Question

Value of positive real parameter 'a' such that the area of the region bounded by the parabolas y = x − ax², ay = x² attains it's maximum value, is equal to

• A. 1/2
• B. 2
• C. 1
• D. None of these

Answer 1

Answer: (C)

1

Answer 2

At point A, = x – ax²

⇒ x = 0, x = $\frac { a } { 1 + a ^ { 2 } }$

Area bounded, $\Delta = \int _ { 0 } ^ { a / 1 + a ^ { 2 } } \left( x - a x ^ { 2 } - \frac { x ^ { 2 } } { a } \right) d x$

$\Rightarrow \frac { d \Delta } { d a } = \int _ { 0 } ^ { a / 1 + a ^ { 2 } } \left( - x ^ { 2 } + \frac { x ^ { 2 } } { a ^ { 2 } } \right) d x + 0 \cdot \frac { d } { d a } \left( \frac { a } { 1 + a ^ { 2 } } \right)$

$= \left. \frac { \left( 1 - a ^ { 2 } \right) } { 3 a ^ { 2 } } x ^ { 3 } \right| _ { 0 } ^ { a / 1 + a ^ { 2 } } = \frac { a \left( 1 - a ^ { 2 } \right) } { 3 \left( 1 + a ^ { 2 } \right) ^ { 3 } }$

$= \frac { a ( 1 - a ) ( 1 + a ) } { \left( 1 + a ^ { 2 } \right) ^ { 3 } }$

⇒ $\frac { d \Delta } { d a }$ > 0 ∀ a ∈ (0, 1) and $\frac { \mathrm { d } \Delta } { \mathrm { da } }$ < 0 ∀ a ∈ (1, ∞)

Thus a = 1 is the point of maxima for ∆