Question

Solve this ques of Jee mains
chapter → Functions​

Answer 1

EXPLANATION.

Function f : R → [-1/2, 1/2] defined as f(x) = x/1 + x².

As we know that,

Let, us find f(1/x).

⇒ f(1/x) = (1/x)/1 + (1/x)².

⇒ f(1/x) = (1/x)/1 + 1/x².

⇒ f(1/x) = (1/x)/x² + 1/x².

⇒ f(1/x) = 1/(x) X (x²)/(x² + 1).

⇒ f(1/x) = (x)/(x² + 1).

⇒ f(1/x) = f(x).

If f(1/x) = f(x) then f(1/2) = f(2) & f(1/3) = f(3) & f(1/4) = f(x) . . . . . So on.

So, f(x) is a many-one functions.

⇒ f(x) = x/1 + x².

⇒ y = f(x) = x/1 + x².

⇒ y = x/1 + x².

⇒ y(1 + x²) = x.

⇒ y + x²y = x.

⇒ y + x²y - x = 0.

⇒ yx² - x + y = 0.

As we know that,

⇒ D ≥ 0.

⇒ b² - 4ac ≥ 0.

⇒ (-1)² - 4(y)(y) ≥ 0.

⇒ 1 - 4y² ≥ 0.

⇒ (1)² - (2y)² ≥ 0.

⇒ (1 - 2y)(1 + 2y) ≥ 0.

⇒ (2y - 1)(2y + 1) ≤ 0.

Zeroes are.

⇒ 2y - 1 = 0.

⇒ 2y = 1.

⇒ y = 1/2. - - - - - (1).

⇒ 2y + 1 = 0.

⇒ 2y = - 1.

⇒ y = - 1/2. - - - - - (2).

Put the value in the wavy curve method, we get.

⇒ y ∈ [-1/2, 1/2].

Range = Co-domain = [-1/2. 1/2].

So, f(x) is a surjective.

Hence, Option [D] is correct answer.

Answer 2

$\Large{\underline{\underline{\red{\bold{➸Problem:}}}}}$

The function $f : \bold{R\;→\;[\frac{-1}{2},\frac{1}{2}]}$ defined as $f(x) = \frac{x}{1+x²}$ is ____________.

• (a) neither injective not surjective.
• (b) invertible.
• (c) injective but not surjective.
• (d) surjective but not injective.

$\Large{\underline{\underline{\bold{\color{salmon}๛Required\; Response:}}}}$

We know that, $f(x) = \frac{x}{1+x²}$

Now let's derive f(x) for f(1/x)

$f( \frac{1}{x} ) = \frac{ \frac{1}{x} }{1 + ({ \frac{1}{x} )}^{2} } \\ f( \frac{1}{x} ) = \frac{ \frac{1}{x} }{1 + \frac{1}{ {x}^{2} } } \\ f( \frac{1}{x} ) = \frac{x}{1 + {x}^{2} }$

$→\;f(\frac{1}{x}) = f(x)$

In The Same Manner

.:. $f( \frac{1}{2} ) = f(2) \: (or) \: f( \frac{1}{3} ) = f(3) \: and \: so \: on...$

Thus We Can Say, f(x) is many-one function.

Now, Again Let's y = f(x)

$y = \frac{ {x} }{1 + {x}^{2} } \\ y(1 + {x}^{2} ) = x \\ y + {x}^{2} y = x \\ {x}^{2} y + y - x = 0$

As, x is a Rational Number [x ∈ R]

• As, D ≥ 0 & b²-4ac ≥ 0

${( - 1)}^{2} - 4(y)(y) ≥ 0$

$1 - 4 {y}^{2} ≥ 0$

$(1)²-(2y)² ≥ 0$

$(1-2y)(1+2y) ≥ 0$

So,

$(2y-1)(1+2y) ≤ 0$

Roots of the Equation are:-

→ 2y-1 = 0

→ 2y = 1

→ y = 1/2 ➝ ①

___________________

→ 1+2y = 0

→ 2y = -1

→ y = -1/2 ➝ ②

Such That, $\bold{y\;∈\;[\frac{-1}{2},\frac{1}{2}]}$

.:. Range = Co-domain $→\;\bold{[\frac{-1}{2},\frac{1}{2}]}$

• Thus, f(x) is surjective.

Hence, $\blue{\sf{f(x) \;is\; surjective\; but \;not\; injective.}}$

• Option D.

$\boxed{\tt{@MrSovereign}}$

Hope This Helps!!