Functions 2M Questions &Solutions || V.S.A.Q.’S||
Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| Functions (2M Questions &Solutions): This note is designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students. Inter Maths – 1A two-mark questions and solutions are very useful in IPE examinations. These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations. Functions QUESTION 1 Find the Domain of the following real-valued functions. (i) f(x) = Sol: given f(x) = It is defined when 6x – x2 – 5 ≠ 0 ⇒ x2 – 6x + 5 ≠ 0 x2 – 5x – x + 5 ≠ 0 x (x – 5) –1(x – 5) ≠ 0 (x – 5) (x – 1) ≠ 0 x ≠ 5 or x ≠ 1 ∴ domain = R – {1, 5} (ii) f(x) = Sol: Given f(x) = It is defined when 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0 ⇒ x ≥ –3, x ≤ 3 and x ≠ 0 ⇒ –3≤ x, x ≤ 3 and x ≠ 0 ⇒ –3≤ x ≤ 3 and x ≠ 0 x ∈ [–3, 3] – {0} ∴ domain = [–3, 3] – {0} (iii) f(x) = Sol: Given f(x) = It is defined when x + 2 ≥ 0, 1 – x > 0 and 1 – x ≠ 0 ⇒ x ≥ –2, x < 1 and x ≠ 0 x ∈ [–2, ∞) ∩ (– ∞, 1) – {0} ⇒ x ∈ [–2, 1) – {0} ∴ domain = [–2, 1) – {0} (iv) f(x) = Sol: Given f(x) = It is defined when 4x – x2 ≥ 0 ⇒ x2 – 4x ≤ 0 x (x – 4) ≤ 0 (x – 0) (x – 4) ≤ 0 x ∈ [0, 4] ∴ domain = [0, 4] (v) f(x) = log (x2 – 4x + 3) Sol: Given f(x) = log (x2 – 4x + 3) It is defined when x2 – 4x + 3 > 0 ⇒ x2 – 3x – x + 3 > 0 x (x – 3) –1(x – 3) > 0 (x – 3) (x – 1) > 0 x ∈ (–∞, 1) ∪ (3, ∞) x ∈ R – [1, 3] ∴ domain = R – [1, 3] (vi) f(x) = Sol: Given f(x) = It is defined when x2 – 1 ≥ 0 and x2 – 3x + 2 > 0 (x + 1)(x – 1) ≥ 0 and x2 – 2x – x + 2 > 0 (x + 1) (x – 1) ≥ 0 and x (x – 2) (x – 1) > 0 x∈ (–∞, –1) ∪ (1, ∞) and x ∈ (–∞, –1) ∪ (2, ∞) ∴ domain = R – (–1, 2] (vii) f(x) = Sol: Given f(x) = It is defined when – x > 0 ⇒ > x ⇒ x < 0 ∴ domain = (–∞, 0) (viii) f(x) = Sol: Given f(x) = It is defined when +x ≠0 ⇒ ≠ – x ⇒ x > 0 ∴ domain = (0, ∞) QUESTION 2 If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x2 + 2 then find (i) (gof) (x) (ii) (gof) () (iii) (fof) (x) (iv) go(fof) (0). Sol: Given f(x) = 4x – 1, g(x) = x2 + 2 (i) (gof) (x) = g (f (x)) = g (4x – 1) = (4x – 1)2 + 2 = 16×2 – 8x + 1 + 2 = 16×2 – 8x + 3 (ii) (gof) ()= (g (f ())
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