MHT CET · Maths · Indefinite Integration
If \(f^{\prime}(x)=k(\cos x-\sin x), f^{\prime}(0)=3, f\left(\frac{\pi}{2}\right)=15\), 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 \(12(\sin x+\cos x)+3\)
Answer & Solution
Correct Answer
(A) \(3(\sin x+\cos x)+12\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=k(\cos x-\sin x)\)
\(f^{\prime}(0)=3 \quad f(\pi / 2)=15\)
\(k=3\) then \(f(x)=8\)
Integrate \(f^{\prime}(x)\)
\(f(x)=k \sin x+k \cos x+c\)
\(f(x)=3 \sin x+3 \cos x+c\)
\(f(\pi / 2)=15\)
\(c+3=15\)
\(c=12\)
\(f(x)=3 \sin x+3 \cos x+12\)
\(f^{\prime}(0)=3 \quad f(\pi / 2)=15\)
\(k=3\) then \(f(x)=8\)
Integrate \(f^{\prime}(x)\)
\(f(x)=k \sin x+k \cos x+c\)
\(f(x)=3 \sin x+3 \cos x+c\)
\(f(\pi / 2)=15\)
\(c+3=15\)
\(c=12\)
\(f(x)=3 \sin x+3 \cos x+12\)
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