Download App

STD 12 - 2. inverse trigonometric function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. inverse trigonometric function4 Marks
MCQ
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1} x$ is equal to:
  • A
    $2$
  • B
    $1$
  • ✓
    $3$
  • D
    $0$
✓

Answer

Correct option: C.
$3$
c
$\sin ^{-1} \frac{3 x}{5}+\sin ^{-1} \frac{4 x}{5}=\sin ^{-1} x$

$\sin ^{-1}\left(\frac{3 x}{5} \sqrt{1-\frac{16 x^{2}}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^{2}}{25}}\right)=\sin ^{-1} x$

$\frac{3 x}{5} \sqrt{1-\frac{16 x^{2}}{25}}+\frac{4 x}{5} \sqrt{1-\frac{9 x^{2}}{25}}=x$

$x=0,3 \sqrt{25-16 x^{2}}+4 \sqrt{25-9 x^{2}}=25$

$4 \sqrt{25-9 x^{2}}=25-3 \sqrt{25-16 x^{2}}$ squaring we get

$16\left(25-9 x^{2}\right)=625+9\left(25-16 x^{2}\right)-150 \sqrt{25-16 x^{2}}$

$400=625+225-150 \sqrt{25-16 x^{2}}$

$\sqrt{25-16 x^{2}}=3 \Rightarrow 25-16 x^{2}=9$

$\Rightarrow x^{2}=1$

Put $x=0,1,-1$ in the original equation We see that all values satisfy the original equation.

Number of solution $=3$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The function $f(x) = {x^2}$ is increasing in the interval
→
The general solution of $sin\, x + sin \,5x = sin\, 2x + sin \,4x$ is :
→
If $A = {\rm{si}}{{\rm{n}}^8}\theta + {\rm{co}}{{\rm{s}}^{14}}\theta ,$then for all real values of ?
→
If $k = \sin \frac{\pi }{{18}}\,.\,\sin \frac{{5\pi }}{{18}}\,.\,\sin \frac{{7\pi }}{{18}},$ then the numerical value of $k$ is
→
If the angle between the lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{{5 - x}}{{ - 2}} = \frac{{7y - 14}}{p} = \frac{{z - 3}}{4}$  is ${\cos ^{ - 1}}\,\left( {\frac{2}{3}} \right),$  then $p$  is equal to
→
Let $A$ and $B$ be two events such that $P\,(A) = 0.3$ and $P\,(A \cup B) = 0.8$. If $A$ and $B$ are independent events, then $P(B) = $
→
Let $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\overrightarrow{ c }=\hat{ i }-\hat{ j }+\hat{ k }$ be three given vectors. Let $\vec{v}$ be a vector in the plane of $\vec{a}$ and $\overrightarrow{ b }$ whose projection on $\overrightarrow{ c }$ is $\frac{2}{\sqrt{3}}$. If $\overrightarrow{ v } . \hat{ j }=7$, then $\overrightarrow{ v } \cdot(\hat{ i }+\hat{ k })$ is equal to
→
The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is
→
The shaded region given in the figure represent the $\ldots \ldots \ldots .$ inequality.
→
If $I _{ n }=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^{ n } x dx ,$ then :
→