NIMCET 2023 Question Paper with solutions

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Question : 41 of 120

Marks: +1, -0

If the maximum value of the function

f(x)=(x-1)^2(x+1)^3

is

\;{2^p 3^q}/{3125}

, then the value of

(p, q)

is:

$(7,3)$
$(4,4)$
$(5,5)$
$(3,7)$

Solution:

f(x)=(x-1)^2(x+1)^3

\;f^{'}(x)=2(x-1)(x+1)^3+3(x+1)^2(x-1)^2

\;=(x-1)(x+1)^2[2(x+1)+3(x-1)]=0

\;=(x-1)(x+1)^2[5 x-1]

\;x=1,-1,1 \/ 5

f (\;{1}/{5}) \;= (\;{1}/{5}-1) ^2 (\;{1}/{5}+1) ^3

\;= (-\;{4}/{5}) ^2 (\;{6}/{5}) ^3

\;=\;{16}/{5^2} × \;{6^3}/{5^3}=\;{2^7 × 3^3}/{5^5}

Hence,

p=7, q=3

.

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