Considering only the principal values of inve | vedclass

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

Question

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

Vedclass · Accepted Answer

is a singleton set.

Vedclass · Answer

(B) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Using 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,$x = \frac{1}{6}$ is the only solution. Therefore,the set $A = \{\frac{1}{6}\}$ is a singleton set.

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

Similar Questions

Which one of the following symbols and its representation,used in human pedigree analysis is correct?

Which of the following statements is correct in the context of observing $DNA$ separated by agarose gel electrophoresis?

If $2 x^2-3 x y+y^2+x+2 y-8=0$, then $\frac{d y}{d x}$ is equal to

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

In an unbiased $p-n$ junction,holes diffuse from the $p-$ region to $n-$ region because

Vedclass Test Series

Exam Paper Generator

Online Exam Module