JEE Main 2022 — Mathematics Calculus

Given,

$(\alpha ,\alpha )$(α,α) lies on $C:({x}^{2}+{y}^{2}-3)+{({x}^{2}-{y}^{2}-1)}^{5}=0$C:(x2+y2−3)+(x2−y2−1)5=0

So on putting $(\alpha ,\alpha )$(α,α) we get,

$$2{\alpha }^{2}-3-{1}^{5}=0$$2α2−3−15=0

$$\Rightarrow \alpha =\sqrt{2}$$⇒α=2​

Now, differentiating the curve $C$C we get,

$$2x+2y\cdot {y}^{'}+5{({x}^{2}-{y}^{2}-1)}^{4}(2x-2y{y}^{'})=0\cdots (1)$$2x+2y⋅y′+5(x2−y2−1)4(2x−2yy′)=0⋯(1)

At $(\sqrt{2},\sqrt{2})$(2​,2​)

$$\sqrt{2}+\sqrt{2}{y}^{'}+5{(-1)}^{4}(\sqrt{2}-\sqrt{2}{y}^{'})=0$$2​+2​y′+5(−1)4(2​−2​y′)=0

$$\Rightarrow {y}^{'}=\frac{3}{2}\cdots (2)$$⇒y′=23​⋯(2)

Again, Diff. $(1)$(1) w.r.t. $x$x we get,

$$1+{({y}^{'})}^{2}+yy{y}^{''}+20{({x}^{2}-{y}^{2}-1)}^{3}{(x-y{y}^{'})}^{2}\cdot 2$$1+(y′)2+yyy′′+20(x2−y2−1)3(x−yy′)2⋅2

$$+5{({x}^{2}-{y}^{2}-1)}^{4}(1-{({y}^{'})}^{2}-yy{y}^{''})=0$$+5(x2−y2−1)4(1−(y′)2−yyy′′)=0

At $(\sqrt{2},\sqrt{2})$(2​,2​) and ${y}^{'}=\frac{3}{2}$y′=23​

We have,

$$(1+\frac{9}{4})+\sqrt{2}{y}^{''}-40{(\sqrt{2}-\sqrt{2}\cdot \frac{3}{2})}^{2}$$(1+49​)+2​y′′−40(2​−2​⋅23​)2

$$+5(1)(1-\frac{9}{4}-\sqrt{2}{y}^{''})=0$$+5(1)(1−49​−2​y′′)=0

$$\Rightarrow 4\sqrt{2}{y}^{''}=-23$$⇒42​y′′=−23

$${y}^{''}=\frac{-23}{4\sqrt{2}}$$y′′=42​−23​

So, value of $3{y}^{'}-{y}^{3}{y}^{''}=\frac{9}{2}+\frac{23}{2}=16$3y′−y3y′′=29​+223​=16