Let $f: R\to R$ be a twice differentiable function such that
$(\sin x \cos y) (f(2x+2y)-f(2x-2y)) = (\cos x \sin y) (f(2x+2y)+f(2x-2y))$, for all $x, y \in R$.
If $f'(0) = \frac{1}{2}$, then the value of $24 f'' (\frac{5\pi}{3})$ is:
Analyzing the Functional Equation
We are given the functional equation:
$(\sin x \cos y) (f(2x+2y)-f(2x-2y)) = (\cos x \sin y) (f(2x+2y)+f(2x-2y))$
Let $u = 2x+2y$ and $v = 2x-2y$. This implies $x = \frac{u+v}{4}$ and $y = \frac{u-v}{4}$.
Using trigonometric product-to-sum identities:
- $\sin A \cos B = \frac{1}{2} (\sin(A+B) + \sin(A-B))$
- $\cos A \sin B = \frac{1}{2} (\sin(A+B) - \sin(A-B))$
Let $A = \frac{u+v}{4}$ and $B = \frac{u-v}{4}$. Then $A+B = u/2$ and $A-B = v/2$. The equation transforms to:
$\frac{1}{2}(\sin(u/2) + \sin(v/2)) (f(u)-f(v)) = \frac{1}{2}(\sin(u/2) - \sin(v/2)) (f(u)+f(v))$
Simplifying this yields:
$(\sin(u/2) + \sin(v/2)) (f(u)-f(v)) = (\sin(u/2) - \sin(v/2)) (f(u)+f(v))$
Expanding and rearranging gives:
$2\sin(v/2)f(u) = 2\sin(u/2)f(v)$
$\sin(v/2)f(u) = \sin(u/2)f(v)$
This implies $\frac{f(u)}{\sin(u/2)} = \frac{f(v)}{\sin(v/2)} = k$ for some constant $k$.
Therefore, the function must be of the form $f(t) = k \sin(t/2)$.
Determining the Constant $k$
We need the first derivative of $f(t)$: $f'(t) = \frac{d}{dt} \left( k \sin(t/2) \right) = k \cos(t/2) \cdot \frac{1}{2} = \frac{k}{2} \cos(t/2)$
We are given the condition $f'(0) = \frac{1}{2}$.
Substituting $t=0$ into the derivative:
$f'(0) = \frac{k}{2} \cos(0) = \frac{k}{2} \cdot 1 = \frac{k}{2}$
Equating this to the given value:
$\frac{k}{2} = \frac{1}{2} \implies k=1$
So, the function is $f(t) = \sin(t/2)$.
Calculating $f''(5\pi/3)$
First, find the second derivative, $f''(t)$.
$f'(t) = \frac{1}{2} \cos(t/2)$
$f''(t) = \frac{d}{dt} \left( \frac{1}{2} \cos(t/2) \right) = \frac{1}{2} \left( -\sin(t/2) \cdot \frac{1}{2} \right) = -\frac{1}{4} \sin(t/2)$
Now, evaluate $f''(t)$ at $t = \frac{5\pi}{3}$:
$f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4} \sin\left(\frac{5\pi/3}{2}\right) = -\frac{1}{4} \sin\left(\frac{5\pi}{6}\right)$
Since $\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$,
$f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4} \cdot \frac{1}{2} = -\frac{1}{8}$
Final Calculation
The question asks for the value of $24 f'' (\frac{5\pi}{3})$.
$24 f''\left(\frac{5\pi}{3}\right) = 24 \cdot \left(-\frac{1}{8}\right) = -\frac{24}{8} = -3$
Similar Questions
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
Let $f(x)= \begin{cases} (1+ax)^{1/x} & , x <0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$
be continuous at $x = 0$. Then $e^{abc}$ is equal to:
Let $f (x) = \int x^3\sqrt{3-x^2} \, dx$. If $5f (\sqrt{2}) = -4$, then $f (1)$ is equal to
Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
The area of the region bounded by the curve $y = \max \{|x|, x|x-2|\}$, the x-axis and the lines $x = -2$ and $x = 4$ is equal to
- If the function $f(x) = 2x^3-9ax^2+12a^2x+1$, where $a > 0$, attains its local maximum and local minimum values at p and q, respectively, such that $p^2 = q$, then $f(3)$ is equal to :
If the area of the region $\{(x, y):|4-x^2|\le y \le x^2, y\le4,x\ge0)$ is $\left(\frac{80\sqrt{2}}{\alpha}-\beta\right)$, $\alpha, \beta\in N$, then $\alpha+\beta$ is equal to _____________.
- Let $a > 0$. If the function $f(x) = 6x^3-45ax^2+108a^2x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1x_2=54$, then $a + x_1 + x_2$ is equal to:
- Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and $2g(x) - 3g\left(\frac{1}{x}\right) = x, x > 0$. If $\alpha = \int_{1}^{2}f(x) dx$, and $\beta = \int_{1}^{2}g(x) dx$, then the value of $9\alpha + \beta$ is:
Important Questions from Calculus
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
Let $f(x)= \begin{cases} (1+ax)^{1/x} & , x <0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$
be continuous at $x = 0$. Then $e^{abc}$ is equal to:
Let $f (x) = \int x^3\sqrt{3-x^2} \, dx$. If $5f (\sqrt{2}) = -4$, then $f (1)$ is equal to
Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
