Question

Question 39 Let f(x)= 3x2 – 2. g(x) = 4x +4. Find the value of the function. (of)(4) o tof)(4) = 6,350 OVO (4) = 6,346 Ovo(4) = 6,348 ovo(4) = 6,347 OVO (4) = 6,345 O None of the above

Answer 1

Answer:- 1. fof(4) = 6346 2. O positive, 1 negative and 8 non-real is a correct option. 3. Therefore, possible rational zeros are, +1, +2, +4, +8 The detailed calculation and explanation is given below:- We are given, f(x) = 3x2 – 2 and g(x) = 4x + 4 Therefore, fof(x) = f(f(x)] = f(3x2 – 2) = 3(3x2 – 2)2 – 2 3[(3x2)2 – 2 · 3x2 • 2 + 2?] – 2 = 3(9x4 – 12x2 + 4) - 2 = = = 27x4 – 36x2 + 12 – 2 = = 27x4 – 36x2 + 10 Now therefore, fof(4) = 27(4)4 – 36(4)2 + 10 = = 27 x 256 – 36 x 16 + 10 = 6912 – 576 + 10 = 6346 Hence, fof(4) = 6346 (Ans.) (2nd option) Descartes' Rule of Sign:- It states that the number of positive real zeros in a polynomial f (x) is equal to the number of changes in sign in front of each term, or less than this by an even number.

To find the number of negative real zeros we need to apply Descartes' Rule of Sign in f(-x) Once we have the number of positive real zeros and negative real zeros and knowing that the total number of zeros of a polynomial is equal to the degree of the polynomial we can now easily calculate the number of non-real zeros of the polynomial. The given polynomial equation is, 8x9 + 11x2 + 3x + 5 = 0 Therefore, as we can see that in the polynomial the number of sign changes is 0 i.e. there is no sign change in front of each term. Hence, the number of possible positive real roots is 0. Now, let us put f(-x) and apply the Descartes' rule of sign. Putting f(-x) in the polynomial we have, 8(-x)º + 11(-x)2 + 3(-x) + 5 = 0 or, -8xº + 11x2 – 3x + 5 = 0 Here, as we can see the sign change happens 3 times. Hence, the number of possible negative real roots are 3 and 1 Total number of zeros Number of possible positive real zeros 0 0 Number of possible negative real zeros 3 Number of possible non-real zeros 6 8 9 9 1 Therefore the correct option is, O positive, 1 negative and 8 non-real is a correct option. Rational zero Theorem:- If a polynomial equation has integer coefficients, then any rational zero must be in the form of +, where p is a factor of the constant term and q is a factor of the leading coefficient. Now the given polynomial is, Q(x) = x4 –4x3 – 5x+8 Now, here the leading coefficient is 1 (it has no other factor) And, the constant term is 8. Therefore, its factors are 8,4,2,1 Therefore, possible rational zeros are, +1, +2, +4, +8