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​Q. GIVEN f( x ) = e x .cos x , x- [0,2pi]. THE SLOPE OF TANGENT OF THE FUNCTION IS MINIMUM FOR- x = pi x = pi/4 x = 3pi/4 x = 3pi/2

​Q. GIVEN f( x ) = ex.cos x , x- [0,2pi]. THE SLOPE OF TANGENT OF THE FUNCTION IS MINIMUM FOR-
  1. x = pi
  2. x = pi/4
  3. x = 3pi/4
  4. x = 3pi/2

Grade:Select Grade

2 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^xcosx
f'(x) = e^xcosx+e^x(-sinx)
f'(x) = e^x(cosx-sinx)
We need to minimize this,
f''(x) = e^x(cosx-sinx)+e^x(-sinx-cosx)
f''(x) = -2e^xsinx
2e^xsinx = 0
x =0, \pi , 2\pi
So
x = \pi
Vicky
15 Points
2 years ago
Pi/4 Will be the answer
Because sinx and cosine x will give minimum value at x=pi/4
So it will be OK for others

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