Step 3: Calculate the binomial coefficient and simplify. $$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ $$(2x)^3 = 2^3 x^3 = 8x^3$$ $$(-1)^3 = -1$$ Step 4: Combine the terms. $$T_4 = 20 \cdot (8x^3) \cdot (-1)$$ $$T_4 = 160x^3 \cdot (-1)$$ $$T_4 = -160x^3$$ The coefficient of $x^3$ is $-160$. The coefficient of $x^3$ is $\boxed{\text{-160}}$. 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: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. 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. $$T_{4+1} = \binom{7}{4} (1)^{7-4} (x)^4$$ $$T_5 = \binom{7}{4} (1)^3 x^4$$ Step 3: Calculate the binomial coefficient and simplify. $$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ $$T_5 = 35 \cdot 1 \cdot x^4$$ $$T_5 = 35x^4$$ The coefficient of $x^4$ is $35$. The coefficient of $x^4$ is $\boxed{\text{35}}$. 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: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. 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. $$T_{0+1} = \binom{4}{0} (1)^{4-0} (-5x)^0$$ $$T_1 = \binom{4}{0} (1)^4 (1)$$ Step 3: Calculate the value. $$\binom{4}{0} = 1$$ $$T_1 = 1 \cdot 1 \cdot 1 = 1$$ The term independent of $x$ is $1$. The term independent of $x$ is $\boxed{\text{1}}$. 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 $\frac{12}{2}=6$-th term and $\frac{12}{2}+1=7$-th term. Step 3: Use the general term formula $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For $(1+x)^{11}$, $a=1$, $b=x$, $n=11$. For the $6$-th term, $k+1=6 \implies k=5$. $$T_6 = \binom{11}{5} (1)^{11-5} (x)^5 = \binom{11}{5} x^5$$ For the $7$-th term, $k+1=7 \implies k=6$. $$T_7 = \binom{11}{6} (1)^{11-6} (x)^6 = \binom{11}{6} x^6$$ The middle terms are $\binom{11}{5} x^5$ and $\binom{11}{6} x^6$. 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 $\boxed{x^5}$. This matches option b) $x^5$. Question 44: The determinant of the $2 \times 2$ matrix $\begin{pmatrix} 3 & 4 \\ 2 & 5 \end{pmatrix}$ is: Step 1: Recall the formula for the determinant of a $2 \times 2$ matrix. For a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad-bc$. Step 2: Substitute the values from the given matrix. Here, $a=3$, $b=4$, $c=2$, $d=5$. $$ \text{Determinant} = (3)(5) - (4)(2) $$ $$ = 15 - 8 $$ $$ = 7 $$ The determinant is $\boxed{\text{7}}$. 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 $\boxed{\text{identity matrix}}$. Got more? Send 'em!