**
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
}
.

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important You are here

Question 13 (Or 2nd)

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd)

Question 27 (Or 1st) Important

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 24, 2021 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
}
.

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