JEE Advance - Mathematics (1979 - No. 1)
Prove that the minimum value of $${{\left( {a + x} \right)\left( {b + x} \right)} \over {\left( {c + x} \right)}},$$
$$a,b > c,x > - c$$ is $${\left( {\sqrt {a - c} + \sqrt {b - c} } \right)^2}$$.
$$a,b > c,x > - c$$ is $${\left( {\sqrt {a - c} + \sqrt {b - c} } \right)^2}$$.
The statement is always true.
The statement is false for all values of a, b, c, and x.
The minimum value is only achieved when a = b.
The minimum value is ${\left( {\sqrt {a - c} - \sqrt {b - c} } \right)^2}$
The minimum value is ${\left( {\sqrt {a - c} + \sqrt {b - c} } \right)^2}$
Comments (0)
