Question

se
For Xe(0,00) if
*(x - 2)dx
=f(x)
x(x’+e")
where
f(1) = log. (1+e) then
Local minimum of
4te²
f(x) = log.
4
f(x)
Lt
1​

Answer 1

Answer:

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Answer 2

Answer:

Given,

f(x)=⎧⎪⎪ex2−ex−1x−e,,,if0≤x≤1if1<x≤2if2<x≤3

and  g(x)=∫x0f(t)dt

⇒ g'(x)=f(x)

Put  g'(x)=0⇒x=1+loge2andx=e.

Also,  g''(x)=⎧⎩⎨⎪⎪ex,−ex−1,1,if0≤x≤1if1<x≤2if2<x≤3

At  x=1+loge2,

g''(1+loge2)=−eloge2<0,g(x) has a local maximum.

Also, at x=e,

g''(e)=1>0,g(x) has a local, minima.

∵f(x) is discontinuous at x=1, then we get local maxima at x=1 and local minima at x=2.

Hence (a) and (b) are correct answers.