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Question
If the function f(x)=2x3−9mx2+12m2x+1, where m>0 attains its maximum and minimum at p and q respectively such that p2=q, then find the value of m.
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Solution
We have
f(x)=2x3−9mx2+12m2x+1
⇒f'(x)=6x2−18mx+12m2
Also, f''(x)=12x−18m
Since, f(x) attains its maximum and minimum values at x = p and x = q, respectively, so f '(p) = 0 and
f '(q) = 0
f '(p) = 0
⇒6p2−18mp+12m2=0
⇒p2−3mp+2m2=0
⇒(p−2m)(p−m)=0
⇒p−2m =0 or p−m=0
⇒p=2m or p=m
Similarly,
f '(q) = 0
⇒q=2m or q=m
Now, consider the following cases:
Case I:
If p = 2m and q = 2m, then
p2=q
⇒4m2=2m
⇒2m2−m=0
⇒m(2m−1)=0
∴m=1/2 (m>0)
But, this gives p = 1 as the point of minima, which is not true.
Case II:
If p = 2m and q = m, then
p2=q
⇒4m2=m
⇒4m2−m=0
⇒m(4m−1)=0
∴m=1/4 (m>0)
But, this gives p = 12 as the point of minima, which is not true.
Case III:
If p = m and q = 2m, then
p2=q
⇒m2=2m
⇒m2−2m=0
⇒m(m−2)=0
∴m=2 (m>0)
For this case, p = 2 and q = 4 are the points of maxima and minima, respectively.
Case IV:
If p = m and q = m, then
p2=q
⇒m2=m
⇒m2−m=0
⇒m(m−1)=0
∴m=1 (m>0)
But, this gives q = 1 as the point of maxima, which is not true.
Hence, the value of m is 2.
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