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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 7 Solution**. Solutions for other problems are available at Exercise 1.2 SolutionsExercise 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.