Now let us solve a few examples of functions to understand functions in maths.

**Q1.** Function f is defined by f(x) = - 3 x 2 + 4 x – 13.Find f(3).

**Answer & Explanation:** f(3) = - 3 (3)^{2} + 4 (3) - 13 f(3) = -27 + 12 -13 f(3) = -28

**Q2.** Function h is defined by h(x) = 7x^{3} + 5x – 2.Find h(-2)

**Answer & Explanation:** h(-2) = 7(-2)^{3} + 5(-2) – 2 h(-2) = 7(-8) -10 -2 h(-2) = -68

**Q3.**If h(2,3) = 17 & h(5,4) = 141.Find h(3,4)

**Answer & Explanation:** The function h(a,b) here represents a^{3} + b^{2} h(2,3) = 2^{3} + 3^{2} = 8+ 9 = 17 h(5,4) = 5^{3} + 4^{2} = 125 + 16 =141 Therefore, h(3,4) = 3^{3} + 4^{2} = 27 + 16 = 43

**Q4.**Defined that x # y = x^{3} + y^{2} – 3xy, then 4 # (5 # 6) = ?

**Answer & Explanation:** 4 # (5 # 6) = 4 # (125 + 16 – 90) = 4 # (51) 4 # 51 = 64 + 2601 – 612 = 2053

**Q5.**Defined that a_{k + 2} = 2a_{k + 1} + a_{k} + a_{k - 1}. If a_{0}=2, a_{1}=3, a_{2}=4, then a_{5}=

**Answer & Explanation:** a_{k + 2} = 2a_{k + 1} + a_{k} + a_{k - 1} Put the value of k = 1, a_{1 + 2} = 2a_{1 + 1} + a_{1} + a_{1 - 1} Therefore, a_{3} = 2a_{2} + a_{1} + a_{0} Substituting values a3 = 2 (4) + 3 + 2 = 13 Using the same approach, a_{4} = 33 & a_{5} = 83.

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**Q6.**Given that, f(x) = x^{2} + 1 ; if x is even if(x) = 3x + 2; if x is odd. Find f(f(2))

**Answer & Explanation:** f(f(2)) =f (2^{2} + 1) = f (5) Now f(5) = 3(5) + 2 = 17

**Q7.**Given f(x) = 2x-5, h(x) = x3 + 3, g(x) = x2 + 3. Find f(h(g(2))).

**Answer & Explanation:** f(h(g(2))) = f(h(22 + 3) = f(h(7)) f(73 + 3) = f(346) = 2(346) – 5 = 687

**Q8.**Given f(y) = y^{3} + 7, g(y) = 2y + 3, h(y) = y – 4. Find f(2) + g(2) – h(2)

**Answer & Explanation:** f(2) = 2^{3} + 7 = 15 g(2) = 2(2) + 3 = 7 h(2) = 2 – 4 = -2 Now, f(2) + g(2) – h(2) = 15 + 7 – (-2) = 24

**Q9.**If k_{1} = 1 and k_{n + 1} – 3k_{n} + 2 = 4n, for every positive integer n, Find k_{25}.

**Answer & Explanation:** Now, k_{1} = 1, K_{2} – 3k_{1} + 2 = 4n K_{2} – 3(1) + 2 = 4(1) K_{2} = 5 By solving this equation, we get k_{2} = 5 Similarly, k_{3} = 21 & k_{4} = 73. Looking at these values we can generalise, that k_{n} = 3^{n} – 2n Therefore, k_{25} = 3^{25} – 50

**Q10.**Function f is defined by f(x) = x^{2} - 3 ; if x is even. Function f is defined by f(x) = x^{2} + 1 ; if x is odd. Find f(f(2).

**Answer & Explanation:** f(f(2)) = f(2^{2} - 3) = f (4-3) = f(1) F(1) = 1^{2} + 1 = 1 + 1 = 2