+x6−2x1+x4dydxx=1=12Also y(1)=0.So the equation of tangent is y−0=12(x−1)

The equation of the tangent to the curvey=∫x2x3 dt1+t2 at x=1 is

dydx=11+t2t=x3ddxx3−11+t2t=x2ddxx2=3x21+x6−2x1+x4dydxx=1=12Also y(1)=0.So the equation of tangent is y−0=12(x−1)

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

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$\sqrt{3} x + 1 + \sqrt{3} = y$

none of these

$\sqrt{3} x + 1 = y$

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

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} = {\frac{1}{\sqrt{1 + t^{2}}}|}_{t = x^{3}} \frac{d}{d x} (x^{3}) - {\frac{1}{\sqrt{1 + t^{2}}}|}_{t = x^{2}} \frac{d}{d x} (x^{2}) \\ = \frac{3 x^{2}}{\sqrt{1 + x^{6}}} - \frac{2 x}{\sqrt{1 + x^{4}}} \\ {\frac{d y}{d x}|}_{x = 1} = \frac{1}{\sqrt{2}} \end{array}$