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
- 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 $[\cdot]$ 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 $f$ 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 $f$ 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 _________.
- If $I_1 = \int_0^1 e^{-x} cos^2x dx$, $I_2 = \int_0^1 e^{-x^2} cos^2x dx$ and $I_3 = \int_0^1 e^{-x^2} dx$; then :
- If $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m} \left( (\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x) \right) + C$ where $C$ is the constant of integration and $m, n \in N$, then $m+n$ is equal to
Let [.] denote the greatest integer function. If $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha - \log_e 2$, then $\alpha^3$ is equal to ____________.
Let $f: R\to R$ be a thrice differentiable odd function satisfying $f'(x)\ge0, f''(x)=f(x), f(0)=0, f'(0)=3$. Then $9f(\log_e 3)$ is equal to ___________.
- Let $y = y (x)$ be the solution curve of the differentialequation $x (x^2 + e^x) dy + (e^x (x-2) y-x^3) dx = 0, x > 0$, passing through the point $(1, 0)$.Then $y (2)$ is equal to
Important Questions from Integral Calculus
- 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 $[\cdot]$ 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 $f$ 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 $f$ 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 _________.
