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Question : 12 of 160
Marks:
+1,
-0
Solution:
f(x)=x2−2(4k−1)x+g(k)>0,∀x∈R⇒D<0⇒4(4k−1)2<4g(k)⇒(4k−1)2<15k2−2k−7⇒16k2−8k+1<15k2−2k−7⇒k2−6k+8<0.⇒(k−2)(k−4)<0⇒2<k<4So,
k∈(2,4)Let
g(k)=15k2−2k−7g′(k)=30k−2
So,
g′(k)>0,∀k∈(2,4)∴g(k) attains no maxima and no minima in
(a,b).
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