MHT CET · Maths · Indefinite Integration
If \(f^{\prime}(x)=k(\cos x+\sin x)\) and \(f(0)=9, f\left(\frac{\pi}{2}\right)=15\), 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 \(3(\cos x+\sin x)-12\)
Answer & Solution
Correct Answer
(A) \(3(\sin x-\cos x)+12\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=k(\cos x+\sin x)\)
On integrating both sides, we get
\(f(x)=k(\sin x-\cos x)+C\)
\(f(0)=k(0-1)+C\)
\(f(0)=-k+C \Rightarrow-k+C=9\) ...(1)
Also \(f\left(\frac{\pi}{2}\right)=k\left(\sin \frac{\pi}{2}-\cos \frac{\pi}{2}\right)+C\)
\(15=k+C\) ...(2)
Adding (1) \(\&(2)\) we get
\(2 C=24 \Rightarrow C=12 \Rightarrow k=3\)
\(\therefore f(x)=3(\sin x-\cos x)+12\)
On integrating both sides, we get
\(f(x)=k(\sin x-\cos x)+C\)
\(f(0)=k(0-1)+C\)
\(f(0)=-k+C \Rightarrow-k+C=9\) ...(1)
Also \(f\left(\frac{\pi}{2}\right)=k\left(\sin \frac{\pi}{2}-\cos \frac{\pi}{2}\right)+C\)
\(15=k+C\) ...(2)
Adding (1) \(\&(2)\) we get
\(2 C=24 \Rightarrow C=12 \Rightarrow k=3\)
\(\therefore f(x)=3(\sin x-\cos x)+12\)
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