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Least value of ( eq in image)

Least value of ( eq in image)

Question Image
Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

f(x) = e^{sinx - 2sin^{2}x}
f'(x) = e^{sinx - 2sin^{2}x}.(cosx - 2.2sinx.cosx)
f'(x) = e^{sinx - 2sin^{2}x}.(cosx - 4sinx.cosx)
f'(x) = 0
f'(x) = e^{sinx - 2sin^{2}x}.(cosx - 4sinx.cosx) = 0
cosx - 4sinx.cosx = 0
cosx(1 - 4sinx) = 0
cosx =0, sinx = \frac{1}{4}
Best thing to find the minimum value here is to go through putting the value corresponding to f’(x) = 0 (Objective question)
For cosx = 0
f(x) = e^{sinx - 2sin^{2}x}
f(x) = e^{1 - 2.1^{2}}
f(x) = e^{-1}
For sinx = 1/4
f(x) = e^{sinx - 2sin^{2}x}
f(x) = e^{\frac{1}{4} - 2.\frac{1}{4}^{2}}
f(x) = e^{\frac{2}{8} - \frac{1}{8}}
f(x) = e^{\frac{1}{8}}
So the least value of f(x) is
\frac{1}{e}

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