Tip:A–D to answerE for explanationV for videoS to reveal answer
The minimum value of the function f(x)=∫02e∣x−t∣dt is:
- A.
e(e−1)
- B.
2(e−1)
(Correct Answer) - C.
2
- D.
2e−1
Explanation
Solving
-
Case for 0≤x≤2:
f(x)=∫0xex−tdt+∫x2et−xdt
f(x)=ex[−e−t]0x+e−x[et]x2
f(x)=ex(1−e−x)+e−x(e2−ex)=ex−1+e2−x−1
f(x)=ex+e2−x−2
-
Finding Minimum: Differentiate f′(x)=ex−e2−x.
-
Set f′(x)=0⟹ex=e2−x⟹x=2−x⟹x=1.
-
Minimum Value: f(1)=e1+e2−1−2=e+e−2=2e−2=2(e−1).
Correct Option: (2)
Explanation
Solving
-
Case for 0≤x≤2:
f(x)=∫0xex−tdt+∫x2et−xdt
f(x)=ex[−e−t]0x+e−x[et]x2
f(x)=ex(1−e−x)+e−x(e2−ex)=ex−1+e2−x−1
f(x)=ex+e2−x−2
-
Finding Minimum: Differentiate f′(x)=ex−e2−x.
-
Set f′(x)=0⟹ex=e2−x⟹x=2−x⟹x=1.
-
Minimum Value: f(1)=e1+e2−1−2=e+e−2=2e−2=2(e−1).
Correct Option: (2)
