Ex 1.3 , 9 - Chapter 1 Class 12 Relation and Functions (Term 1)
Last updated at Jan. 28, 2020 by
Last updated at Jan. 28, 2020 by
Transcript
Ex 1.3, 9 Consider f: R+ → [-5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = ((√(y +6)) − 1)/3 . f(x) = 9x2 + 6x – 5 f is invertible 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 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0 9[(x1)2 – (x2)2 ]+ 6[x1 – x2] = 0 9[(x1 – x2) (x1 + x2) ]+ 6[x1 – x2] = 0 3(x1 – x2) [3(x1 + x2) + 2] = 0 (x1 – x2) [3x1 + 3x2 + 2] = 0/3 (x1 – x2) [3x1 + 3x2 + 2] = 0 (x1 – x2) = 0 ⇒ x1 = x2 (3x1 + 3x2 + 2) = 0 ⇒ 3x1 = – 3x2 – 2 Since f: R+ → [-5,∞ ) So x ∈ R+ 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 Check onto f(x) = 9x2 + 6x – 5 Let f(x) = y such that y ∈ [-5, ∞) Putting in equation 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 + 4(9)(5 + 𝑦) ))/18 = (− 6 ± √(36 + 36(5 + 𝑦)))/18 = (− 6 ± √(36(1 + (5 + 𝑦)) ))/18 = (− 6 ± √(36(6 + 𝑦) ))/18 = (− 6 ± √36 √((6 + 𝑦)))/18 = (− 6 ± √(6^2 ) √((6 + 𝑦)))/18 = (− 6 ± 6 √((6 + 𝑦)))/18 = 6[− 1 ± √((6 + 𝑦) )]/18 = (− 1 ± √((6 + 𝑦) ))/3 So, x = (− 1 − √((6 + 𝑦) ))/3 or (− 1 + √((6 + 𝑦) ))/3 As x ∈ R+ , i.e., x is a positive real number x cannot be equal to (−1 − √((6 + 𝑦) ))/3 Hence, x = (−𝟏 + √((𝟔 + 𝒚) ))/𝟑 Since f: R+ → [−5,∞ ) So y ∈ [−5,∞ ) i.e. y is greater than or equal to −5 i.e. y ≥ −5 y + 5 ≥ 0 Hence the value inside root is positive Hence √(𝑦+6) ≥ 0 x ≥ 0 Hence x is a real number which is greater than or equal to 0. ∴ x ∈ R+ Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((−𝟏 + √((𝟔 + 𝒚) ))/𝟑) = 〖9((−𝟏 + √((𝟔 + 𝒚) ))/𝟑)〗^2+6((−𝟏 + √((𝟔 + 𝒚) ))/𝟑)−5 = (−𝟏+√((𝟔 + 𝒚) ))^2+2(−𝟏+√((𝟔 + 𝒚) ))−5 = 1+(6+𝑦)−2√((𝟔 + 𝒚) )−2+2√((𝟔 + 𝒚) )−5 = 7+𝑦−7 = 𝑦 Thus, for every y ∈ [−5, ∞) , there exists x ∈ R+ such that f(x) = y Hence, f is onto Since f(x) is one-one and onto, So, f(x) is invertible And Inverse of x = 𝑓^(−1) (𝑦) = (−𝟏 + √((𝟔 + 𝒚) ))/𝟑
Finding Inverse
Inverse of a function
How to check if function has inverse?
Example 22 Deleted for CBSE Board 2022 Exams
Ex 1.3, 5 (i) Deleted for CBSE Board 2022 Exams
How to find Inverse?
Example 28 (a) Deleted for CBSE Board 2022 Exams
Misc 11 (i) Important Deleted for CBSE Board 2022 Exams
Ex 1.3, 11 Deleted for CBSE Board 2022 Exams
Example 27 Important Deleted for CBSE Board 2022 Exams
Misc 1 Deleted for CBSE Board 2022 Exams
Ex 1.3 , 6 Deleted for CBSE Board 2022 Exams
Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams
Example 23 Important Deleted for CBSE Board 2022 Exams
Misc 2 Deleted for CBSE Board 2022 Exams
Ex 1.3 , 4 Deleted for CBSE Board 2022 Exams
Example 24 Deleted for CBSE Board 2022 Exams
Ex 1.3 , 8 Important Deleted for CBSE Board 2022 Exams
Example 25 Important Deleted for CBSE Board 2022 Exams
Ex 1.3 , 9 Important Deleted for CBSE Board 2022 Exams You are here
Finding Inverse
About the Author