Let a , b are positive real numbers such that a - b =10, then the smallest value of the constant K for which √( x 2+a x)-√( x 2+b x)0, is

Question

Let a , b are positive real numbers such that a - b =10, then the smallest value of the constant K for which √( x 2+a x)-√( x 2+b x)< K for all x >0, is (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

f ( x )=√ x 2+ ax -√ x 2+ bx ∴ f ( x )=(( a - b )/√1+( a / x )+√1+( b / x )) (on rationalising) ∴ f ( x ) max =( a - b /2) Hence, k=5

Q. Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $(x^{2} + a x) - (x^{2} + b x) < K$ for all $x > 0$ , is

2

3

4

5

$f (x) = x^{2} + a x - x^{2} + b x$
$∴ f (x) = \frac{( a - b )}{1 + ( a / x ) + 1 + ( b / x )}$ (on rationalising)
$∴ f (x)_{m a x} = \frac{a - b}{2}$
Hence, $k = 5$

Q. Let a,b are positive real numbers such that a−b=10, then the smallest value of the constant K for which (x2+ax)​−(x2+bx)​<K for all x>0, is

2

3

4

5

f(x)=x2+ax​−x2+bx​ ∴f(x)=1+(a/x)​+1+(b/x)​(a−b)​ (on rationalising) ∴f(x)max​=2a−b​ Hence, k=5

Q. Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $(x^{2} + a x) - (x^{2} + b x) < K$ for all $x > 0$ , is

$f (x) = x^{2} + a x - x^{2} + b x$
$∴ f (x) = \frac{( a - b )}{1 + ( a / x ) + 1 + ( b / x )}$ (on rationalising)
$∴ f (x)_{m a x} = \frac{a - b}{2}$
Hence, $k = 5$