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If F : R → R Be the Function Defined By F(X) = 4x3 + 7, Show That F Is a Bijection.

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प्रश्न

If f : R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.

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उत्तर

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)

⇒ 4x3+7 = 4y3+ 7

⇒ 4x3= 4y3

⇒ x3= y3

⇒ x = y

So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain)

f(x) = y

⇒ 4x3+7 = y

⇒ 4x3= y −7

⇒ `x^3 = (y - 7)/4`

⇒  `x = 3sqrt(y-7)/4 in R`

So, for every element in the co-domain, there exists some pre-image in the domain. - f is onto.
Since, f is both one-to-one and onto, it is a bijection.

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अध्याय 2: Functions - Exercise 2.1 [पृष्ठ ३२]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 2 Functions
Exercise 2.1 | Q 11 | पृष्ठ ३२

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