Verified Solution • 2025–26 Session moderate

QJEEM17Ma30d0: JEE Main 2017 Mathematics - Class 11 JEEM17Ma30d0 Solution: The function defined as , is

The function f:R→[−12,12]f:R \to \left[ { - {1 \over 2},{1 \over 2}} \right] defined as

f(x)=x1+x2f\left( x \right) = {x \over {1 + {x^2}}} , is

Multiple Choice Options

  • A invertible
  • B injective but not surjective.
  • C surjective but not injective
  • D neither injective nor surjective.

Step-by-Step Verified Solution

f(x)=x1+x2f\left( x \right) = {x \over {1 + {x^2}}}

∴\therefore f(1x)=1x1+1x2=x1+x2=f(x)f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)

∴\therefore f(x) is many-one function.

Now let y = f(x) = x1+x2{x \over {1 + {x^2}}}

⇒\Rightarrow y + x2y = x

⇒\Rightarrow yx - x + y = 0

As x ∈\in R

∴\therefore (-1)2 - 4(y)(y) ≥\ge 0

⇒\Rightarrow 1 - 4y2 ≥\ge 0

⇒\Rightarrow y ∈\in [−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right]

∴\therefore Range = Codomain = [−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right]

So, f(x) is surjective.

∴\therefore f(x) is surjective but not injective

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