Question

Let $f(x)$ and $g(x)$ be two differentiable functions defined from $\mathbb{R} \to \mathbb{R}$ such that $f(x)= \cos x - \cos^3 x$ and $g(x) = x^2 - \frac{\pi^2}{4}$. If area of the region bounded by $y = f(x)$ and $y = g(x)$ is $\left( \frac{2}{3} + \frac{\pi^3}{a} \right)$. Then:

If the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{36} = 1$ is $\frac{\sqrt{p}}{q}$, where $p$ and $q$ are relatively prime positive integers, then value of $(p+q)$ is equal to:

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (A)

3

Answer 2

The problem consists of two parts. The first part involves calculating the area between two curves to find the value of a parameter 'a'. The second part uses this parameter to define a hyperbola and asks for a value derived from its eccentricity.

Part 1: Area Calculation The functions are $f(x) = \cos x - \cos^3 x = \cos x (1 - \cos^2 x) = \cos x \sin^2 x$ and $g(x) = x^2 - \frac{\pi^2}{4}$. We find the intersection points by setting $f(x) = g(x)$. We note that $f(\pm \pi/2) = 0$ and $g(\pm \pi/2) = (\pm \pi/2)^2 - \frac{\pi^2}{4} = 0$. Thus, $x = -\pi/2$ and $x = \pi/2$ are intersection points. For $x \in (-\pi/2, \pi/2)$, $f(x) = \cos x \sin^2 x \ge 0$ and $g(x) = x^2 - \frac{\pi^2}{4} < 0$. Hence, $f(x) > g(x)$ in this interval. The area $A$ is given by: $A = \int_{-\pi/2}^{\pi/2} (f(x) - g(x)) dx = \int_{-\pi/2}^{\pi/2} (\cos x \sin^2 x - (x^2 - \frac{\pi^2}{4})) dx$ $A = \int_{-\pi/2}^{\pi/2} \cos x \sin^2 x dx - \int_{-\pi/2}^{\pi/2} x^2 dx + \int_{-\pi/2}^{\pi/2} \frac{\pi^2}{4} dx$

Evaluating the integrals:

1. $\int_{-\pi/2}^{\pi/2} \cos x \sin^2 x dx = \left[\frac{\sin^3 x}{3}\right]_{-\pi/2}^{\pi/2} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.
2. $\int_{-\pi/2}^{\pi/2} x^2 dx = \left[\frac{x^3}{3}\right]_{-\pi/2}^{\pi/2} = \frac{(\pi/2)^3}{3} - \frac{(-\pi/2)^3}{3} = \frac{\pi^3}{24} + \frac{\pi^3}{24} = \frac{\pi^3}{12}$.
3. $\int_{-\pi/2}^{\pi/2} \frac{\pi^2}{4} dx = \frac{\pi^2}{4} [x]_{-\pi/2}^{\pi/2} = \frac{\pi^2}{4} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{\pi^2}{4} (\pi) = \frac{\pi^3}{4}$.

So, $A = \frac{2}{3} - \frac{\pi^3}{12} + \frac{\pi^3}{4} = \frac{2}{3} + \pi^3 \left(\frac{1}{4} - \frac{1}{12}\right) = \frac{2}{3} + \pi^3 \left(\frac{3-1}{12}\right) = \frac{2}{3} + \frac{\pi^3}{6}$. Given that the area is $\left( \frac{2}{3} + \frac{\pi^3}{a} \right)$, we compare the two expressions to find $a=6$.

Part 2: Hyperbola Eccentricity The hyperbola equation is $\frac{x^2}{a^2} - \frac{y^2}{36} = 1$. Substituting $a=6$, we get $\frac{x^2}{6^2} - \frac{y^2}{36} = 1$, which is $\frac{x^2}{36} - \frac{y^2}{36} = 1$. This is a hyperbola of the form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, with $A^2 = 36$ and $B^2 = 36$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{B^2}{A^2}}$. $e = \sqrt{1 + \frac{36}{36}} = \sqrt{1+1} = \sqrt{2}$. We are given that the eccentricity is $\frac{\sqrt{p}}{q}$, where $p$ and $q$ are relatively prime positive integers. So, $\sqrt{2} = \frac{\sqrt{p}}{q}$. Squaring both sides gives $2 = \frac{p}{q^2}$, or $p = 2q^2$. For $p$ and $q$ to be relatively prime positive integers, the only solution is $q=1$, which yields $p = 2(1)^2 = 2$. Thus, $p=2$ and $q=1$. They are positive and $\text{gcd}(2,1)=1$. The problem asks for the value of $(p+q)$. $p+q = 2+1 = 3$.