The set of values of x such that tan ^{-1}lef | vedclass

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

Question

(D) We use the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A-B}{1+AB} ight)$.Given equation: $	an ^{-1}\left(\frac{x}{x-2}ight

Vedclass · Accepted Answer

$\left\{\frac{1}{3}, 4\right\}$

Vedclass · Answer

(D) We use the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A-B}{1+AB} ight)$.Given equation: $	an ^{-1}\left(\frac{x}{x-2}ight)-	an ^{-1}\left(\frac{x}{2 x-1}ight)=	an ^{-1}\left(\frac{2}{3}ight)$.Applying the formula,we get:$	an ^{-1} \left( \frac{\frac{x}{x-2} - \frac{x}{2x-1}}{1 + \frac{x}{x-2} \cdot \frac{x}{2x-1}} ight) = 	an ^{-1} \left( \frac{2}{3} ight)$.Simplifying the expression inside the $	an ^{-1}$:$\frac{x(2x-1) - x(x-2)}{(x-2)(2x-1) + x^2} = \frac{2}{3}$.$\frac{2x^2 - x - x^2 + 2x}{2x^2 - x - 4x + 2 + x^2} = \frac{2}{3}$.$\frac{x^2 + x}{3x^2 - 5x + 2} = \frac{2}{3}$.Cross-multiplying:$3(x^2 + x) = 2(3x^2 - 5x + 2)$.$3x^2 + 3x = 6x^2 - 10x + 4$.$3x^2 - 13x + 4 = 0$.Factoring the quadratic equation:$(3x - 1)(x - 4) = 0$.Thus,$x = \frac{1}{3}$ or $x = 4$.Both values satisfy the domain of the original expression. Therefore,the set of values is $\left\{\frac{1}{3}, 4ight\}$.

The set of values of $x$ such that $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ is

Similar Questions

Given $0 \leq x \leq \frac{1}{2}$,then the value of $\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$ is:

$\sin^{-1} \frac{1}{\sqrt{5}} + \cot^{-1} 3$ is equal to

$\sin \left[3 \sin ^{-1}(0.4)\right] = \ldots$

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$,then $\tan S$ is equal to

$\cos \left[ {{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module