The slope of the tangent to the curve y=∫0x 11+t3dt Then, x1+y1=a at the point where x = 1, is

The slope of the tangent to the curve $y = \int_{0}^{x} \frac{1}{1 + t^{3}} d t$ Then, $\sqrt{x_{1}} + \sqrt{y_{1}} = \sqrt{a}$ at the point where $x = 1$ , is

A
$\frac{1}{2}$
B
1
C
$\frac{1}{4}$
D
non-existent

Solution:

We have,

$y = \int_{0}^{x} \frac{1}{1 + t^{3}} d t \Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{3}} \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \frac{1}{2}$

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