Question

The area bounded by the curve y = ƒ(x) = x⁴ – 2x³ + x² + 3, x-axis and ordinates corresponding to minimum of the function ƒ(x) is -

• A. 1
• B. 91/30
• C. 30/9
• D. 4

Answer 1

Answer: (B)

91/30

Answer 2

ƒ¢ (x) = 4x³ – 6x² + 2x = 2x = 2x (2x² – 3x + 1) = 2x (2x – 1) (x – 1).Since ƒ is a differentiable function, so extremum points of ƒ(x), we must have ƒ¢(x) = 0 so x = 0, 1/2, 1. Now ƒ¢¢(x) = 12x² – 12x + 2, ƒ¢¢(0) = 2 ƒ¢¢(1) = 2 and ƒ¢¢(1/2) = 3 – 6 + 2 = –1. Thus the function has minimum at x = 0 and x = 1. Therefore, the required area = $\int _ { 0 } ^ { 1 } \left( x ^ { 4 } \right.$– 2x³ + x² + 3) dx

= $\left. \left( \frac { x ^ { 5 } } { 5 } - \frac { x ^ { 4 } } { 2 } + \frac { x ^ { 3 } } { 3 } + 3 x \right) \right| _ { 0 } ^ { 1 }$ = $\frac { 1 } { 5 }$ – $\frac { 1 } { 2 }$ + $\frac { 1 } { 3 }$ + 3 = $\frac { 91 } { 30 }$.