Let cos ^{-1}left(frac{y}{b}right)=log _eleft | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\cos ^{-1}\left(\frac{y}{b}\right)=n \log _e\left(\frac{x}{n}\right)$.Differentiating with respect to $x$:$-\frac{1}{\sqrt{1-\frac{y^2}{b^2

Vedclass · Accepted Answer

$A=x^2, B=x, C=n^2$

Vedclass · Answer

(B) Given $\cos ^{-1}\left(\frac{y}{b}ight)=n \log _e\left(\frac{x}{n}ight)$.Differentiating with respect to $x$:$-\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} y_1 = n \cdot \frac{n}{x} \cdot \frac{1}{n} = \frac{n}{x}$.$-\frac{y_1}{\sqrt{b^2-y^2}} = \frac{n}{x} \implies x y_1 = -n \sqrt{b^2-y^2}$.Squaring both sides: $x^2 y_1^2 = n^2 (b^2-y^2)$.Differentiating again with respect to $x$:$2 x y_1^2 + x^2 \cdot 2 y_1 y_2 = -n^2 \cdot 2 y y_1$.Dividing by $2 x y_1$ (assuming $x 
eq 0, y_1 
eq 0$):$y_1 + x y_2 = -\frac{n^2 y}{x}$.$x y_1 + x^2 y_2 + n^2 y = 0$.Comparing with $A y_2 + B y_1 + C y = 0$, we get $A=x^2, B=x, C=n^2$.

Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log _e\left(\frac{x}{n}\right)^n$. Then $A y_2+B y_1+C y=0$ is possible for:

Similar Questions

If $f(x) = (x^2 - 1)^7$, then $f^{(14)}(x)$ is equal to

If $y(\theta) = \frac{2 \cos \theta + \cos 2 \theta}{\cos 3 \theta + 4 \cos 2 \theta + 5 \cos \theta + 2}$,then at $\theta = \frac{\pi}{2}$,$y'' + y' + y$ is equal to:

If $y=A \cos n x+B \sin n x$,then $y_2+n^2 y$ is equal to

If $y = \sin x + e^x,$ then $\frac{d^2x}{dy^2} = $

If $y = \sin(m \sin^{-1} x)$, then $(1-x^2) y_2 - x y_1$ is equal to (Here, $y_n$ denotes $\frac{d^n y}{dx^n}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module