Question

Assertion (A) : Let \(f: R \rightarrow R\) be a smooth function such that \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(1-\mathrm{x})\) for all \(\mathrm{x}\) and \(\mathrm{f}(0)=1 .\) Then \(\mathrm{f}(\mathrm{x})=\cos\) \(x+(\sec 1+\tan 1) \sin x\) Reason (R) : Differentiating the given equation gives $f^{\prime \prime}(x)=-f(x)\(. This has solution of the form \)f(x)=A \cos x+$ \(\mathrm{B} \sin \mathrm{x}\), when \(\mathrm{A} \& \mathrm{~B}\) are determined by the boundary conditions.

Short answer

Question: Show that the smooth function f(x) with f'(x) = f(1-x) for all x and f(0) = 1 is given by f(x) = cos(x) + (sec(1) + tan(1)) sin(x). Answer: Following the given steps, we find that the function f(x) is indeed given by f(x) = cos(x) + (sec(1) + tan(1)) sin(x).

Step-by-step solution

Finding the second derivative

Differentiate the given equation, f'(x) = f(1-x), with respect to x: \(f^{\prime\prime}(x)=-f^\prime(1-x)\) Using the original equation f'(x) = f(1-x) to replace f'(1-x), we get \(f^{\prime\prime}(x)=-f(x)\)

Solving the second order differential equation

We now solve the second-order linear ordinary differential equation: \(f^{\prime\prime}(x) + f(x) = 0\) The general solution for this ODE is given by: \(f(x) = A\cos(x) + B\sin(x)\) for constants A and B.

Using the boundary condition to find A

Use the boundary condition f(0) = 1 to find the constant A: \(f(0) = A\cos(0) + B\sin(0) = 1\) Since \(\cos(0) = 1\) and \(\sin(0) = 0\): \(A = 1\)

Using the derivative to find B

Compute the derivative of f(x): \(f^\prime(x) = -A\sin(x) + B\cos(x)\) Now, plug the given condition f'(x) = f(1-x) into the equation: \(f^\prime(x) = A\cos(1-x) + B\sin(1-x)\) Next, use the fact that \(f^\prime(1) = f(0)\): \(-A\sin(1) + B\cos(1) = A\cos(1-x) + B\sin(1-x)\) Substitute the known values \(A = 1\): \(-1\sin(1) + B\cos(1) = 1\) Since the condition f'(1) = f(0) must hold, we can solve for B: \(B\cos(1) = 1 + \sin(1)\) Thus, \(B = \frac{1 + \sin(1)}{\cos(1)} = \sec(1) + \tan(1)\)

Combining the results

Finally, we substitute A = 1 and B = (sec(1) + tan(1)): \(f(x) = \cos(x) + (\sec(1) + \tan(1))\sin(x)\) This verifies the assertion.