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Question: Let f(x) be a function defined by ƒ(x) = \(\int_{1}^{x}t\) (t<sup>2</sup> – 3t + 2)dt, 1 £ x £ 3. Th...

Let f(x) be a function defined by ƒ(x) = ∫1xt\int_{1}^{x}t (t2 – 3t + 2)dt, 1 £ x £ 3. Then the range of f(x) is –

A

[0, 2]

B

[−14,4]\left\lbrack - \frac{1}{4},4 \right\rbrack

C

[−14,2]\left\lbrack - \frac{1}{4},2 \right\rbrack

D

None of these

Answer

[−14,2]\left\lbrack - \frac{1}{4},2 \right\rbrack

Explanation

Solution

f¢(x) = x(x2 – 3x + 2) = x (x – 1) (x – 2). The sign scheme for f¢(x) is as shown in figure.

\ f¢(x) £ 0 in 1 £ x £ 2 and f¢(x) ³ 0 in 2 £ x £ 3

\ f¢(x) is decreasing in [1, 2] and increasing in [2, 3]

\ min. f(x) = f(2) = ∫12x(x2−3x+2)\int_{1}^{2}{x(x^{2} - 3x + 2)}dx

= [x44−x3+x2]12\left\lbrack \frac{x^{4}}{4} - x^{3} + x^{2} \right\rbrack_{1}^{2} = −14- \frac{1}{4}

max. f(x) = the greatest among [f(1), f(3)]

f(1) =

∫11x(x2−3x+2)dx=0\int_{1}^{1}{x(x^{2} - 3x + 2)dx = 0}

f(3) = ∫13x(x2−3x+2)dx=2\int_{1}^{3}{x(x^{2} - 3x + 2)dx = 2}

\ max f(x) = 2, so the range =[−14,2]\left\lbrack - \frac{1}{4},2 \right\rbrack