JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 5)
Let f(x) = $${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$$ x $$\, \in $$ R, where a, b and d are non-zero real constants. Then :
f is an increasing function of x
f is neither increasing nor decreasing function of x
f ' is not a continuous function of x
f is a decreasing function of x
Explanation
$$f\left( x \right) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }}$$
$$f'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R$$
f(x) is an increasing function.
$$f'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R$$
f(x) is an increasing function.
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