Question

Let f: R→R defined by f(x)=2x³-7 for x∈R. Then:
(A) f is one-one function
(B) f is many to one function
(C) f is bijective function
(D) f is into function
Choose the correct answer from the options given below:

• (A) and (D) only
• (B) and (D) only
• (B) and (C) only
• (A) and (C) only

Answer 1

The Correct Option is D

To determine if the function \( f(x) = 2x^3 - 7 \) is one-one (injective) and onto (surjective), we will analyze its properties.

Step 1: Checking for injectivity (one-one)
A function is one-one if different elements in the domain map to different elements in the co-domain. Suppose \( f(a) = f(b) \) for \( a, b \in \mathbb{R} \). Then:
\[ 2a^3 - 7 = 2b^3 - 7 \]
Solving, we have:
\[ 2a^3 = 2b^3 \]
\[ a^3 = b^3 \]
\[ a = b \]
This shows that \( f \) is injective since \( a = b \) implies that no two different domain elements map to the same co-domain element.

Step 2: Checking for surjectivity (onto)
A function is onto if for every element \( y \) in the co-domain, there exists an element \( x \) in the domain such that \( f(x) = y \). For function \( f(x)=2x^3 - 7 \) where \( y \in \mathbb{R} \), consider:
\[ y = 2x^3 - 7 \]
Solving for \( x \), we have:
\[ 2x^3 = y + 7 \]
\[ x^3 = \frac{y + 7}{2} \]
\[ x = \sqrt[3]{\frac{y + 7}{2}} \]
Since \( f(x) \) can acquire any real number \( y \), the function is surjective.

Since the function is both injective and surjective, it is bijective.

The correct answer is thus (A) and (C) only.