Question

Which of the following points lies on the tangent to the curve $x^4{e}^y+2\sqrt{y+1}=3$ at the point $(1,0)$?

• A. $(2,6)$
• B. $(2,2)$
• C. $(-2,6)$
• D. $(-2,4)$

Answer 1

The correct option is C $(-2,6)$

$x^4{e}^y+2\sqrt{y+1}=3$
Differentiate w.r.t $x$
$4x^3{e}^y+x^4{e}^y{y}^{{}^{\prime }}+\frac{2y^{{}^{\prime }}}{2\sqrt{y+1}}=0$
at $P(1,0)$
$4+y_P^{{}^{\prime }}+\frac{2y_P^{{}^{\prime }}}{2}=0$
$\Rightarrow 2y_P^{{}^{\prime }}=-4\Rightarrow y_P^{{}^{\prime }}=-2$
Equation of tangent at $P(1,0)$:
$y-0=-2(x-1)$
$\Rightarrow 2x+y=2$

From the given options, only point $(-2,6)$ lies on $2x+y=2$.