Question 13 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at March 23, 2023 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
}
.

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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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