Let \(f: \mathrm{R}\rightarrow \mathrm{R}\) defined by \(f(x)=2x^3-7\) for \(x\in \mathrm{R}\). 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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
One-one: suppose \(f(x_1) = f(x_2)\):
\(2x_1^3 - 7 = 2x_2^3 - 7\)
\(x_1^3 = x_2^3\)
\(x_1 = x_2\)
since every real number has exactly one real cube root.
Onto: for any real k, solve f(x) = k:
\(2x^3 - 7 = k\)
\( \displaystyle x = \left(\frac{k+7}{2}\right)^{1/3} \)
which is a real number, so every real k is a value of f.
So f is one-one and onto, that is, bijective.
Checking the statements
(A) f is one-one function — Correct. Equal values \(x_1^3 = x_2^3\) force \(x_1 = x_2\).
(B) f is many to one function — Incorrect. Different x give different values, so f is one-one.
(C) f is bijective function — Correct. f is both one-one and onto.
(D) f is into function — Incorrect. Every real k is reached at \(x = \left(\frac{k+7}{2}\right)^{1/3}\), so f is onto.