Question

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
• C. 3
• D. 0

Answer 1

Answer: (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 $= 3$