If the area in sq-unitsof bounded region enclosed between curves y-x-bx2 and y-x2b is maximum- then the possible positive value of b is

For b gt;0, the bounded region shown in the figure : So, A=∫_0^b/(1+b^2)[x bx^2 x^2b]dx =[x^22 (b+1b)x^33]_0^b/(1+b^2) A=b^26(1+b^2)^2 ⇒dAdb=b(1 b^2)3(1+b^2 ...

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Standard XII

Mathematics

Sufficient Condition for an Extrema

If the area i...

Question

If the area (in $sq.units$ )of bounded region enclosed between curves $y = x - b x^{2}$ and $y = \frac{x^{2}}{b}$ is maximum, then the possible positive value of $b$ is

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Solution

For $b > 0,$ the bounded region shown in the figure :
So,
$A = b / (1 + b^{2}) \int 0 [x - b x^{2} - \frac{x^{2}}{b}] d x = {[\frac{x^{2}}{2} - (b + \frac{1}{b}) \frac{x^{3}}{3}]}_{0}^{b / (1 + b^{2})} A = \frac{b^{2}}{6 (1 + b^{2})^{2}}$
$\Rightarrow \frac{d A}{d b} = \frac{b (1 - b^{2})}{3 (1 + b^{2})^{3}} = 0,$ gives $b = - 1, 0, 1$
and at $b = 1,$ $A$ is maximum.

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If the area (in sq.units)of bounded region enclosed between curves y=x−bx2 and y=x2b is maximum, then the possible positive value of b is

For b>0, the bounded region shown in the figure : So, A=b/(1+b2)∫0[x−bx2−x2b]dx=[x22−(b+1b)x33]b/(1+b2)0A=b26(1+b2)2 ⇒dAdb=b(1−b2)3(1+b2)3=0, gives b=−1,0,1 and at b=1, A is maximum.

If the area (in $sq.units$ )of bounded region enclosed between curves $y = x - b x^{2}$ and $y = \frac{x^{2}}{b}$ is maximum, then the possible positive value of $b$ is