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Let f:[1/2,1]vecR (the set of all real n...

Let `f:[1/2,1]vecR` (the set of all real numbers) be a positive, non-constant, and differentiable function such that `f^(prime)(x)<2f9x)a n df(1/2)=1` . Then the value of `int_(1/2)^1f(x)dx` lies in the interval `(2e-1,2e)` (b) `(3-1,2e-1)` `((e-1)/2,e-1)` (d) `(0,(e-1)/2)`

A

(a) `(2e-1,2e)`

B

(b) (e-1,2e-1)

C

(c) `((e-1)/2,e-1)`

D

(d) `(0,(e-1)/2)`

Text Solution

Verified by Experts

PLAN Whenever we have linear differential equation containing
inequality,we should always check for increasing or
decreasing,
`i.e. for dy/dx+ Py lt0 rArr dydx +Py gt0`
Multiply by integrating factor, `i.e.e^(intPdx)`nand convert into
total differential equation
Here, `f'(x)lt2f(x),` multiplying by`e^(-int2dx)`
`f'(x) cdote^(-2x) -2e^(-2x)f(x) lt0 rArr d/dx(f(x) cdote^(-2x))lt0`
`therefore phi (x)=f(x)e^(-2x)` is decreasing for `x in[1/2,1]` Thus, when `xgt1/2`
`phi (x) lt phi(1/2) rArr e^(-2x)f(x) lt e^(-1) cdot f(1/2)`
`rArr f(x)lte^(2x-1)cdot1, given f(1/2)=1`
`rArr 0ltint_(1//2)^(1) f(x) dx lt int _(1//2)^(1) e^(2x 1) dx`
`rArr 0ltint_(1//2)^(1) f(x)dx lt((e^(2x -1))/2)_(1//2)^(1) `
`rArr 0ltint_(1//2)^(1) f(x)dx lt(e -1)/2`
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