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 ___________.
Solving the Differential Equation
We are given the second-order linear differential equation $f''(x) = f(x)$, which can be rewritten as $f''(x) - f(x) = 0$.
The characteristic equation is $m^2 - 1 = 0$. The roots are $m = 1$ and $m = -1$.
The general solution is of the form $f(x) = c_1 e^x + c_2 e^{-x}$, where $c_1$ and $c_2$ are constants.
Using Initial Conditions
We are given that $f(x)$ is an odd function, meaning $f(-x) = -f(x)$. This implies $f(0) = 0$.
Using $f(0)=0$: $c_1 e^0 + c_2 e^0 = 0 \implies c_1 + c_2 = 0 \implies c_2 = -c_1$.
Substituting $c_2 = -c_1$ into the general solution, we get $f(x) = c_1 e^x - c_1 e^{-x} = c_1 (e^x - e^{-x})$.
Now, we find the derivative: $f'(x) = c_1 (e^x - (-e^{-x})) = c_1 (e^x + e^{-x})$.
Using the condition $f'(0) = 3$: $f'(0) = c_1 (e^0 + e^0) = c_1 (1 + 1) = 2c_1$. So, $2c_1 = 3 \implies c_1 = \frac{3}{2}$.
The specific solution is $f(x) = \frac{3}{2} (e^x - e^{-x})$.
Verifying Function Properties
Let's check the given properties:
- Odd Function: $f(-x) = \frac{3}{2} (e^{-x} - e^{-(-x)}) = \frac{3}{2} (e^{-x} - e^x) = -\frac{3}{2} (e^x - e^{-x}) = -f(x)$. Verified.
- Non-decreasing derivative: $f'(x) = \frac{3}{2} (e^x + e^{-x})$. Since $e^x > 0$ and $e^{-x} > 0$ for all real $x$, $f'(x) > 0$, so $f'(x) \ge 0$. Verified.
- Second derivative: $f''(x) = \frac{d}{dx} \left(\frac{3}{2} (e^x + e^{-x})\right) = \frac{3}{2} (e^x - e^{-x}) = f(x)$. Verified.
- Initial conditions: $f(0) = \frac{3}{2}(e^0 - e^0) = 0$. $f'(0) = \frac{3}{2}(e^0 + e^0) = 3$. Verified.
Calculating the Final Value
We need to find $9f(\log_e 3)$. Let $x = \log_e 3$. This means $e^x = 3$.
Then, $e^{-x} = e^{-\log_e 3} = e^{\log_e (3^{-1})} = 3^{-1} = \frac{1}{3}$.
Now, substitute into the function $f(x)$: $f(\log_e 3) = \frac{3}{2} (e^{\log_e 3} - e^{-\log_e 3}) = \frac{3}{2} \left(3 - \frac{1}{3}\right)$.
Calculate the value inside the parenthesis: $3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}$.
So, $f(\log_e 3) = \frac{3}{2} \times \frac{8}{3} = \frac{8}{2} = 4$.
Finally, calculate $9f(\log_e 3)$: $9f(\log_e 3) = 9 \times 4 = 36$.
Similar Questions
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 ____________.
- 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 $y = y (x)$ 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
The integral $\int_{-1}^{2} (\pi^2 x \sin (\pi x))dx$ is equal to :
If $\int \frac{2x+5}{\sqrt{7-6x-x^2}} dx$ = $A\sqrt{7-6x-x^2} + Bsin^{-1} \left( \frac{x+3}{4} \right) + C$
(Where C is a constant of integration), then the ordered pair (A,B) is equal to :-
- 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 :
$4\int_{0}^{1} (\frac{1}{\sqrt{3+x^2} + \sqrt{1+x^2}}) dx - 3\log_e (\sqrt{3})$ is equal to :
- Let $(a, b)$ be the point of intersection of the curve $x^2 = 2y$ and the straight line $y -2x-6=0$ in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9x^2}{1 + 5^x} dx$ is equal to :
- Let $f: [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{x} f(t)dt = 5xf(x) - x^5 - 9$ for all $x\ge1$, then the value of $f(3)$ is :
- 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 __________.
Important Questions from Integral Calculus
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 ____________.
- 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 $y = y (x)$ 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
The integral $\int_{-1}^{2} (\pi^2 x \sin (\pi x))dx$ is equal to :
If $\int \frac{2x+5}{\sqrt{7-6x-x^2}} dx$ = $A\sqrt{7-6x-x^2} + Bsin^{-1} \left( \frac{x+3}{4} \right) + C$
(Where C is a constant of integration), then the ordered pair (A,B) is equal to :-
