Question

If \(f'(x) = k(\cos x + \sin x)\) and \(f(0) = 9,\ f\!\left(\dfrac{\pi}{2}\right) = 15\), then \(f(x)=\)

• A. \(3(\sin x - \cos x) + 12\)
• B. \(3(\sin x - \cos x) - 12\)
• C. \(3(\sin x + \cos x) + 12\)
• D. \(3(\cos x + \sin x) - 12\)

Hint:

When \(f'(x)\) is given with constants, first integrate to get \(f(x)\), then use the given values to find the constants.

Answer 1

The Correct Option is A

Step 1: Integrate \(f'(x)\) to get \(f(x)\).
Given \[ f'(x) = k(\cos x + \sin x) \] Integrating both sides w.r.t. \(x\): \[ f(x) = k\int(\cos x + \sin x)\,dx \] \[ f(x) = k(\sin x - \cos x) + C \]
Step 2: Use the condition \(f(0)=9\) to find \(C\).
Put \(x=0\): \[ f(0) = k(\sin 0 - \cos 0) + C = k(0-1) + C = -k + C \] Given \(f(0)=9\), so \[ -k + C = 9 \Rightarrow C = 9 + k \]
Step 3: Use the condition \(f\!\left(\dfrac{\pi}{2}\right)=15\) to find \(k\).
Put \(x=\dfrac{\pi}{2}\): \[ f\!\left(\frac{\pi}{2}\right) = k\left(\sin\frac{\pi}{2} - \cos\frac{\pi}{2}\right) + C \] \[ = k(1-0) + C = k + C \] Given \(f\!\left(\dfrac{\pi}{2}\right)=15\), so \[ k + C = 15 \] Now substitute \(C = 9 + k\): \[ k + (9 + k) = 15 \Rightarrow 2k + 9 = 15 \Rightarrow 2k = 6 \Rightarrow k = 3 \]
Step 4: Write the final function \(f(x)\).
\[ f(x) = k(\sin x - \cos x) + C \] \[ f(x) = 3(\sin x - \cos x) + (9+3) \] \[ f(x) = 3(\sin x - \cos x) + 12 \]