Considering only the principal values of the | vedclass

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

Question

(A) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Applying the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{

Vedclass · Accepted Answer

is a singleton set.

Vedclass · Answer

(A) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Applying the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$,we get: $	an^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} ight) = \frac{\pi}{4}$ $\frac{5x}{1 - 6x^2} = 	an \left( \frac{\pi}{4} ight) = 1$ $5x = 1 - 6x^2$ $6x^2 + 5x - 1 = 0$ Factoring the quadratic equation: $(6x - 1)(x + 1) = 0$ This gives $x = \frac{1}{6}$ or $x = -1$. Since the condition is $x \geq 0$,we reject $x = -1$. Thus,the only solution is $x = \frac{1}{6}$. Therefore,the set contains only one element,which is a singleton set.

Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

Similar Questions

If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$,then $\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C=$

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

Considering the principal values of inverse trigonometric functions,the value of the expression $\tan\left(2 \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to:

Identify the pair$(s)$ of functions which are identical.

If $\alpha$ and $\beta$ are roots of the equation $x^2+5|x|-6=0$,then the value of $|\tan^{-1} \alpha - \tan^{-1} \beta|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module