The maximum value of the expression y -2 a - x x - x 2- b 2 isA- 2 a 2- b 2B- -2 a2-b2-C- a2-b2D- a2-2 b2

The correct option is C a^2+b^2 Let t = x + √(x^2+b^2) ⇒ (√(x^2+b^2) x)(√(x^2+b^2)+x)=b^2 ⇒ √(x^2+b^2) x =b^2/t ⇒ 2x=t b^2/t⇒ x=1/2(t^2 b^2/t ) Now let y = ...

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Byju's Answer

Standard XIII

Mathematics

Range of Quadratic Expression

The maximum v...

Question

The maximum value of the expression $y = 2 (a - x) (x + \sqrt{x^{2} + b^{2}})$ is

$2 a^{2} + b^{2}$

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$a^{2} + 2 b^{2}$

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$a^{2} + b^{2}$

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$| 2 a^{2} - b^{2} |$

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Solution

The correct option is C
$a^{2} + b^{2}$

Let $t = x + \sqrt{x^{2} + b^{2}}$
$\Rightarrow (\sqrt{x^{2} + b^{2}} - x) (\sqrt{x^{2} + b^{2}} + x) = b^{2} \Rightarrow \sqrt{x^{2} + b^{2}} - x = \frac{b^{2}}{t} \Rightarrow 2 x = t - \frac{b^{2}}{t} \Rightarrow x = \frac{1}{2} (\frac{t^{2} - b^{2}}{t})$
Now let y = 2(a - x)t
$= 2 (a - (\frac{t^{2} - b^{2}}{2 t})) t = 2 a t - t^{2} + b^{2} = b^{2} - t^{2} + 2 a t + a^{2} - a^{2} = a^{2} + b^{2} - (t - a)^{2} ∴ y \leq a^{2} + b^{2}$

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The maximum value of the expression y=2(a−x)(x+√x2+b2) is

The correct option is C a2+b2 Let t=x+√x2+b2 ⇒ (√x2+b2−x)(√x2+b2+x)=b2⇒ √x2+b2−x=b2t ⇒ 2x=t−b2t⇒x=12(t2−b2t) Now let y = 2(a - x)t =2(a−(t2−b22t))t=2at−t2+b2=b2−t2+2at+a2−a2=a2+b2−(t−a)2∴ y≤a2+b2

The maximum value of the expression $y = 2 (a - x) (x + \sqrt{x^{2} + b^{2}})$ is