Question

Let f(x) = ax² + bx + c, where a∈R⁺ and b² – 4ac < 0. Area bounded by y = f(x), x-axis and the lines x = 0, x = 1 is equal to

• A. (3f(1) + f(-1) + 2f(0))
• B. (5f(1) + f(-1) + 8f(0))
• C. (3f(1) – f(-1) + 2f(0))
• D. $rac { 1 } { 12 }$ (5f(1) – f(-1) + 8f(0)

Answer 1

Answer: (D)

rac { 1 } { 12 } (5f(1) – f(-1) + 8f(0)

Answer 2

Clearly f(x) > 0 V x ∈ R. Thus, required area,

$\Delta = \int _ { 0 } ^ { 1 } \left( a x ^ { 2 } + b x + c \right) d x = \frac { a } { 3 } + \frac { b } { 2 } + c$

= $\frac { 1 } { 6 }$ (2a + 3b + 6c)

Now, f(0) = c, f(1) = a + b + c, f(-1) =a - b + c

⇒ a = $\frac { 1 } { 2 }$(f(1) + f(−1) − 2f(0)), b = $\frac { 1 } { 2 }$ (f(1) − f(-1)

Thus ∆ = $\frac { 1 } { 12 }$ (5f(1) − f(-1) + 8f(0))