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 __________
We are given the equation defining the function $f(x)$: $f(x) = \int_{0}^{x} \tan(t - x)dt - \int_{0}^{x} f(t)\tan t dt$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. We need to find the value of $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$.
Simplifying the Integral Equation
First, simplify the integral $\int_{0}^{x} \tan(t - x)dt$. Let $u = t - x$. Then $du = dt$. The limits change from $t=0 \to u=-x$ and $t=x \to u=0$. The integral becomes $\int_{-x}^{0} \tan(u)du$. The antiderivative of $\tan(u)$ is $-\ln|\cos u|$. So, $\int_{-x}^{0} \tan(u)du = [-\ln|\cos u|]_{-x}^{0} = (-\ln|\cos 0|) - (-\ln|\cos(-x)|) = -\ln(1) + \ln|\cos x| = \ln(\cos x)$ (since $\cos x > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$). Thus, the equation simplifies to: $f(x) = \ln(\cos x) - \int_{0}^{x} f(t)\tan t dt$.
Finding the Differential Equation
Differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$f'(x) = \frac{d}{dx}(\ln(\cos x)) - \frac{d}{dx}\left(\int_{0}^{x} f(t)\tan t dt\right)$
$f'(x) = \frac{-\sin x}{\cos x} - f(x)\tan x$
$f'(x) = -\tan x - f(x)\tan x = -\tan x (1 + f(x)) \quad (*)$
Differentiate again to find $f''(x)$:
$f''(x) = \frac{d}{dx}(-\tan x (1 + f(x)))$
Using the product rule:
$f''(x) = (-\sec^2 x)(1 + f(x)) - \tan x (f'(x))$
Substitute $f'(x)$ from equation $(*)$:
$f''(x) = -\sec^2 x (1 + f(x)) - \tan x (-\tan x (1 + f(x)))$
$f''(x) = -\sec^2 x (1 + f(x)) + \tan^2 x (1 + f(x))$
$f''(x) = (\tan^2 x - \sec^2 x)(1 + f(x))$
Using the identity $\sec^2 x - \tan^2 x = 1$, we have $\tan^2 x - \sec^2 x = -1$.
$f''(x) = (-1)(1 + f(x)) = -1 - f(x)$
Rearranging gives the differential equation: $f''(x) + f(x) = -1$
Solving the Differential Equation
The general solution to $f''(x) + f(x) = -1$ is $f(x) = A\cos x + B\sin x - 1$. We need to find the constants $A$ and $B$. First, find $f(0)$. From the original integral equation, setting $x=0$: $f(0) = \int_{0}^{0} \tan(t)dt - \int_{0}^{0} f(t)\tan t dt = 0 - 0 = 0$. Substitute $f(0)=0$ into the general solution: $0 = A\cos 0 + B\sin 0 - 1 = A(1) + B(0) - 1 = A - 1$. So, $A = 1$. The solution becomes $f(x) = \cos x + B\sin x - 1$. Now, use the relationship $f'(x) = -\tan x (1 + f(x))$. The derivative is $f'(x) = -\sin x + B\cos x$. Substitute $f(x)$ and $f'(x)$ into the relation: $-\sin x + B\cos x = -\tan x (1 + (\cos x + B\sin x - 1))$ $-\sin x + B\cos x = -\tan x (\cos x + B\sin x)$ $-\sin x + B\cos x = -\frac{\sin x}{\cos x}(\cos x + B\sin x)$ Multiply by $\cos x$: $(-\sin x + B\cos x)\cos x = -\sin x (\cos x + B\sin x)$ $-\sin x \cos x + B\cos^2 x = -\sin x \cos x - B\sin^2 x$ $B\cos^2 x = -B\sin^2 x$ $B(\cos^2 x + \sin^2 x) = 0$ $B(1) = 0$, which implies $B = 0$. Therefore, the function is $f(x) = \cos x - 1$.
Calculating the Final Value
We need to evaluate $f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$. Using $f(x) = \cos x - 1$: $f'(x) = -\sin x$ $f''(x) = -\cos x$ Evaluate the terms:
- $f\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) - 1 = \frac{\sqrt{3}}{2} - 1$
- $f''\left(\frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
- $f'\left(-\frac{\pi}{6}\right) = -\sin\left(-\frac{\pi}{6}\right) = - \left(-\sin\left(\frac{\pi}{6}\right)\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
Substitute these values into the expression:
$f''\left(\frac{\pi}{6}\right) + 12f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right) = \left(-\frac{\sqrt{3}}{2}\right) + 12\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2} - 1\right)$
$= -\frac{\sqrt{3}}{2} + 6 + \frac{\sqrt{3}}{2} - 1$
$= 6 - 1 = 5$
The value of the expression is 5.
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 $f : \mathbf{R} \rightarrow \mathbf{R}$ be a function such that $f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x$, $x \in \mathbf{R}$. Let the maximum value of $f$ on $\mathbf{R}$ be $\alpha$. If the area of the region bounded by the curves $g(x) = x^{2}$ and $h(x) = \beta x^{3}, \beta > 0$, is $\alpha^{2}$, then $30\beta^{3}$ 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 __________.
