Home
Class 11
MATHS
Statement-1: The function f:R to R defin...

Statement-1: The function `f:R to R` defined by `f(x)=x^(3)+4x-5` is a bijection.
Statement-2: Every odd degree has at least one real root.

A

1

B

2

C

3

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^3 + 4x - 5 \) is a bijection, we need to show that it is both injective (one-to-one) and surjective (onto). We will also use the fact that every odd degree polynomial has at least one real root. ### Step 1: Show that \( f(x) \) is a bijection #### 1.1: Check if \( f(x) \) is injective To check if \( f(x) \) is injective, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \). Assume \( f(x_1) = f(x_2) \): \[ x_1^3 + 4x_1 - 5 = x_2^3 + 4x_2 - 5 \] This simplifies to: \[ x_1^3 - x_2^3 + 4(x_1 - x_2) = 0 \] Factoring the left-hand side: \[ (x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2 + 4) = 0 \] This gives us two cases: 1. \( x_1 - x_2 = 0 \) which implies \( x_1 = x_2 \). 2. \( x_1^2 + x_1x_2 + x_2^2 + 4 = 0 \). Since \( x_1^2 + x_1x_2 + x_2^2 \) is always non-negative (as it is a sum of squares and a positive constant), the second case cannot hold. Therefore, \( f(x) \) is injective. #### 1.2: Check if \( f(x) \) is surjective To show that \( f(x) \) is surjective, we need to show that for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \). Since \( f(x) \) is a cubic polynomial (odd degree), it will approach \( -\infty \) as \( x \to -\infty \) and \( +\infty \) as \( x \to +\infty \). By the Intermediate Value Theorem, since \( f(x) \) is continuous, it must take every value in \( \mathbb{R} \). Thus, \( f(x) \) is surjective. ### Conclusion Since \( f(x) \) is both injective and surjective, we conclude that \( f(x) \) is a bijection. ### Step 2: Verify Statement 2 Statement 2 states that every odd degree polynomial has at least one real root. Since \( f(x) \) is a cubic polynomial (odd degree), it must have at least one real root, which is consistent with our findings. ### Final Conclusion Both statements are true: - Statement 1: The function \( f(x) = x^3 + 4x - 5 \) is a bijection. - Statement 2: Every odd degree polynomial has at least one real root.

To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^3 + 4x - 5 \) is a bijection, we need to show that it is both injective (one-to-one) and surjective (onto). We will also use the fact that every odd degree polynomial has at least one real root. ### Step 1: Show that \( f(x) \) is a bijection #### 1.1: Check if \( f(x) \) is injective To check if \( f(x) \) is injective, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \). ...
Promotional Banner

Topper's Solved these Questions

  • FUNCTIONS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Exercise|48 Videos
  • FUNCTIONS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Chapter Test|29 Videos
  • FUNCTIONS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Section I - Solved Mcqs|49 Videos
  • DISCRETE PROBABILITY DISTRIBUTIONS

    OBJECTIVE RD SHARMA ENGLISH|Exercise Exercise|40 Videos
  • HYPERBOLA

    OBJECTIVE RD SHARMA ENGLISH|Exercise Chapter Test|29 Videos

Similar Questions

Explore conceptually related problems

The function f:R→R defined by f(x)=(x−1)(x−2)(x−3) is

Show f:R rarr R defined by f(x)=x^(2)+4x+5 is into

Show that the function f: R->R given by f(x)=x^3+x is a bijection.

Show that the function f: R->R defined by f(x)=3x^3+5 for all x in R is a bijection.

Show that f: R->R , defined as f(x)=x^3 , is a bijection.

Consider a function f:R rarr R defined by f(x)=x^(3)+4x+5 , then

Show that the function f: RvecR defined by f(x)=3x^3+5 for all x in R is a bijection.

The function f : R -> R is defined by f (x) = (x-1) (x-2) (x-3) is

Show that the function f: R-{3}->R-{1} given by f(x)=(x-2)/(x-3) is a bijection.

The function f:R rarr R defined as f(x)=(3x^2+3x-4)/(3+3x-4x^2) is :