100+ Free MET Practice Questions

Explanation: From the given equation, the sum of roots is $\alpha + \beta = 5$ and the product of roots is $\alpha\beta = 6$. Using the algebraic identity, we have $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$. Substituting the values gives $5^3 - 3(6)(5) = 125 - 90 = 35$.

Explanation: The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = 1/3$. Since $|r| < 1$, the sum of the infinite series is given by $S = a / (1 - r)$. Substituting the values yields $S = 1 / (1 - 1/3) = 1 / (2/3) = 3/2$.

Explanation: The number of terms in the multinomial expansion of $(x_1 + x_2 + \dots + x_r)^n$ is given by the formula $^{n+r-1}C_{r-1}$. For $(x + y + z)^{10}$, we have $n = 10$ and $r = 3$. The number of terms is $^{10+3-1}C_{3-1} = ^{12}C_2 = (12 \times 11) / 2 = 66$.

Explanation: Using the combination property $^nC_x = ^nC_y \implies x = y$ or $x + y = n$, we compare the subscripts. Since $8 \neq 2$, we must have $8 + 2 = n$. Therefore, $n = 10$.

Explanation: The $r$-th term of the series is $T_r = r(r+1) = r^2 + r$. The sum of the series is $S_n = \sum T_r = \sum r^2 + \sum r = [n(n+1)(2n+1)/6] + [n(n+1)/2]$. Factoring out $n(n+1)/2$, we get $[n(n+1)/2] \cdot [(2n+1)/3 + 1] = [n(n+1)/2] \cdot [(2n+4)/3] = n(n+1)(n+2)/3$.

Explanation: The general term in the trinomial expansion of $(1 + x - 2x^2)^6$ is given by $T = [6! / (p! q! r!)] \cdot (1)^p (x)^q (-2x^2)^r$, where $p + q + r = 6$. For the power of $x$ to be 4, we must have $q + 2r = 4$. Possible non-negative integer solutions are: (1) $r=0, q=4 \implies p=2$, term is $[6! / (2!4!0!)] \cdot (-2)^0 = 15$; (2) $r=1, q=2 \implies p=3$, term is $[6! / (3!2!1!)] \cdot (-2)^1 = -120$; (3) $r=2, q=0 \implies p=4$, term is $[6! / (4!0!2!)] \cdot (-2)^2 = 60$. The sum of these coefficients is $15 - 120 + 60 = -15$.

Explanation: We can write $\sin(75^\circ)$ as $\sin(45^\circ + 30^\circ)$. Using the angle sum identity $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$, we have $\sin(75^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) = (1/\sqrt{2})(\sqrt{3}/2) + (1/\sqrt{2})(1/2) = (\sqrt{3} + 1)/(2\sqrt{2})$.

Explanation: We use the tangent sum identity $\tan(A + B) = [\tan(A) + \tan(B)] / [1 - \tan(A)\tan(B)]$. Substituting the given values gives $\tan(A + B) = (1/2 + 1/3) / [1 - (1/2)(1/3)] = (5/6) / (1 - 1/6) = (5/6) / (5/6) = 1$. Therefore, $A + B = \pi/4$ (or $45^\circ$).

Explanation: The principal value of $x$ for which $\cos(x) = -1/2$ is $2\pi/3$. The general solution for $\cos(x) = \cos(\alpha)$ is $x = 2n\pi \pm \alpha$, where $n \in \mathbb{Z}$. Thus, the general solution is $x = 2n\pi \pm 2\pi/3$.

Explanation: Using the Taylor series expansion of $\sin(x) = x - x^3/6 + x^5/120 - \dots$, the expression becomes $[(x - x^3/6 + \dots) - x] / x^3 = -x^3/(6x^3) + O(x^2) = -1/6 + O(x^2)$. Taking the limit as $x \to 0$, the value is $-1/6$. Alternatively, applying L'Hopital's rule three times yields the same result.