Question

Question 1. (a) Solve the following in equation and write down the solution set : [3] 11x – 4 < 15x + 4 ≤ 3x + 14, x ∈ W Represent the solution on a real number line. (b) A man invests 4500 in shares of a company which is paying 7.5% dividend. [3] If 100 shares are available at a discount of 10%. Find : (i) Number of shares he purchases. (ii) His annual income. (c) In a class of 40 students, marks obtained by the students in a class test (out of 10) are given below : [4]  Calculate the following for the given distribution : (i) Median (ii) Mode Solution :  Total investment = ₹ 4500 Face value of a share = ₹ 100 Discount = 10% ∴ Market value of a share = ₹ (100 – 10) = ₹ 90 Now, Number of shares purchased = 450090=50 Annual income =  = ₹ 375  Here, Marks corresponding to cumulative frequency 20 is 6 Thus, the required median is 6. Clearly, 6 occurs 10 times which is maximum. Hence, mode is 6. Question 2. (a) Using the factor theorem, show that (x – 2) is a factor of x3 + x2 – 4x – 4. [3] Hence, factorise the polynomial completely. (b) Prove that : (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 [3] (c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the : [4] (i) first term (ii) common difference (iii) sum of the first 20 terms. Solution : (a) Given polynomial is p(x) = x3 + x2 – 4x – 4 x – 2 is its factor, if p(2) = 0 p(2) = (2)3 + (2)2 – 4(2) – 4 = 8 + 4 – 8 – 4 = 0 Thus, x – 2 is a factor of p(x). Now, x3 + x2 – 4x + 4 = x2(x +1) – 4(x + 1) = (x + 1) (x2 – 4) = (x + 1) (x + 2) (x – 2) Hence, the required factors are (x + 1), (x + 2) and (x – 2). L.H.S. = (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ)   Hence, first term is – 1, common difference is 3 and sum of the first 20 terms is 550.​

Answer 1

Answer:

(a), this is the first one