This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 3: Calculate the binomial coefficient and simplify. 63 = (6!)/(3!(6-3)!) = (6!)/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 20 (2x)^3 = 2^3 x^3 = 8x^3 (-1)^3 = -1 Step 4: Combine the terms. T_4 = 20 · (8x^3) · (-1) T_4 = 160x^3 · (-1) T_4 = -160x^3 The coefficient of x^3 is -160. The coefficient of x^3 is -160. This is not among options a) 60, b) -60, c) 120, d) -120. The correct option is e) None of the above. Question 41: The coefficient of x^4 in the expansion of (1+x)^7 is: Step 1: Use the general term formula: T_k+1 = nk a^n-k b^k. Here, a=1, b=x, and n=7. We want the coefficient of x^4, so we set k=4. Step 2: Substitute k=4 into the formula. T_4+1 = 74 (1)^7-4 (x)^4 T_5 = 74 (1)^3 x^4 Step 3: Calculate the binomial coefficient and simplify. 74 = (7!)/(4!(7-4)!) = (7!)/(4!3!) = (7 × 6 × 5)/(3 × 2 × 1) = 35 T_5 = 35 · 1 · x^4 T_5 = 35x^4 The coefficient of x^4 is 35. The coefficient of x^4 is 35. This matches option b) 35. Question 42: The term independent of x in the expansion of (1-5x)^4 is: Step 1: Use the general term formula: T_k+1 = nk a^n-k b^k. Here, a=1, b=-5x, and n=4. The term independent of x is the term where the power of x is 0. This occurs when k=0. Step 2: Substitute k=0 into the formula. T_0+1 = 40 (1)^4-0 (-5x)^0 T_1 = 40 (1)^4 (1) Step 3: Calculate the value. 40 = 1 T_1 = 1 · 1 · 1 = 1 The term independent of x is 1. The term independent of x is 1. This is not among options a) -50, b) 50, c) -10, d) 10. The correct option is e) None of the above. Question 43: The middle term in the expansion of (1+x)^11 is: Step 1: Determine the number of terms in the expansion. For (a+b)^n, there are n+1 terms. Here n=11, so there are 11+1=12 terms. Step 2: Find the position of the middle terms. Since there are 12 terms (an even number), there are two middle terms. Their positions are (12)/(2)=6-th term and (12)/(2)+1=7-th term. Step 3: Use the general term formula T_k+1 = nk a^n-k b^k. For (1+x)^11, a=1, b=x, n=11. For the 6-th term, k+1=6 k=5. T_6 = 115 (1)^11-5 (x)^5 = 115 x^5 For the 7-th term, k+1=7 k=6. T_7 = 116 (1)^11-6 (x)^6 = 116 x^6 The middle terms are 115 x^5 and 116 x^6. The options provided are powers of x. Option b) x^5 corresponds to the variable part of one of the middle terms. The middle term (referring to the power of x) is x^5. This matches option b) x^5. Question 44: The determinant of the 2 × 2 matrix 3 & 4 \\ 2 & 5 is: Step 1: Recall the formula for the determinant of a 2 × 2 matrix. For a matrix a & b \\ c & d , the determinant is ad-bc. Step 2: Substitute the values from the given matrix. Here, a=3, b=4, c=2, d=5. Determinant = (3)(5) - (4)(2) = 15 - 8 = 7 The determinant is 7. This matches option b) 7. Question 45: The product of a square matrix and its inverse is always: The product of a square matrix and its inverse is, by definition, the identity matrix. The identity matrix is a square matrix with ones on the main diagonal and zeros elsewhere. The product of a square matrix and its inverse is always the identity matrix. Got more? Send 'em!