मराठी

Consider F : R+ → [−5, ∞) Given By F(X) = 9x2 + 6x − 5. Show That F Is Invertible with `F^-1 (X) = Sqrt (X +6-1)/3 .`

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प्रश्न

Consider f : R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with `f^-1 (x) = (sqrt (x +6)-1)/3 .`

बेरीज
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उत्तर

Injectivity of f :
Let x and y be two elements of domain` (R^+)`, such that
f(x)=f(y)

⇒ 9x2+6x−5=9y2+ 6y − 5

⇒ 9x2+6x=9y2+6y

⇒ x = y (As, x, y ∈ `R^+`)

So, f is one-one.

Surjectivity of f:
Let y is in the co domain (Q) such that f(x) = y

⇒ 9x2 + 6x - 5 = y

⇒ 9x2 +6x = y + 5

⇒ 9x2 + 6x +1 = y +6 (Adding 1 on both sides )

⇒ (3x +1)2 = y + 6

⇒ `3x +1 = sqrt(y + 6)`

⇒ `3x = sqrt (y + 6) -1`

⇒ `x = (sqrt (y + 6)-1)/3 in R^+` (domain)

f is onto.
So, f is a bijection and hence, it is invertible.

Finding `f^-1`

Let f−1(x) = y                 ...(1)

⇒ x = f (y)

⇒ x = 9y2+ 6y − 5

⇒ x + 5 = 9y2+6y

⇒ x + 6= 9y2+ 6y + 1         (adding 1 on both sides)

⇒ x + 6 = ( 3y + 1 )2

⇒3y+1=`sqrt(x +6)`

⇒ `3y = sqrt (x +6) -1`

⇒ `y = (sqrt (x+6)-1)/3`

`So, f^-1  (x)  (sqrt (x-6)-1)/3 ` [from (1)]

 

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पाठ 2: Functions - Exercise 2.4 [पृष्ठ ६८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 2 Functions
Exercise 2.4 | Q 9 | पृष्ठ ६८

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