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.

 

Answer 1

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.