Question 6. Evaluate \lim_{x \to \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos\left(\frac{1}{t^{3}}\right)\,dt\right)
(1) \frac{3\pi}{8}
(2) \frac{3\pi^{2}}{4}
(3) \frac{3\pi^{2}}{8}
(4) \frac{3\pi}{4}
Answer (3)
Explanation:
As x \to \frac{\pi}{2}, the numerator and denominator both tend to zero.
So, apply L’Hôpital’s rule.
Differentiate numerator and denominator with respect to x,
= \lim_{x \to \frac{\pi}{2}^{-}} \frac{0-\cos(x)\cdot 3x^{2}}{2\left(x-\frac{\pi}{2}\right)}Rewrite using trigonometric identity,
= \lim_{x \to \frac{\pi}{2}^{-}} \frac{\sin\left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3\pi^{2}}{4}Now, \lim_{x \to 0} \frac{\sin x}{x} = 1
Hence, the required limit is \frac{3\pi^{2}}{8}
In a triangle \triangle ABC, suppose the equation of the bisector of angle B is y = x
The equation of side AC is 2x – y = 2