The real value of x that satisfies the equati | vedclass

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

Question

(D) Given the equation: $	an ^{-1} x+	an ^{-1} 2 x=\frac{\pi}{4}$ Using the identity $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-A

Vedclass · Accepted Answer

$\frac{\sqrt{17}-3}{4}$

Vedclass · Answer

(D) Given the equation: $	an ^{-1} x+	an ^{-1} 2 x=\frac{\pi}{4}$ Using the identity $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$,we get: $	an ^{-1} \left( \frac{x+2x}{1-x(2x)} ight) = \frac{\pi}{4}$ $\frac{3x}{1-2x^2} = 	an \left( \frac{\pi}{4} ight)$ $\frac{3x}{1-2x^2} = 1$ $3x = 1 - 2x^2$ $2x^2 + 3x - 1 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$ Since $x$ must be positive for $	an ^{-1} x + 	an ^{-1} 2x$ to be $\frac{\pi}{4}$ (as $	an ^{-1} x$ is an increasing function),we reject the negative root. Thus,$x = \frac{\sqrt{17}-3}{4}$.

The real value of $x$ that satisfies the equation $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$ is

Similar Questions

Prove $\tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

If $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$,then $\frac{dy}{dx} = $

The differential coefficient of ${\tan ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} \right)$ with respect to ${\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} \right)$ is

If $0 \leqslant \cos ^{-1} x \leqslant \pi$ and $\frac{-\pi}{2} \leqslant \sin ^{-1} x \leqslant \frac{\pi}{2}$,then at $x=\frac{1}{5}$ the value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ is

If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module