Let y=yx be a function of x satisfying y√(1-x^2)=k-x√(1-y^2) where k is a constant and y1/2=-1/4.Then dy/dx at x=1/2 , is equal to

Correct answer: (B) -(√5)/2. x=1/2,y=(-1)/4⇒xy=(-1)/8 y.(-2x)/(2√(1-x^2))+y^(').√(1-x^2)=0-1.√(1-y^2)+(x.-2y)/(2√(1-y^2))y^(') -(xy√(1-x^2)+y^(')√(1-x^2)=-√(1-y^2)+xy.y^(')√(1-y^2)y^(')√(1-x^2)-xy√(1-y^2)=xy√(1-x^2)-√(1-y^2)y^(')√32+18.√154=-1)/(8.√34)-√154 y^(')(√45+1)/(2√15)=-(1+√45)/(4√3) y^(')=-√152√3=-√52

Q42 / 75 · JEE Main 2020 (07 Jan Shift 2) · Mathematics

Let $y = y (x)$ be a function of $x$ satisfying $y \sqrt{1 - x^{2}} = k - x \sqrt{1 - y^{2}}$ where $k$ is a constant and $y (\frac{1}{2}) = - \frac{1}{4}$ .Then $\frac{d y}{d x}$ at $x = \frac{1}{2}$ , is equal to

A
$- \frac{\sqrt{5}}{4}$
Correct answer
B
$- \frac{\sqrt{5}}{2}$
C
$\frac{2}{\sqrt{5}}$
D
$\frac{\sqrt{5}}{2}$

Solution

$x = \frac{1}{2}, y = \frac{- 1}{4} \Rightarrow x y = \frac{- 1}{8}$

$y . \frac{(- 2 x)}{2 \sqrt{1 - x^{2}}} + y^{'} . \sqrt{1 - x^{2}} = 0 - \{1 . \sqrt{1 - y^{2}} + \frac{x . (- 2 y)}{2 \sqrt{1 - y^{2}}} y^{'}\}$

$- \frac{x y}{\sqrt{1 - x^{2}}} + y^{'} \sqrt{1 - x^{2}} = - \sqrt{1 - y^{2}} + \frac{x y . y^{'}}{\sqrt{1 - y^{2}}}$

$y^{'} (\sqrt{1 - x^{2}} - \frac{x y}{\sqrt{1 - y^{2}}}) = \frac{x y}{\sqrt{1 - x^{2}}} - \sqrt{1 - y^{2}}$

$y^{'} (\frac{\sqrt{3}}{2} + \frac{1}{8 . \frac{\sqrt{15}}{4}}) = \frac{- 1}{8 . \sqrt{\frac{3}{4}}} - \frac{\sqrt{15}}{4}$

$y^{'} (\frac{\sqrt{45} + 1}{2 \sqrt{15}}) = - \frac{(1 + \sqrt{45})}{4 \sqrt{3}}$

$y^{'} = - \frac{\sqrt{15}}{2 \sqrt{3}} = - \frac{\sqrt{5}}{2}$

Hence, the correct answer is option: 'B'.

Let y=yx be a function of x satisfying y1-x2=k-x1-y2 where k is a constant and y12=-14.Then dydx at x=12 , is equal to

Solution

More Differentiation Questions

Let $y = y (x)$ be a function of $x$ satisfying $y \sqrt{1 - x^{2}} = k - x \sqrt{1 - y^{2}}$ where $k$ is a constant and $y (\frac{1}{2}) = - \frac{1}{4}$ .Then $\frac{d y}{d x}$ at $x = \frac{1}{2}$ , is equal to