JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q26

← JEE Main 2022 · 27 Jul · Shift 2

MathematicsInteger+4 / -1Question 26 of 90

For the curve $C:\left(x^2+y^2-3\right)+\left(x^2-y^2-1\right)^5=0$ , the value of $3y'-y^3y''$ , at the point $(\alpha,\alpha)$ , $\alpha>0$ , on C, is equal to ______.

Your answer

Answer and worked solution

Official answer: 16

Point. $(\alpha,\alpha)$ on C gives $2\alpha^2-3-1=0$ , so $\alpha=\sqrt2$ .

First derivative. Differentiating $x+yy'+5u^4(x-yy')=0$ with $u=x^2-y^2-1=-1$ gives $y'=\frac32$ .

Second derivative. Differentiating again gives $yy''=-\frac{23}4$ .

Value. $3y'-y^3y''=\frac92+\frac{23}2=16$ .

Answer: 16

Topics:Chain rule and implicit differentiation Higher-order derivatives

← Previous question Next question →

JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 26