The function f(x)=∫1x[2(t−1)(t−2)3+3(t−1)2(t−2)2] dt has:

f(x)=∫1x{2(t−1)(t−2)3+3(t−1)2(t−2)2} dt ⇒f1(x)=2(x−1)(x−2)3+3(x−1)2(x−2)2 For maximum (or) minimum f1(x)=0 ⇒2(x−1)(x−2)3+3(x−1)2(x−2)2=0 ⇒(x−1)(x−2)2{2(x−2)+3(x−1)}=0 ⇒(x−1)(x−2)2{5x−7}=0 x=1, 2, 7/5 now f1(x)=(x−1)(x−2)2(5x−7) ⇒f1(x)=(x−2)2(5x−7)+2(x−1)(x−2)(5x−3)+(x−1)(x−2)25........(2) Putting x=1,x=2,x=7/5 is(2) we get f11(1) 0

The function f(x)=∫1x[2(t−1)(t−2)3+3(t−1)2(t−2)2] dt has:

f(x)=∫1x{2(t−1)(t−2)3+3(t−1)2(t−2)2} dt ⇒f1(x)=2(x−1)(x−2)3+3(x−1)2(x−2)2 For maximum (or) minimum f1(x)=0 ⇒2(x−1)(x−2)3+3(x−1)2(x−2)2=0 ⇒(x−1)(x−2)2{2(x−2)+3(x−1)}=0 ⇒(x−1)(x−2)2{5x−7}=0 x=1, 2, 7/5 now f1(x)=(x−1)(x−2)2(5x−7) ⇒f1(x)=(x−2)2(5x−7)+2(x−1)(x−2)(5x−3)+(x−1)(x−2)25........(2) Putting x=1,x=2,x=7/5 is(2) we get f11(1) 0

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The function $f (x) = \int_{1}^{x} [2 (t - 1) {(t - 2)}^{3} + 3 {(t - 1)}^{2} {(t - 2)}^{2}] d t$ has:

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minimum at $x = \frac{7}{5}$

neither maximum nor minimum at x = 2

all the above

maximum at x = 1

answer is D.

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Detailed Solution

$f (x) = \int_{1}^{x} {2 (t - 1) {(t - 2)}^{3} + 3 {(t - 1)}^{2} {(t - 2)}^{2}} d t$

$\Rightarrow f^{1} (x) = 2 (x - 1) {(x - 2)}^{3} + 3 {(x - 1)}^{2} {(x - 2)}^{2}$

For maximum (or) minimum $f^{1} (x) = 0$

$\Rightarrow 2 (x - 1) {(x - 2)}^{3} + 3 {(x - 1)}^{2} {(x - 2)}^{2} = 0$

$\Rightarrow (x - 1) {(x - 2)}^{2} {2 (x - 2) + 3 (x - 1)} = 0$

$\Rightarrow (x - 1) {(x - 2)}^{2} {5 x - 7} = 0$

$x = 1, 2, 7 / 5$

now $f^{1} (x) = (x - 1) {(x - 2)}^{2} (5 x - 7)$

$\Rightarrow f^{1} (x) = {(x - 2)}^{2} (5 x - 7) + 2 (x - 1) (x - 2) (5 x - 3) + (x - 1) {(x - 2)}^{2} 5........ (2)$ Putting $x = 1, x = 2, x = 7 / 5 i s (2) w e g e t$