Download App

Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to :
  • A
    $\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}$
  • ✓
    $\frac{\text{y}}{\text{x}}$
  • C
    $\frac{\text{x}}{\text{y}}$
  • D
    None of these.
✓

Answer

Correct option: B.
$\frac{\text{y}}{\text{x}}$
$\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$
$\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\sin(\log\text{a})=\text{k}$
$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$
$\text{x}^2-\text{y}^2=\text{kx}^2+\text{ky}^2$
$\text{x}^2-\text{kx}^2=\text{ky}^2+\text{y}^2$
$(1-\text{k})\text{x}^2=(\text{x}+1)\text{y}^2$
$\frac{1-\text{x}}{\text{k}+1}=\frac{\text{y}^2}{\text{x}^2}\ .....(\text{i})$
Consider,
$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$
Differentaiting with resepect to $x,$
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{kx}+2\text{ky}\frac{\text{dy}}{\text{dx}}$
$2\text{x}-2\text{kx}=2\text{ky}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}$
$2\text{x}(1-\text{k})=2\text{y}(\text{k}+1)\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(1-\text{k})}{\text{y}(\text{k}+1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\times\frac{\text{y}^2}{\text{x}^2}\ .....(\because$ from $(i))$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Continuity and Differentiability→Continuity and Differentiability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}=\text{ax}^4+\text{bx}^3+\text{cx}^2+\text{dx}+\text{e}$ be an identity in $x,$ where $a, b, c, d, e$ are independent of $x$. Then the value of $e$ is:
→
$\int_{\pi /4}^{\pi /2} {\cos \theta \,{\rm{cose}}{{\rm{c}}^{\rm{2}}}\theta \,d\theta = } $
→
$ \begin{bmatrix}1 & \text{x} & \text{x}^2 \\1 & \text{y} & \text{y}^2 \\1 & \text{z} & \text{z}^2\end{bmatrix}$
→
The area enclosed between the curves $y=x|x|$ and $\mathrm{y}=\mathrm{x}-|\mathrm{x}|$ is :
→
If $A = [1, 2, 3],$ then the set of elements of $A$ is :
→
A box contains $3$ orange balls, $3$ green balls and $2$ blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing $22$ green balls and one blue ball is
→
Find the cofactor of element $-3$ in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is:
→
Difference between the greatest and the least values of the function$f (x) = x(ln x - 2)$ on $[1, e^2]$ is
→
The equation of the curve whose slope is given by $\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}};\text{x}>0,\text{y}>0$ and which passes through the point (1, 1) is:
→
If $f(x) = max(sinx, sin^{-1}(cosx))$, then
→