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Question : 71 of 160
Marks:
+1,
-0
Solution:
x4−x2−2x+5=f(x) (say)
Then,
f′(x)=4x3−2x−2 Equate
f′(x)=0⇒4x3−2x−2=0 ⇒2x3−x−1=0 ⇒2x(x2−1)+(x−1)=0 ⇒2x(x−1)(x+1)+(x−1)=0 ⇒(x−1)[2x(x+1)+1]=0 ⇒(x−1)(2x2+2x+1)=0 ⇒x−1=0 and
2x2+2x+1=0 ⇒x=1 and
x=− (Imaginary)
⇒x=1 and
x= f"(x)=12x2−2 f"=1=10>0 ⇒f(x) is minimum at
x=1 Now,
f(1)=(1)4−(1)2−2+5=3 ∴ Absolute minimum value is 3
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