**
Question 13 (OR 1
**
**
st
**
**
question)
**

Prove that the function f:[0, ∞) → R given by f(x) = 9x
^{
2
}
+ 6x – 5 is not invertible. Modify the codomain of the function f to make it invertible, and hence find f
^{
–1
}
.

Last updated at Dec. 12, 2018 by Teachoo

**
Question 13 (OR 1
**
**
st
**
**
question)
**

Prove that the function f:[0, ∞) → R given by f(x) = 9x
^{
2
}
+ 6x – 5 is not invertible. Modify the codomain of the function f to make it invertible, and hence find f
^{
–1
}
.

Transcript

Question 13 (OR 1st question) Prove that the function f:[0, ∞) → R given by f(x) = 9x2 + 6x – 5 is not invertible. Modify the codomain of the function f to make it invertible, and hence find f–1 . If f(x) is invertible f(x) is one-one f(x) is onto First, let us check if f(x) is onto Let f(x) = y y = 9x2 + 6x – 5 0 = 9x2 + 6x – 5 – y 9x2 + 6x – 5 – y = 0 9x2 + 6x – (5 + y) = 0 Comparing equation with ax2 + bx + c = 0 a = 9, b = 6 , c = – (5 + y) x = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎 Putting values x = (− 6 ± √(6^2 − 4(9) (−(5 + 𝑦)) ))/2(9) x = (− 6 ± √(36 + 36(5 + 𝑦)))/18 x = (− 6 ± √(36(1 + (5 + 𝑦)) ))/18 x = (− 6 ± √(36(6 + 𝑦) ))/18 x = (− 6 ± 6 √((6 + 𝑦)))/18 x = 6[− 1 ± √((6 + 𝑦) )]/18 x = (− 1 ± √((6 + 𝑦) ))/3 So, x = (− 1 − √((6 + 𝑦) ))/3 or (− 1 + √((6 + 𝑦) ))/3 As x ∈ [0, ∞) , i.e., x is a positive real number x cannot be equal to (−1 − √((6 + 𝑦) ))/3 Hence, x = (−1 + √((6 + 𝑦) ))/3 Since y ∈ R For y = –6 x = (−1 + √((6 − 6) ))/3 = (−1)/3 Since x = (−1)/3 ∉ [0, ∞) So, f is not invertible To make it invertible, x ≥ 0 (−1 + √((6 + 𝑦) ))/3 ≥ 0 –1 + √((6 + 𝑦) ) ≥ 0 √((6 + 𝑦) ) ≥ 0 + 1 √((6 + 𝑦) ) ≥ 1 6 + y ≥ 1 y ≥ 1 – 6 y ≥ –5 So, our range should be [–5, ∞) Now, our function is f: [0,∞) → [–5, ∞) f(x) = 9x2 + 6x – 5 We need to first check if it is one-one and onto Checking one-one f (x1) = 9(x1)2 + 6x1 – 5 f (x2) = 9(x2)2 + 6x2 – 5 Putting f (x1) = f (x2) 9(x1)2 + 6x1 – 5 = 9(x2)2 + 6x2 – 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = – 5 + 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0 Now, our function is f: [0,∞) → [–5, ∞) f(x) = 9x2 + 6x – 5 We need to first check if it is one-one and onto Checking one-one f (x1) = 9(x1)2 + 6x1 – 5 f (x2) = 9(x2)2 + 6x2 – 5 Putting f (x1) = f (x2) 9(x1)2 + 6x1 – 5 = 9(x2)2 + 6x2 – 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = – 5 + 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0 (x1 – x2) = 0 ⇒ x1 = x2 (3x1 + 3x2 + 2) = 0 ⇒ 3x1 = –3x2 – 2 Since x ∈ [0, ∞) i.e. x is always positive, Hence 3x1 = – 3x2 – 2 is not true Hence, if f (x1) = f (x2) , then x1 = x2 ∴ f is one-one Now, Checking onto f(x) = 9x2 + 6x – 5 Putting f(x) = y, We get x = (−1 + √((6 + 𝑦) ))/3 Since for y ∈ [–5, ∞) x ∈ [0, ∞) Thus, f is onto Since the function is one-one and onto ∴ It is invertible Calculating inverse For finding inverse, we put f(x) = y and find x in terms of y We have done that while proving onto x = (−1 + √((6 + 𝑦) ))/3 Let g(y) = (−1 + √((6 + 𝑦) ))/3 where g: [–5,∞ ) → [0, ∞) So, inverse of f = f–1 = (−𝟏 + √((𝟔 + 𝒚) ))/𝟑

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4 (Or 1st)

Question 4 (Or 2nd)

Question 5

Question 6

Question 7

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st)

Question 10 (Or 2nd)

Question 11

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) You are here

Question 13 (Or 2nd)

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17

Question 18

Question 19

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23

Question 24 (Or 1st)

Question 24 (Or 2nd)

Question 25

Question 26 (Or 1st)

Question 26 (Or 2nd)

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28

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Class 12

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.