Match the statements in Column I with the values in Column II. Column I Column II (A) The function f ( x ) = ( x − 1 ) 2 + ( x − 2 ) 2 + ( x − 3 ) 2 has a minimum for x = (P) 0 (B) The least value of

Question

Match the statements in Column I with the values in Column II.

Column I	Column II
(A) The function $f (x) = (x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2}$ has a minimum for $x =$	(P) $0$
(B) The least value of the function $f (x) = 2. 3^{3 x} - 3^{2 x} .4 + 2. 3^{x}$ in $[- 1, 1]$ is	(Q) $2$
(C) Let $f (x) = \frac{4}{3} x^{3} - 4 x, 0 \leq x \leq 2$ , then the global minimum value of the function is	(R) $\frac{8}{27}$
(D) Let $f (x) = 6 - 12 x + 9 x^{2} - 2 x^{3}, x \in [1, 4]$ , then absolute maximum value of $f (x)$ is	(S) $\frac{- 8}{3}$

TestHub · Accepted Answer

A) f ( x ) = 3 x 2 − 12 x + 14 , f ′ ( x ) = 0 ⇒ 6 x − 12 = 0 ⇒ x = 2 . B) Put 3 x = t ⇒ f ( t ) = 2 t 3 − 4 t 2 + 2 t , t ∈ [ 1/3 , 3 ] . f ′ ( t ) < 0 in (1/3, 1) and f ′ ( t ) > 0 in ( 1 , 3 ) . f ( t ) minimum = f ( 1 ) = 0 . C) f ′ ( x ) = 4 ( x 2 − 1 ) ⇒ f is decreasing for 0 < x < 1 and f is increasing for 1 < x ≤ 2. ∴ f minimum is f ( 1 ) = 4/3 − 4 = − 8/3 . D) f D ′ ​ ( x ) = − 12 + 18 x − 6 x 2 = − 6 ( x − 1 ) ( x − 2 ) is increasing in ( 1 , 2 ) and decreasing in ( 2 , 4 ) . ∴ absolute maximum = max { f ( 1 ) , f ( 2 )} = max { 1 , 2 } = 2 .

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