Solving the Functional Equation
We are given the functional equation:
$ f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x \quad \quad (1) $
Substitute $x$ with $\frac{\pi}{2} - x$:
$ f\left(\frac{\pi}{2} - x\right) + 3f\left(\frac{\pi}{2} - \left(\frac{\pi}{2} - x\right)\right) = \sin\left(\frac{\pi}{2} - x\right) $
$ f\left(\frac{\pi}{2} - x\right) + 3f(x) = \cos x \quad \quad (2) $
From equation (2), we get $f\left(\frac{\pi}{2} - x\right) = \cos x - 3f(x)$. Substitute this into equation (1):
$ f(x) + 3(\cos x - 3f(x)) = \sin x $
$ f(x) + 3\cos x - 9f(x) = \sin x $
$ -8f(x) = \sin x - 3\cos x $
$ f(x) = \frac{3\cos x - \sin x}{8} $
Finding the Maximum Value ($\alpha$)
The function is $f(x) = \frac{3\cos x - \sin x}{8}$.
To find the maximum value, we consider the numerator $3\cos x - \sin x$. This is in the form $a\cos x + b\sin x$, where $a=3$ and $b=-1$. The maximum value of this form is $\sqrt{a^2 + b^2}$.
Maximum value of $3\cos x - \sin x = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
Therefore, the maximum value of $f(x)$ is:
$ \alpha = \frac{\sqrt{10}}{8} $
Calculating the Area Between Curves
The curves are $g(x) = x^2$ and $h(x) = \beta x^3$, with $\beta > 0$. First, find the intersection points:
$ x^2 = \beta x^3 $
$ x^2 - \beta x^3 = 0 $
$ x^2(1 - \beta x) = 0 $
The intersection points are $x=0$ and $x = \frac{1}{\beta}$.
Between $x=0$ and $x=\frac{1}{\beta}$, $x^2$ is the upper curve (since $x^2 > \beta x^3$ for $0 < x < \frac{1}{\beta}$ because $1 > \beta x$).
The area is given by the integral:
$ \text{Area} = \int_{0}^{1/\beta} (x^2 - \beta x^3) dx $
Evaluate the integral:
$ \text{Area} = \left[ \frac{x^3}{3} - \frac{\beta x^4}{4} \right]_{0}^{1/\beta} $
$ \text{Area} = \left( \frac{(1/\beta)^3}{3} - \frac{\beta (1/\beta)^4}{4} \right) - (0 - 0) $
$ \text{Area} = \frac{1}{3\beta^3} - \frac{1}{4\beta^3} = \frac{4 - 3}{12\beta^3} = \frac{1}{12\beta^3} $
Relating Area to $\alpha$ and Finding $\beta$
We are given that the Area = $\alpha^2$. First, calculate $\alpha^2$:
$ \alpha^2 = \left(\frac{\sqrt{10}}{8}\right)^2 = \frac{10}{64} = \frac{5}{32} $
Now, equate the calculated area to $\alpha^2$:
$ \frac{1}{12\beta^3} = \frac{5}{32} $
Solve for $\beta^3$:
$ 12\beta^3 = \frac{32}{5} $
$ \beta^3 = \frac{32}{5 \times 12} = \frac{32}{60} = \frac{8}{15} $
Final Calculation
We need to find the value of $30\beta^3$. Substitute the value of $\beta^3$ we found:
$ 30\beta^3 = 30 \times \frac{8}{15} $
$ 30\beta^3 = \frac{30}{15} \times 8 = 2 \times 8 = 16 $
Similar Questions
- If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________
- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
- Let $f(x) = \int \frac{dx}{x^{\left(\frac{2}{3}\right)} + 2x^{\left(\frac{1}{2}\right)}}$ be such that $f(0) = -26 + 24\log_{e}(2)$. If $f(1) = a + b\log_{e}(3)$, where $a, b \in \mathbf{Z}$, then $a + b$ is equal to :
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
- Let $f(x) = \int \frac{(2 - x^2) \cdot e^x}{(\sqrt{1 + x})(1 - x)^{3/2}} dx$. If $f(0) = 0$, then $f\left(\frac{1}{2}\right)$ is equal to :
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Then $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to __________- Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0, x \geq 1$. If $y(1) = 1$, then the greatest integer less than $y\left(\sqrt{5}\right)$ is _________.
Important Questions from Integral Calculus
- If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________
- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to
Let $f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} \,dx$, $x > 0$, $\lim_{x \rightarrow 0} f(x) = 0$ and $f(1) = \frac{1}{4}$.
If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
- Let $[\cdot]$ denote the greatest integer function and $f(x) = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right]$. Then $12 \sum_{j=1}^\infty f(j)$ is equal to __________.
