For the least possible value of n in Z, the s | vedclass

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

Question

(A) Let $a = \cos^{-1} x$ and $b^2 = (\sin^{-1} y)^2$. The given equations are $a + b^2 = \frac{n \pi^2}{4}$ and $a \cdot b^2 = \frac{\pi^4}{16}$.Thes

Vedclass · Accepted Answer

$(\cos(\frac{\pi^2}{4}), \pm 1)$

Vedclass · Answer

(A) Let $a = \cos^{-1} x$ and $b^2 = (\sin^{-1} y)^2$. The given equations are $a + b^2 = \frac{n \pi^2}{4}$ and $a \cdot b^2 = \frac{\pi^4}{16}$.These are roots of the quadratic equation $t^2 - (\frac{n \pi^2}{4})t + \frac{\pi^4}{16} = 0$.For real roots, the discriminant $D \ge 0$, so $(\frac{n \pi^2}{4})^2 - 4(\frac{\pi^4}{16}) \ge 0$.$\frac{n^2 \pi^4}{16} - \frac{\pi^4}{4} \ge 0 \Rightarrow n^2 \ge 4 \Rightarrow n \ge 2$ (since $n \in Z$ and $n > 0$ for real $x, y$).For the least value $n = 2$, the quadratic becomes $t^2 - \frac{2 \pi^2}{4}t + \frac{\pi^4}{16} = 0$, which is $(t - \frac{\pi^2}{4})^2 = 0$.Thus, $a = \frac{\pi^2}{4}$ and $b^2 = \frac{\pi^2}{4}$.Since $a = \cos^{-1} x = \frac{\pi^2}{4}$, we have $x = \cos(\frac{\pi^2}{4})$.Since $b^2 = (\sin^{-1} y)^2 = \frac{\pi^2}{4}$, we have $\sin^{-1} y = \pm \frac{\pi}{2}$, so $y = \sin(\pm \frac{\pi}{2}) = \pm 1$.The solution is $(\cos(\frac{\pi^2}{4}), \pm 1)$.

For the least possible value of $n \in Z$, the solution $(x, y)$ of the equations $\cos ^{-1} x + (\sin ^{-1} y)^2 = \frac{n \pi^2}{4}$ and $(\cos ^{-1} x)(\sin ^{-1} y)^2 = \frac{\pi^4}{16}$ is

Similar Questions

If $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$, then $\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}$ for $0 < a < \frac{1}{2}$ is

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

Find $\frac{dx}{dy}$ if $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ for $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

For the equation $\cos^{-1} x + \cos^{-1} 2x + \pi = 0$,the number of real solutions is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module