Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Which one of the following functions is not one-one?

Options:

$f:(-1,∞) → R$ given by $f (x) = x^2 + 2x$

$g:(1,∞) → R$ given by $g(x) = e^{x^3-3x+2}$

$h: R→ R$ given by $h(x) = 2^{x(x-1)}$

$\phi: (-∞, 0) → R$ given by $\phi(x)=\frac{x^2}{x^2+1}$

Correct Answer:

$h: R→ R$ given by $h(x) = 2^{x(x-1)}$

Explanation:

We have,

$f(x) = x^2 + 2x, x ∈ (-1, ∞)$

$⇒f'(x)=2(x+1)>0$ for all $x ∈(-1,∞)$

⇒ f(x) is one-one.

It is given that $g(x) = e^{x^3-3x+2}, x ∈(1,∞)$

$∴g'(x)=e^{x^3-3x+2} × 3 (x^2-1) > 0$ for all $x ∈(1,∞)$

⇒ g(x) is one-one.

We have,

$⇒ h(x) = 2^{x(x-1)}$

$⇒ h' (x) = 2^{x (x-1)} (2x-1)$

$⇒h' (x) > 0$ for all $x >\frac{1}{2}$ and $h' (x) <0$ for all $x <\frac{1}{2}$

⇒ h(x) is not one-one.

$∵\phi(x)=\frac{x^2}{x^2+1}$ for all $x ∈ (-∞, 0)$

$∴\phi'(x)=\frac{(x^2+1)2x-x^2×2x}{(x^2+1)^2}$

$⇒f'(x)=\frac{2x}{(x^2+1)^2}<0$ for all $x∈ (-∞, 0)$

$⇒\phi(x)$ is one-one.