This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 7 Solution. Solutions for other problems are available at Exercise 1.2 Solutions
Exercise 1.2 Problem 7 Solution
7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
i.
f : R → R defined by {f(x) = 3 – 4x}
ii.
f : R → R defined by {f(x) = 1 + x^2}
To check whether f is one-one where f : R → R defined by {f(x) = 3 - 4x}
Consider two elements x_1, x_2 ∈ R in the domain such that {f(x_1) = f(x_2)}
⇒ {3 - 4x_1 = 3 - 4x_2}
⇒ {x_1 = x_2}
∴ f is one-one.
To check whether f is onto where f : R → R defined by {f(x) = 3 - 4x}
For every element y ∈ R should be an image of atleast one element x in the domain such that
{y = 3 - 4x}
⇒ {4x = 3 - y}
{⇒ x = \dfrac{3 - y}{4}}
⇒ Every element y ∈ R in the co-domain will be an image of an the element {\dfrac{3 - y}{4}} in the domain R
∴ f is onto.
∴ As f is both one-one and onto, f is bijective.
To check whether f is one-one where f : R → R defined by {f(x) = 1 + x^2}
Consider two elements x_1, x_2 ∈ R in the domain such that {f(x_1) = f(x_2)}
⇒ {1 + x_1^2 = 1 + x_2^2}
⇒ {x_1^2 = x_2^2}
⇒ {x_1 = \pm x_2}
⇒ Two distinct elements have the same image.
∴ f is not one-one.
To check whether f is onto where f : R → R defined by {f(x) = 1 + x^2}
For every element y ∈ R should be an image of atleast one element x in the domain such that
{y = 1 + x^2}
As we know, x^2 is always positive. Also, 1 is a positive number.
So, the sum of two positive numbers i.e. 1 and x^2 is always positive.
⇒ y is always positive and can not be negative.
As we know, the co-domain R contains both positive and negative real numebers.
⇒ The negative numbers in the co-domain can not be images of any elements in the domain.
In otherwords, if you consider a random number -7 in the co-domain, there exists no element x in the domain R such that f(x) = -7.
∴ f is not onto.
∴ As f is neither one-one nor onto, f is not bijective.
