Analyze \(R_1\): \((a, b) R_1 (c, d) \iff ad - bc \ge 0\).


Consider the elements \(A=(1, -1)\), \(B=(-1, 2)\), and \(C=(-2, 1)\).


Check \(A R_1 B\): \(1(2) - (-1)(-1) = 2 - 1 = 1 \ge 0\). True.


Check \(B R_1 C\): \((-1)(1) - 2(-2) = -1 + 4 = 3 \ge 0\). True.


Check \(A R_1 C\): \(1(1) - (-1)(-2) = 1 - 2 = -1 < 0\). False.


Thus, \(R_1\) is not transitive.


Analyze \(R_2\): \((a, b) R_2 (c, d) \iff a + d \ge b + c\).


Rearrange the inequality: \(a - b \ge c - d\).


Let \(f(x, y) = x - y\). The relation is \(f(A) \ge f(B)\).


For any \(A, B, C\), if \(f(A) \ge f(B)\) and \(f(B) \ge f(C)\), then by the transitive property of real numbers, \(f(A) \ge f(C)\).


Thus, \(R_2\) is transitive.
Quick Tip: To test for transitivity in relations defined by inequalities, try to map the condition to a standard order relation like \(\ge\) on real numbers (as in \(R_2\)). If that's not possible, search for a counterexample (as in \(R_1\)).

The roots of \(x^2 - 3x + 9 = 0\) are \(x = \frac{3 \pm \sqrt{9 - 36}}{2} = \frac{3 \pm 3\sqrt{3}i}{2} = 3\left(\frac{1 \pm \sqrt{3}i}{2}\right)\).


Recognize that \(\frac{1 + \sqrt{3}i}{2} = -\omega^2\) and \(\frac{1 - \sqrt{3}i}{2} = -\omega\), where \(\omega\) is a cube root of unity.


Thus, the roots are \(\alpha = -3\omega^2\) and \(\beta = -3\omega\). Also \(\alpha\beta = 9\).


The expression is \(E = \left( \frac{\beta^{30}}{(\alpha\beta \cdot \alpha)^{10}} + \frac{\alpha^{30}}{(\alpha\beta \cdot \beta)^{10}} \right)^2 = \left( \frac{\beta^{30}}{\alpha^{20}\beta^{10}} + \frac{\alpha^{30}}{\alpha^{10}\beta^{20}} \right)^2\).


Simplify terms: \(\frac{\beta^{20}}{\alpha^{20}} + \frac{\alpha^{20}}{\beta^{20}} = \left(\frac{\beta}{\alpha}\right)^{20} + \left(\frac{\alpha}{\beta}\right)^{20}\).


Calculate ratio: \(\frac{\alpha}{\beta} = \frac{-3\omega^2}{-3\omega} = \omega\). So \(\frac{\beta}{\alpha} = \frac{1}{\omega} = \omega^2\).


Substitute back: \(E = \left( (\omega^2)^{20} + (\omega)^{20} \right)^2 = (\omega^{40} + \omega^{20})^2\).


Using \(\omega^3 = 1\), \(\omega^{40} = \omega\) and \(\omega^{20} = \omega^2\).

\(E = (\omega + \omega^2)^2 = (-1)^2 = 1\).
Quick Tip: When quadratic roots involve \(\sqrt{3}i\), express them in terms of complex cube roots of unity \(\omega\). Remember \(1 + \omega + \omega^2 = 0\) and \(\omega^3 = 1\).

Given \(z = 2 + 5i\), we find the quadratic equation for \(z\).

\(z - 2 = 5i \implies (z - 2)^2 = -25 \implies z^2 - 4z + 29 = 0\).


Let \(P(z) = 2z^3 + 21z^2 - 58z + 4\). Divide \(P(z)\) by \((z^2 - 4z + 29)\).

\(2z^3 + 21z^2 - 58z + 4 = 2z(z^2 - 4z + 29) + 8z^2 - 58z + 21z^2 - 58z + 4\).

\(= 2z(0) + 29z^2 - 116z + 4\).


Substitute \(z^2 - 4z + 29 = 0 \implies z^2 = 4z - 29\). No, simpler to divide again.

\(29z^2 - 116z + 4 = 29(z^2 - 4z + 29) - 29(29) + 4\).

\(= 29(0) - 841 + 4 = -837\).


The value of the polynomial is \(-837\).


The modulus is \(|-837| = 837\).
Quick Tip: To evaluate a polynomial \(P(z)\) for a complex number \(z\), divide \(P(z)\) by the quadratic polynomial formed by \(z\) and its conjugate. The remainder gives the value.

For infinitely many solutions, the third equation must be a linear combination of the first two.


Let \(Eq_3 = A \cdot Eq_1 + B \cdot Eq_2\).


Compare constant terms: \(5\sqrt{21} + 3\sqrt{55} = A\sqrt{7} + B\sqrt{11}\).


Since \(\sqrt{21} = \sqrt{3}\sqrt{7}\) and \(\sqrt{55} = \sqrt{5}\sqrt{11}\), we guess \(A = 5\sqrt{3}\) and \(B = 3\sqrt{5}\).


Verify with y-coefficient: \(A(-\sqrt{2}) + B(\sqrt{3}) = -5\sqrt{6} + 3\sqrt{15}\). Matches.


Verify with z-coefficient: \(A(\sqrt{5}) + B(-\sqrt{2}) = 5\sqrt{15} - 3\sqrt{10}\). Matches.


Now equate x-coefficients: \(A(K) + B(\sqrt{5}) = 30\).

\(5\sqrt{3}K + 3\sqrt{5}(\sqrt{5}) = 30\).

\(5\sqrt{3}K + 15 = 30 \implies 5\sqrt{3}K = 15 \implies K = \frac{3}{\sqrt{3}} = \sqrt{3}\).


Thus, \(K^2 = 3\).
Quick Tip: In a consistent dependent system (infinitely many solutions), check the RHS constants to determine the linear combination scalars quickly.

Calculate \(C = A * B = A^2 + B^2\). Since matrices are diagonal:

\(A^2 = diag(\alpha^2, \beta^2)\), \(B^2 = diag(\alpha^2, \alpha^2)\).

\(C = diag(2\alpha^2, \alpha^2 + \beta^2)\).


Next, calculate \(C * 2I = C^2 + (2I)^2\).

\(C^2 = diag(4\alpha^4, (\alpha^2 + \beta^2)^2)\).

\((2I)^2 = (2I)(2I) = 4I = diag(4, 4)\).


The result is \(diag(4\alpha^4 + 4, (\alpha^2 + \beta^2)^2 + 4)\).


Given this equals \(20I = diag(20, 20)\).

\(4\alpha^4 + 4 = 20 \implies \alpha^4 = 4 \implies \alpha^2 = 2\).

\((\alpha^2 + \beta^2)^2 + 4 = 20 \implies (\alpha^2 + \beta^2)^2 = 16\).


Since \(\alpha^2, \beta^2 \ge 0\), \(\alpha^2 + \beta^2 = 4\).


Substitute \(\alpha^2 = 2\): \(2 + \beta^2 = 4 \implies \beta^2 = 2\).

\(|\alpha\beta| = \sqrt{\alpha^2 \beta^2} = \sqrt{2 \cdot 2} = 2\).
Quick Tip: Operations on diagonal matrices are performed element-wise on the diagonal entries. This reduces matrix equations to simple scalar equations.

The general term is \(T_{n} = \frac{n}{1 + n^2 + n^4}\).


Factor the denominator: \(1 + n^2 + n^4 = (n^2 + 1)^2 - n^2 = (n^2 - n + 1)(n^2 + n + 1)\).


Notice that \((n^2 + n + 1) - (n^2 - n + 1) = 2n\).


Rewrite \(T_{n}\): \(T_{n} = \frac{1}{2} \left[ \frac{2n}{(n^2 - n + 1)(n^2 + n + 1)} \right] = \frac{1}{2} \left[ \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right]\).


Let \(f(n) = n^2 - n + 1\). Then \(n^2 + n + 1 = (n+1)^2 - (n+1) + 1 = f(n+1)\).


Sum \(S_{11} = \sum_{n=1}^{11} \frac{1}{2} [\frac{1}{f(n)} - \frac{1}{f(n+1)}]\).


This is a telescoping series: \(S_{11} = \frac{1}{2} \left[ \frac{1}{f(1)} - \frac{1}{f(12)} \right]\).

\(f(1) = 1\).

\(f(12) = 12^2 - 12 + 1 = 144 - 11 = 133\).

\(S_{11} = \frac{1}{2} \left[ 1 - \frac{1}{133} \right] = \frac{1}{2} \cdot \frac{132}{133} = \frac{66}{133}\).
Quick Tip: Always check if the rational term can be split into partial fractions that form a telescoping series, especially when the denominator factors into polynomials like \(n^2 \pm n + 1\).

The limit is of the form \(1^\infty\).


Using the formula \(\lim_{x \to 0} f(x)^{g(x)} = e^{\lim_{x \to 0} (f(x)-1)g(x)}\):


Here \(f(x) = 1+3x\) and \(g(x) = \frac{x+2}{x}\).


Exponent \(L = \lim_{x \to 0} ((1+3x) - 1) \cdot \frac{x+2}{x} = \lim_{x \to 0} 3x \cdot \frac{x+2}{x}\).

\(L = \lim_{x \to 0} 3(x+2) = 3(0+2) = 6\).


The limit is \(e^{L} = e^6\).
Quick Tip: For limits of the form \(1^\infty\), use the identity \(\lim f(x)^{g(x)} = e^{\lim (f(x)-1)g(x)}\).

Let the line be L. A point on L is \(A(5, 12, 10)\) and direction vector is \(\vec{d} = (3, 1, 2)\).


Vector \(\vec{AQ} = (1-5, 2-12, 0-10) = (-4, -10, -10)\).


Projection of \(\vec{AQ}\) on \(\vec{d}\) gives the distance from A to the foot of perpendicular F.

\(proj = \frac{\vec{AQ} \cdot \vec{d}}{|\vec{d}|^2} \vec{d}\).

\(\vec{AQ} \cdot \vec{d} = -4(3) - 10(1) - 10(2) = -12 - 10 - 20 = -42\).

\(|\vec{d}|^2 = 3^2 + 1^2 + 2^2 = 9 + 1 + 4 = 14\).


Scalar t for foot \(F = A + t\vec{d}\) is \(-42/14 = -3\).

\(F = (5, 12, 10) - 3(3, 1, 2) = (5-9, 12-3, 10-6) = (-4, 9, 4)\).


Distance \(QF^2 = (-4-1)^2 + (9-2)^2 + (4-0)^2 = (-5)^2 + 7^2 + 4^2 = 25 + 49 + 16 = 90\).


Since P is the image of Q, F is the midpoint of PQ. Thus, \(PQ = 2QF\).

\((PQ)^2 = 4(QF)^2 = 4(90) = 360\).
Quick Tip: The square of the distance between a point and its image is 4 times the square of the distance from the point to the line (or plane).

Total balls to draw = 8.


Constraints: Red \(\ge 3\), White \(\ge 3\).


Possible cases for (Red, White) summing to 8:


Case 1: (3 Red, 5 White). Ways = \(\binom{7}{3} \times \binom{9}{5} = 35 \times 126 = 4410\).


Case 2: (4 Red, 4 White). Ways = \(\binom{7}{4} \times \binom{9}{4} = 35 \times 126 = 4410\).


Case 3: (5 Red, 3 White). Ways = \(\binom{7}{5} \times \binom{9}{3} = 21 \times 84 = 1764\).


Total Ways = \(4410 + 4410 + 1764 = 10584\).
Quick Tip: List all possible compositions that satisfy the "at least" condition and sum the combinations for each mutually exclusive case.

Let \(f(x) = 1 + x^2 + x^4\). For \(x \in [0, 1]\), \(f(x)\) increases from 1 to 3.


We find where \(f(x)\) crosses integer values. \(f(x)=2 \implies x^4+x^2+1=2 \implies x^4+x^2-1=0\).


Solving for \(x^2\): \(x^2 = \frac{-1+\sqrt{5}}{2}\). Let \(x_1 = \sqrt{\frac{\sqrt{5}-1}{2}}\).


In interval \([0, x_1)\), \(1 \le f(x) < 2\), so \([f(x)] = 1\).


In interval \([x_1, 1)\), \(2 \le f(x) < 3\), so \([f(x)] = 2\).

\(a = \int_{0}^{x_1} 1 dx + \int_{x_1}^{1} 2 dx = x_1 + 2(1 - x_1) = 2 - x_1\).


Thus \(x_1 = 2 - a\).


Since \(x_1^4 + x_1^2 - 1 = 0\), substitute \(x_1 = 2-a\):

\((2-a)^4 + (2-a)^2 - 1 = 0\).


Expand: \((a^4 - 8a^3 + 24a^2 - 32a + 16) + (a^2 - 4a + 4) - 1 = 0\).

\(a^4 - 8a^3 + 25a^2 - 36a + 19 = 0\).


Rearrange: \(36a - 25a^2 + 8a^3 - a^4 = 19\).
Quick Tip: Instead of calculating the numerical value of 'a', use the polynomial equation satisfied by the boundary point \(x_1\) to find the relationship involving 'a'.

Since PL and QM are parallel tangents at points P and Q, PQ must be a diameter of the circle. Let radius be \(r\), so \(PQ = 2r\).


Set coordinates: \(Q(0,0)\) and \(P(0, 2r)\) (assuming vertical alignment for diameter perpendicular to horizontal tangents).


Line PL is \(y=2r\). L is at \((8, 2r)\) (length 8).


Line QM is \(y=0\). M is at \((2, 0)\) (length 2, on the same side to intersect correctly on circle).


Line PM connects \((0, 2r)\) and \((2, 0)\): \(y - 0 = \frac{2r - 0}{0 - 2}(x - 2) \implies y = -r(x-2)\).


Line QL connects \((0,0)\) and \((8, 2r)\): \(y = \frac{2r}{8}x = \frac{r}{4}x\).


Intersection R: \(\frac{r}{4}x = -r(x-2) \implies \frac{x}{4} = -x + 2 \implies \frac{5x}{4} = 2 \implies x = \frac{8}{5}\).


y-coordinate: \(y = \frac{r}{4}(\frac{8}{5}) = \frac{2r}{5}\).


R lies on the circle with diameter PQ. Equation: \(x^2 + y(y-2r) = 0\).


Substitute R: \((\frac{8}{5})^2 + \frac{2r}{5}(\frac{2r}{5} - 2r) = 0\).

\(\frac{64}{25} + \frac{2r}{5}(\frac{-8r}{5}) = 0\).

\(\frac{64}{25} - \frac{16r^2}{25} = 0 \implies 16r^2 = 64 \implies r^2 = 4 \implies r = 2\).
Quick Tip: For parallel tangents at P and Q, PQ is the diameter. Using coordinate geometry with P and Q on the y-axis simplifies the intersection calculation.

Let \(R(h, k)\). Centroid \(G(x, y) = (\frac{2+\alpha+h}{3}, \frac{5-11+k}{3})\).

\(h = 3x - \alpha - 2\) and \(k = 3y + 6\).


R lies on \(l_1\), so \(9(3x - \alpha - 2) + 7(3y + 6) + \alpha = 0\).

\(27x + 21y - 9\alpha - 18 + 42 + \alpha = 0 \implies 27x + 21y - 8\alpha + 24 = 0\).


Dividing by 3: \(l_2: 9x + 7y + \frac{24-8\alpha}{3} = 0\).


Distance between \(l_1\) and \(l_2\): \(\frac{|\alpha - \frac{24-8\alpha}{3}|}{\sqrt{9^2+7^2}} = \frac{|11\alpha - 24|}{3\sqrt{130}}\).


Given distance is \(\frac{20}{3\sqrt{130}}\). So \(|11\alpha - 24| = 20\).

\(11\alpha - 24 = 20 \implies 11\alpha = 44 \implies \alpha = 4\). (Since \(\alpha \in \mathbb{N}\)).


Line \(l_2\) becomes \(27x + 21y - 8(4) + 24 = 0 \implies 27x + 21y - 8 = 0\).


Distance of \(Q(4, -11)\) from \(l_2\): \(\frac{|27(4) + 21(-11) - 8|}{\sqrt{27^2 + 21^2}}\).


Or using simpler form \(9x+7y-8/3=0\): \(d = \frac{|9(4) + 7(-11) - 8/3|}{\sqrt{130}} = \frac{|36 - 77 - 2.66|}{\sqrt{130}} = \frac{|-41 - 8/3|}{\sqrt{130}}\).

\(d = \frac{131/3}{\sqrt{130}} = \frac{131}{3\sqrt{130}}\).


Given \(d = \frac{k}{3\sqrt{130}}\), so \(k = 131\).
Quick Tip: The locus of the centroid of a triangle where one vertex moves on a line \(L\) is a line parallel to \(L\).

\(A_1(x, y)\) with \(x, y > 0\). \(A_2(-x, y)\) (image in y-axis).

\(A_1 A_2 = 2 \implies 2x = 2 \implies x = 1\).

\(1^2 + y^2 = 10 \implies y = 3\). So \(A_1(1, 3)\) and \(A_2(-1, 3)\).

\(A_3(u, v)\) on circle. \(A_1 A_3 = 2 \implies (u-1)^2 + (v-3)^2 = 4\).

\(u^2+v^2 - 2u - 6v + 10 = 4\). Since \(u^2+v^2=10\), \(20 - 2u - 6v = 4 \implies u + 3v = 8\).


Substitute \(u = 8 - 3v\) into \(u^2+v^2=10\):

\((8-3v)^2 + v^2 = 10 \implies 64 - 48v + 10v^2 = 10 \implies 10v^2 - 48v + 54 = 0\).

\(5v^2 - 24v + 27 = 0 \implies (5v - 9)(v - 3) = 0\).

\(v = 3\) (gives point \(A_2\)) or \(v = 9/5\). So \(A_3\) has \(y\)-coord \(9/5\).


Base \(A_1 A_2 = 2\). Height = \(y_{A_1} - y_{A_3} = 3 - 1.8 = 1.2\).


Area = \(\frac{1}{2} \times 2 \times 1.2 = 1.2\).

\(20 \times Area = 20 \times 1.2 = 24\).
Quick Tip: For questions involving distances between points on a circle, combining the distance formula with the circle equation leads to linear constraints (chords).

Compute powers of 7 modulo 15:

\(7^1 \equiv 7 \pmod{15}\).

\(7^2 = 49 \equiv 4 \pmod{15}\).

\(7^3 \equiv 7 \times 4 = 28 \equiv 13 \equiv -2 \pmod{15}\).

\(7^4 \equiv 7 \times 13 = 91 \equiv 1 \pmod{15}\).


The cycle length is 4.


Power \(89 = 4 \times 22 + 1\).

\(7^{89} = (7^4)^{22} \cdot 7^1 \equiv 1^{22} \cdot 7 \equiv 7 \pmod{15}\).
Quick Tip: Find the smallest power \(k\) such that \(a^k \equiv 1 \pmod n\). Then \(a^m \equiv a^{m \pmod k} \pmod n\).

Rewrite the plane equation: \(\alpha x - y - \beta z + \gamma = 0\). Normal \(\vec{n} = (\alpha, -1, -\beta)\).


It is perpendicular to normals \(\vec{n}_1 = (2, 1, 1)\) and \(\vec{n}_2 = (3, -2, 2)\).

\(\vec{n}\) is parallel to \(\vec{n}_1 \times \vec{n}_2\).

\(\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} i & j & k
2 & 1 & 1
3 & -2 & 2 \end{vmatrix} = i(2 - (-2)) - j(4 - 3) + k(-4 - 3) = (4, -1, -7)\).


Compare \((\alpha, -1, -\beta)\) with \((4, -1, -7)\). Since y-component is \(-1\) in both:

\(\alpha = 4\) and \(-\beta = -7 \implies \beta = 7\).


Plane is \(4x - y - 7z + \gamma = 0\).


Passes through \((1, -1, 3)\): \(4(1) - (-1) - 7(3) + \gamma = 0\).

\(4 + 1 - 21 + \gamma = 0 \implies -16 + \gamma = 0 \implies \gamma = 16\).


Sum \(\alpha + \beta + \gamma = 4 + 7 + 16 = 27\).
Quick Tip: The cross product of the normal vectors of two planes gives the direction vector of their line of intersection, which is also the normal vector to any plane perpendicular to both.

From \(\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}\), we have \((\vec{r} - (\vec{b} + \vec{c})) \times \vec{a} = \vec{0}\).


This implies \(\vec{r} - (\vec{b} + \vec{c}) = \lambda \vec{a}\) for some scalar \(\lambda\).

\(\vec{r} = \vec{b} + \vec{c} + \lambda \vec{a}\).


Given \(\vec{r} \cdot (\vec{b} - \vec{c}) = 0\), substitute \(\vec{r}\):

\((\vec{b} + \vec{c}) \cdot (\vec{b} - \vec{c}) + \lambda \vec{a} \cdot (\vec{b} - \vec{c}) = 0\).

\(|\vec{b}|^2 - |\vec{c}|^2 + \lambda (\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}) = 0\).


Calculate values: \(|\vec{b}|^2 = 1+1+4=6\), \(|\vec{c}|^2 = 4+1+16=21\).

\(\vec{a} \cdot \vec{b} = 1 - 2 + 6 = 5\).

\(\vec{a} \cdot \vec{c} = 2 + 2 - 12 = -8\).

\(6 - 21 + \lambda (5 - (-8)) = 0 \implies -15 + 13\lambda = 0 \implies \lambda = \frac{15}{13}\).


Now \(\vec{r} = \vec{b} + \vec{c} + \frac{15}{13}\vec{a}\).


We need \(\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k})\). Let \(\vec{d} = \hat{i} + \hat{j} - \hat{k}\).

\(\vec{b} + \vec{c} = 3\hat{i} + 0\hat{j} - 2\hat{k}\).

\((\vec{b} + \vec{c}) \cdot \vec{d} = 3(1) + 0(1) - 2(-1) = 3 + 2 = 5\).

\(\vec{a} \cdot \vec{d} = 1(1) + 2(1) + 3(-1) = 1 + 2 - 3 = 0\).


So, \(\vec{r} \cdot \vec{d} = 5 + \frac{15}{13}(0) = 5\).
Quick Tip: If \(\vec{x} \times \vec{a} = \vec{y} \times \vec{a}\), then \(\vec{x} = \vec{y} + \lambda \vec{a}\). Use the scalar product condition to find \(\lambda\).

The equation is \(\frac{x^{119} - 1}{x - 1} = 0\), so roots are \(x = e^{i \frac{2k\pi}{119}}\) for \(k = 1, 2, ..., 118\).


Total number of roots \(n(S) = 118\).


We want roots satisfying \(x^7 = 1\). These are of the form \(e^{i \frac{2m\pi}{7}}\).


Equating arguments: \(\frac{2k\pi}{119} = \frac{2m\pi}{7} \implies \frac{k}{119} = \frac{m}{7} \implies k = 17m\).


Since \(1 \le k \le 118\), we have \(1 \le 17m \le 118\).

\(m\) can take integer values \(1, 2, 3, 4, 5, 6\) (since \(17 \times 6 = 102\) and \(17 \times 7 = 119 > 118\)).


So there are 6 favorable roots.


Probability = \(\frac{6}{118} = \frac{3}{59}\).
Quick Tip: Roots of \(x^n=1\) are common with \(x^m=1\) if the powers of the complex exponentials match. This reduces to finding common multiples of the angle denominators.

Mean \(np = 3\). Variance \(npq = 2\).


Dividing gives \(q = \frac{2}{3}\). So \(p = 1 - q = \frac{1}{3}\).


Then \(n(\frac{1}{3}) = 3 \implies n = 9\).


We need \(P(X < \frac{9}{4}) = P(X < 2.25)\).


Since X is integer-valued, this is \(P(X=0) + P(X=1) + P(X=2)\).


Formula: \(P(X=k) = \binom{n}{k} p^k q^{n-k}\).

\(P(0) = \binom{9}{0} (\frac{1}{3})^0 (\frac{2}{3})^9 = \frac{2^9}{3^9}\).

\(P(1) = \binom{9}{1} (\frac{1}{3})^1 (\frac{2}{3})^8 = \frac{9 \cdot 2^8}{3^9}\).

\(P(2) = \binom{9}{2} (\frac{1}{3})^2 (\frac{2}{3})^7 = \frac{36 \cdot 2^7}{3^9}\).


Total = \(\frac{1}{3^9} (2^9 + 9 \cdot 2^8 + 36 \cdot 2^7)\).


Factor out \(2^7\): \(\frac{2^7}{3^9} (2^2 + 18 + 36) = \frac{2^7}{3^9} (4 + 18 + 36) = \frac{2^7 \cdot 58}{3^9}\).

\(58 = 29 \times 2\). So result is \(\frac{29 \times 2^8}{3^9}\).
Quick Tip: For Binomial distribution, Mean = \(np\) and Variance = \(np(1-p)\). Always find \(n\) and \(p\) first.

For \(\cos^{-1}(t)\), we need \(-1 \le t \le 1\).


Let \(t = \frac{x^2 - 3x + 2}{x^2 + 2x - 1}\).


Inequality 1: \(t \le 1 \implies \frac{x^2 - 3x + 2 - (x^2 + 2x - 1)}{x^2 + 2x - 1} \le 0 \implies \frac{-5x + 3}{x^2 + 2x - 1} \le 0\).


Roots of denominator: \(x = -1 \pm \sqrt{2}\). Root of numerator: \(x = \frac{3}{5}\).


Sign analysis for \(\frac{-(5x-3)}{(x - (-1-\sqrt{2}))(x - (-1+\sqrt{2}))} \le 0\):


Regions: \((-\infty, -1-\sqrt{2})\), \((-1-\sqrt{2}, -1+\sqrt{2})\), \((-1+\sqrt{2}, 3/5)\), \([3/5, \infty)\).


Valid regions for \(t \le 1\): \((-1-\sqrt{2}, -1+\sqrt{2}) \cup [3/5, \infty)\).


Inequality 2: \(t \ge -1 \implies \frac{x^2 - 3x + 2 + x^2 + 2x - 1}{x^2 + 2x - 1} \ge 0 \implies \frac{2x^2 - x + 1}{x^2 + 2x - 1} \ge 0\).


Numerator discriminant \(< 0\), so numerator is always positive.


We need denominator \(> 0\): \(x \in (-\infty, -1-\sqrt{2}) \cup (-1+\sqrt{2}, \infty)\).


Intersection of valid regions from both inequalities:

\(((-1-\sqrt{2}, -1+\sqrt{2}) \cup [3/5, \infty)) \cap ((-\infty, -1-\sqrt{2}) \cup (-1+\sqrt{2}, \infty))\).


The first part \((-1-\sqrt{2}, -1+\sqrt{2})\) has no overlap with the second set.


The second part \([3/5, \infty)\) is fully contained in \((-1+\sqrt{2}, \infty)\) because \(0.6 > 0.414\).


Thus, the domain is \([3/5, \infty)\).
Quick Tip: Solve \(-1 \le \frac{P(x)}{Q(x)} \le 1\) by splitting into two inequalities and finding the intersection of their solution sets.

Let's analyze option (C): \(S = ((p \land q) \land (\sim q)) \implies p\).


LHS: \((p \land q) \land (\sim q) \equiv p \land (q \land \sim q) \equiv p \land F \equiv F\).


Statement: \(F \implies p\).


In logic, a statement with a false hypothesis is always true.


Therefore, statement (C) is always true (a tautology).
Quick Tip: Check for contradictions in the antecedent (LHS) of an implication. If the LHS is always false, the implication is a tautology.

The region is bounded by \(y = 3 - |x|\) (top) and \(y = |x - 1|\) (bottom).


Intersection points: \(3-|x| = |x-1|\).

\(x \ge 0: 3-x = |x-1|\). If \(x \ge 1\), \(3-x=x-1 \implies x=2\). If \(0 \le x < 1\), \(3-x=1-x\) (False).

\(x < 0: 3+x = 1-x \implies 2x = -2 \implies x = -1\).


Region interval: \(x \in [-1, 2]\). Total Area A.


Split into integrals:


1. \([-1, 0]\): Top \(3+x\), Bottom \(1-x\). Area = \(\int_{-1}^0 (2+2x)dx = 1\).


2. \([0, 1]\): Top \(3-x\), Bottom \(1-x\). Area = \(\int_0^1 2dx = 2\).


3. \([1, 2]\): Top \(3-x\), Bottom \(x-1\). Area = \(\int_1^2 (4-2x)dx = 1\).


Total Area = \(1 + 2 + 1 = 4\).


Parabola \(y = x^2+1\).


In \([0, 1]\), \(x^2+1\) lies between \(1-x\) and \(3-x\). It divides the rectangular region of area 2.


Area below parabola in region = Area \([-1, 0]\) (none, parabola is below) + Area \([0, 1]\) (parabola is upper bound) + Area \([1, 2]\) (full, parabola is above).


Wait, in \([0, 1]\), parabola is above bottom (\(1-x\)). Area segment = \(\int_0^1 ((x^2+1) - (1-x)) dx = \int_0^1 (x^2+x) dx = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}\).


Area above parabola in region = Total - Area Below.


Area "below" curve wrt region means intersection of region with \(y < x^2+1\).


For \(x \in [-1, 0]\), \(x^2+1 < 1-x\) is False (\(x^2+x < 0\) in \((-1, 0)\)). So parabola is below region. Area intersection is 0.


For \(x \in [1, 2]\), \(x^2+1 > 3-x\) is True. So parabola is above region. Area intersection is Full = 1.


Total Area below curve = \(0 + \frac{5}{6} + 1 = \frac{11}{6}\).


Total Area above curve = \(4 - \frac{11}{6} = \frac{13}{6}\).


Ratio \(11:13\). Coprime. \(m=11, n=13\).

\(m+n = 11+13 = 24\).
Quick Tip: Carefully check the relative position of the dividing curve with respect to the boundary curves in each sub-interval to determine which part of the area is "under" or "above".

12 students. \(x_i \ge 1\).


Exactly 2 students get \(\ge 2\) pens. The other 10 must get exactly 1 pen.


Step 1: Choose the 2 students. \(\binom{12}{2} = 66\).


Step 2: Distribute pens. Let the chosen students have \(x_1, x_2\).

\(x_1 \ge 2, x_2 \ge 2\). Other 10 have \(x_i = 1\).


Total pens = 30.

\(x_1 + x_2 + 10(1) = 30 \implies x_1 + x_2 = 20\).


Let \(x_1 = 2 + a, x_2 = 2 + b\) where \(a, b \ge 0\).

\(2+a + 2+b = 20 \implies a+b = 16\).


Number of non-negative solutions = \(\binom{16+2-1}{2-1} = \binom{17}{1} = 17\).


Total ways = \(66 \times 17 = 1122\).
Quick Tip: "Exactly k students satisfy condition P" implies the other n-k students must NOT satisfy P. This fixes their allocation often to a single value.

Let \(P(x) = (1 + x^2 - x^4)^{12}\).


Sum of all even coefficients \(S = a_0 + a_2 + ... + a_{48} = \frac{P(1) + P(-1)}{2}\).

\(P(1) = (1+1-1)^{12} = 1\). \(P(-1) = 1\). \(S = 1\).


We need \(S' = S - a_{46} - a_{48}\).

\(a_{48}\) is coeff of \(x^{48}\). Highest power term \((-x^4)^{12} = x^{48}\). Coeff is \(\binom{12}{12}(-1)^{12} = 1\).

\(a_{46}\) is coeff of \(x^{46}\). Term \(x^{46}\) comes from \(\frac{12!}{k!m!p!} (1)^k (x^2)^m (-x^4)^p\).

\(2m + 4p = 46 \implies m + 2p = 23\). Also \(k+m+p=12\).


Substitute \(m = 23-2p\). \(k + 23 - 2p + p = 12 \implies p - k = 11\).


Since \(p \le 12\), \(k \ge 0\), only solution is \(p=11, k=0, m=1\).


Coeff \(a_{46} = \frac{12!}{0!1!11!} (1)^0 (1)^1 (-1)^{11} = 12 \times (-1) = -12\).


Required Sum \(= 1 - (-12) - 1 = 12\).
Quick Tip: Use \(f(1) + f(-1)\) to find the sum of even coefficients. Don't forget to subtract the high-order terms excluded from the truncated sum.

Differences: \(3, 6, 9, 12...\) are in AP.

\(T_n\) is quadratic \(an^2 + bn + c\).

\(2a = 3 \implies a = 3/2\).

\(3a + b = 3 \implies 9/2 + b = 3 \implies b = -3/2\).

\(a + b + c = 7 \implies 3/2 - 3/2 + c = 7 \implies c = 7\).

\(T_n = \frac{3}{2}(n^2 - n) + 7\).

\(S_{30} - S_{20} = \sum_{n=21}^{30} T_n\).

\(= \frac{3}{2} [ (\sum_{1}^{30} n^2 - \sum_{1}^{20} n^2) - (\sum_{1}^{30} n - \sum_{1}^{20} n) ] + 7(10)\).

\(\sum_{21}^{30} n = 255\). \(\sum_{21}^{30} n^2 = 9455 - 2870 = 6585\).


Sum \(= \frac{3}{2} [ 6585 - 255 ] + 70 = \frac{3}{2} [ 6330 ] + 70 = 9495 + 70 = 9565\).
Quick Tip: When differences of terms are in AP, the general term is a quadratic \(an^2+bn+c\). \(2a\) is the common difference of the differences.

\(f(x)\) is sum of three terms. Non-differentiability comes from discontinuity or corners.


Term 1: \(h(x) = [x + x^3]\). Let \(g(x) = x + x^3\). \(g'(x) > 0\), strictly increasing.


In \((-10, 10)\), \(g(x)\) ranges from \(-1010\) to \(1010\).

\(h(x)\) is discontinuous (and non-diff) whenever \(g(x)\) is an integer.


Number of integers strictly between limits: \(1009 - (-1009) + 1 = 2019\).


Term 2: \(|x - x^3|\) has corners at \(x = 0, 1, -1\). At these points \(g(x) \in \mathbb{Z}\), so they are already counted.


Term 3: \(|x + 0.5|\) has corner at \(x = -0.5\).

\(g(-0.5) = -0.5 - 0.125 = -0.625 \notin \mathbb{Z}\). This point is new.


Total points = \(2019 + 1 = 2020\).
Quick Tip: For \(f(x) = [g(x)]\), if \(g(x)\) is monotonic, points of non-differentiability correspond to integer values of \(g(x)\). Check overlap with modulus corners.

Let \(x = \frac{1}{t}\). Integral reduces to \(\int \frac{-du}{5u^2 - 29}\) where \(u = \sqrt{7t^2+10}\).


Result is \(\frac{1}{2\sqrt{29 \cdot 5/25}} \ln \left| \frac{\sqrt{29} - \sqrt{5}u}{\sqrt{29} + \sqrt{5}u} \right|\).


Matches coeff \(-\frac{1}{\sqrt{580}}\).

\(f(x) = \frac{\sqrt{29} - \sqrt{5}u}{\sqrt{29} + \sqrt{5}u}\) where \(u = \frac{\sqrt{10x^2+7}}{x}\).


As \(x \to 0\), \(u \to \infty\).

\(\lim_{u \to \infty} \frac{\sqrt{29} - \sqrt{5}u}{\sqrt{29} + \sqrt{5}u} = -1\).
Quick Tip: For integrals of type \(\frac{1}{(ax^2+b)\sqrt{cx^2+d}}\), substitute \(x = 1/t\).

Center \((1, 2)\). \(2ae = 6 \implies ae = 3\). Given \(e = \sqrt{3}\), so \(a = \sqrt{3}\).

\(b^2 = a^2(e^2-1) = 3(2) = 6\).


Equation: \(\frac{(y-2)^2}{3} - \frac{(x-1)^2}{6} = 1\).

\(2(y^2 - 4y + 4) - (x^2 - 2x + 1) = 6\).

\(-x^2 + 2y^2 + 2x - 8y + 1 = 0\).

\(x^2 - 2y^2 - 2x + 8y - 1 = 0\).

\(a = -2, b = 8, c = -1\).

\(|a+b+c| = |-2+8-1| = 5\).
Quick Tip: Identify the center and axis orientation from foci coordinates. \(b^2 = a^2(e^2-1)\) relates the parameters.

Distance formula: \(\frac{|6 + 4 + 3\alpha + 10|}{\sqrt{10+\alpha^2}} = \sqrt{35}\).

\((3\alpha+20)^2 = 35(10+\alpha^2) \implies 13\alpha^2 - 60\alpha - 25 = 0\).


Roots are \(5, -5/13\).


Vertices: \(A(5, -5/13), B(-5/13, 5), C(164/13, 5)\).


Base BC is on \(y=5\). Length \(164/13 - (-5/13) = 13\).


Height is \(|y_A - 5| = |-5/13 - 5| = 70/13\).


Area = \(\frac{1}{2} \times 13 \times \frac{70}{13} = 35\).
Quick Tip: If two vertices share a coordinate (like y=5), use that side as the base for a simple \(\frac{1}{2} \times base \times height\) calculation.

\(\vec{N} = \vec{OA} \times \vec{OB} = -56\hat{i} - 34\hat{j} + 14\hat{k}\).


Volume \(V = |\vec{u} \cdot \vec{N}|\). \(\vec{u} = x\hat{i} + y\hat{j}\) with \(x^2+y^2=1\).


Max value is magnitude of projection of \(\vec{N}\) on xy-plane.

\(V_{max} = \sqrt{(-56)^2 + (-34)^2} = \sqrt{3136 + 1156} = \sqrt{4292}\).

\(4292 = 4 \times 1073\). So \(V_{max} = 2\sqrt{1073}\).

\(\alpha = 1073\).
Quick Tip: Maximum value of \(\vec{u} \cdot \vec{v}\) where \(\vec{u}\) is a unit vector is \(|\vec{v}|\). If \(\vec{u}\) is restricted to a plane, project \(\vec{v}\) onto that plane first.

Put \(x+y=u\). Equation reduces to \(2(y-x) = \ln \frac{(x+y)^2-2}{C}\).


Using \((\sqrt{2}, \sqrt{2})\), constant determined.

\(2(y-x) - \ln((x+y)^2-2) = -\ln 6\).


Substitute \((2, \alpha)\): \(2(\alpha-2) - \ln((\alpha+2)^2-2) = -\ln 6\).

\(2\alpha - 4 = \ln \frac{\alpha^2+4\alpha+2}{6}\).

\(2\alpha - \ln \dots = 4\).
Quick Tip: For \(dy/dx = f(x+y)\), substitute \(u = x+y\) to separate variables.

Adobe is a building material made from earth and organic materials.


It essentially refers to sun-dried mud bricks, which are one of the oldest building materials used by humans.


The mixture typically contains sand, clay, water, and some kind of fibrous or organic material (sticks, straw, and/or manure), which the builders shape into bricks using frames and dry in the sun.


Therefore, "Type of Brick" is the correct classification.
Quick Tip: Associate "Adobe" with "sun-dried mud brick". It is distinct from kiln-fired bricks used in modern construction. Adobe architecture is famously associated with Pueblo style buildings in the American Southwest and arid regions globally.

Analyze Statement I: Chandigarh was indeed the first planned city in India after independence in 1947, built to serve as the capital of Punjab (and later Haryana). This statement is factually correct.


Analyze Statement II: The master plan of the city was prepared by Le Corbusier, a Swiss-French architect. He designed the Capitol Complex and established the city's grid system. This statement is factually correct.


Since both statements are true, option (A) is the correct answer.
Quick Tip: While Albert Mayer and Maciej Nowicki were the initial planners, Le Corbusier is the principal architect credited with the final realized plan of Chandigarh. It is known for its sector-based grid planning.

Analyze Statement I: Glass is an amorphous solid that acts as a thermal insulator. Compared to metals (conductors), it has very low thermal conductivity. This statement is correct.


Analyze Statement II: Interaction of light with glass includes transmission (passing through, like in windows), refraction (bending, like in lenses), and absorption (taking in energy, like in tinted glass). This statement is correct.


Thus, both statements describe valid physical properties of glass.
Quick Tip: Properties of building materials are frequent exam topics. Remember that while glass insulates against conductive heat transfer, ordinary glass is transparent to solar radiation (radiative heat transfer), which creates the greenhouse effect.

A. PMUY stands for Pradhan Mantri Ujjwala Yojana. Matches with III.


B. PMAY stands for Pradhan Mantri Awas Yojana. Matches with IV.


C. PMKVY stands for Pradhan Mantri Kaushal Vikas Yojana. Matches with I.


D. PMJDY stands for Pradhan Mantri Jan Dhan Yojana. Matches with II.


The correct sequence is A-III, B-IV, C-I, D-II.
Quick Tip: Government acronyms often contain the key Hindi word defining the scheme's purpose: 'Awas' (Housing), 'Ujjwala' (Light/LPG), 'Kaushal' (Skill), 'Jan Dhan' (Public Wealth/Banking).

B. Louis I Kahn is the world-renowned architect of IIM Ahmedabad. (B matches I).


D. Achyut Kanvinde designed the campus of IIT Kanpur. (D matches III).


C. B.V. Doshi is the architect of IIM Bengaluru. (C matches IV).


A. C.P. Kukreja designed the Jawaharlal Nehru University (JNU) campus in New Delhi. (A matches II).


The sequence is A-II, B-I, C-IV, D-III.
Quick Tip: Louis Kahn = IIM Ahmedabad (Brick arches). B.V. Doshi = IIM Bangalore (Stone pergolas/greenery). Achyut Kanvinde = IIT Kanpur (Functionalism). Charles Correa = IIT Bombay/Kanchanjunga.

Complementary colors are pairs of colors which cancel each other out (produce grayscale) when combined or are opposite each other on the color wheel.


Standard complementary pairs are:


1. Red and Green (Opposite).


2. Blue and Orange (Opposite).


3. Yellow and Violet (Opposite).


Blue and Green are adjacent (analogous) colors on the wheel, not complementary.


Therefore, Blue-Green is the correct answer.
Quick Tip: Remember the primary color pairs: Red-Green, Blue-Orange, Yellow-Violet. Any pair consisting of adjacent colors (like Blue-Green) is analogous, not complementary.

Robber's Cave, locally known as Guchhupani, is a river cave formation located in Dehradun.


Dehradun is the capital city of the state of Uttarakhand.


Hence, it is situated in Uttarakhand.
Quick Tip: Robber's Cave is a popular tourist spot in Dehradun where a river flows through a narrow cave-like formation.

The Petronas Twin Towers are iconic skyscrapers.


They are located in Kuala Lumpur, the capital city of Malaysia.


Designed by Cesar Pelli, they were the tallest buildings in the world from 1998 to 2004.
Quick Tip: Iconic Skyscrapers: Petronas (Kuala Lumpur), Burj Khalifa (Dubai), Empire State (New York), Taipei 101 (Taipei), The Shard (London).

The Konark Sun Temple is a 13th-century CE Sun temple.


It is located at Konark about 35 kilometres northeast from Puri on the coastline of Odisha, India.


It is a UNESCO World Heritage Site attributed to King Narasimhadeva I.
Quick Tip: Konark Sun Temple is famous for its Kalinga architecture and the stone chariot wheels which act as sundials.

The Lotus Temple, located in New Delhi, India, is a Baha'i House of Worship.


It was designed by the Iranian-Canadian architect Fariborz Sahba.


Completed in 1986, it is notable for its flower-like shape.
Quick Tip: The structure is expressionist architecture composed of 27 free-standing marble-clad "petals" arranged in clusters of three to form nine sides.

A "Dumbwaiter" is a small freight elevator or lift intended to carry objects rather than people.


Historically, they were used to move food and dishes between a kitchen (often in the basement) and dining rooms on upper floors.


Since they carried food (orders) but could not speak, the term "dumb" (silent) "waiter" was coined.


In modern construction, they are used in hospitals, restaurants, and libraries to transport small loads vertically.
Quick Tip: Remember the origin: It "waits" on you (serves food) but is "dumb" (silent/mechanical). It is strictly for goods, not passengers.

Cuttack and Bhubaneswar are the two major cities of Odisha located very close to each other (approx. 27 km apart).


Cuttack is the former capital and judicial capital (High Court), while Bhubaneswar is the current administrative capital.


Due to their proximity and integrated urban spread, they are collectively known as the "Twin Cities" of Odisha.
Quick Tip: Twin cities usually share geographic proximity and grow into each other over time. Famous Indian examples: Hyderabad-Secunderabad, Cuttack-Bhubaneswar, Hubli-Dharwad.

The approximate timelines for the emergence of major ancient river valley civilizations are:


1. Mesopotamia (Sumerian): c. 4000-3500 BCE (Tigris-Euphrates valley). Oldest known civilization.


2. Egyptian: c. 3100 BCE (Nile valley). Unification of Upper and Lower Egypt.


3. Harappan (Indus Valley): c. 2600 BCE (Mature phase). Early roots go back further, but urbanization peaked here.


4. Chinese (Shang Dynasty): c. 1600 BCE (Yellow River valley). Earliest confirmed dynasty with writing.


Ordering these gives: Mesopotamia \(\to\) Egyptian \(\to\) Harappa \(\to\) Chinese.
Quick Tip: Remember the order of river valleys: Tigris/Euphrates (First) \(\to\) Nile \(\to\) Indus \(\to\) Yellow River. Sumer (Mesopotamia) is universally acknowledged as the "Cradle of Civilization".

The construction of Qutub Minar was initiated by Qutb ud-Din Aibak, the founder of the Delhi Sultanate, around 1192 AD.


He only constructed the basement (first storey).


His successor, Iltutmish, added three more storeys.


The final storey was later added/repaired by Firoz Shah Tughlaq.


However, the question asks who "built" it (implying the founder/initiator), which is Qutb ud-Din Aibak.
Quick Tip: The name of the minaret itself "Qutub" Minar corresponds to its builder "Qutub" ud-din Aibak.

In the Southern Hemisphere, the sun path is predominantly in the Northern sky.


The sun rises in the East, travels through the North, and sets in the West.


Therefore, direct sunlight enters mainly from the North, East, and West facing windows.


South-facing windows receive indirect, reflected skylight which is consistent, glare-free, and diffused.


(Note: In the Northern Hemisphere, it is the North direction that provides glare-free light).
Quick Tip: Solar Orientation Rule: The "pole-facing" side of a building receives diffused light. In the Southern Hemisphere, the South Pole is the reference, so South is the diffused side.

The Jawahar Kala Kendra (referred to as Jawaharlal Kala Complex in the question) in Jaipur is a famous multi-arts center.


It was designed by the renowned Indian architect Charles Correa in 1986 and completed in 1991.


The design is based on the 'Vastu Purusha Mandala' of nine squares, reflecting the plan of the city of Jaipur itself (which was based on the Navagraha mandala).
Quick Tip: Charles Correa is famous for integrating traditional concepts (like Vastu and open-to-sky spaces) with modern architecture. JKK is his masterpiece referencing the 9-square mandala.

Dhajji-Dewari is a traditional timber-frame construction method.


"Dhajji" means patchwork quilt and "Dewari" means wall.


It consists of a braced timber frame filled with stone or brick masonry set in mud mortar.


This style is indigenous to the Western Himalayas, particularly Kashmir and parts of Himachal Pradesh (Mountainous Regions).


It is highly earthquake-resistant due to its flexibility.
Quick Tip: Dhajji-Dewari and Kath-Kuni are distinct vernacular styles of the Himalayas designed to withstand earthquakes.

In mountainous regions, roofing materials need to be locally available and capable of splitting into thin, flat sheets.


Slate, which is a metamorphic rock derived from Shale, is the most common roofing stone.


Shale (sedimentary) has a property called fissility, allowing it to split into thin layers. While Slate is the specific roofing term, technically it originates from Shale/Mudstone.


Among the options provided, Shale/Slate is the correct category for roofing tiles (often called "pathal" in local dialects). Marble, Granite, and Sandstone are generally too heavy and hard to split into thin roofing tiles compared to slate/shale.
Quick Tip: Slate (metamorphosed shale) is the standard natural stone for roofing due to its ability to be split into thin, waterproof sheets (cleavage).

Chandigarh was planned by Le Corbusier based on a rectangular grid system.


The city is divided into numbered sectors (each approximately \(800m \times 1200m\)).


The roads form a hierarchy (The 7Vs) that intersect at right angles, creating a classic Grid-iron pattern.


This contrasts with Delhi (Radio-centric) or old medieval cities (Organic).
Quick Tip: Grid-iron planning involves streets running at right angles to each other, forming a grid. It is the hallmark of planned cities like Chandigarh, Jaipur, and Manhattan.

Goa is located on the western coast of India (Konkan coast).


The Arabian Sea lies to the West of Goa.


Therefore, to view the sea from a building in Goa, the rooms must face West.
Quick Tip: Visualize the map of India. The West Coast (Mumbai, Goa, Kerala) faces the Arabian Sea to the West. The East Coast (Chennai, Odisha) faces the Bay of Bengal to the East.

The National Institute of Fashion Technology (NIFT) in New Delhi is a prominent architectural project.


It was designed by the firm Stein, Doshi & Bhalla.


The design is widely attributed to the Pritzker Prize-winning architect B.V. Doshi, who incorporated elements like the sunken courtyard and traditional Indian architectural concepts into the modern institutional form.
Quick Tip: B.V. Doshi is India's first Pritzker Prize laureate. His works, like NIFT Delhi and IIM Bangalore, often feature fusion of modernism with traditional Indian space-making elements like kunds (steps) and jaalis.

Image A shows a massive domed structure with minarets. This is the **Hagia Sophia** in Istanbul. (Matches II).


Image B shows a classic Art Deco skyscraper with a needle spire. This is the **Empire State Building** in New York. (Matches I).


Image C shows an ancient Roman amphitheater with arches. This is the **Colosseum** in Rome. (Matches IV).


Image D shows a structure with white sail-like shells. This is the **Sydney Opera House**. (Matches III).


The correct sequence is A-II, B-I, C-IV, D-III.
Quick Tip: Visual recognition of world heritage sites is crucial. Key features: Hagia Sophia (Dome + Minarets), Empire State (Spire + Setbacks), Colosseum (Arched Rings), Sydney Opera House (Shells).

Logo A consists of three overlapping ellipses. This is the logo of **Toyota** (IV).


Logo B is a stylized 'T' that looks like a cross-section of an electric motor. This is the logo of **Tesla** (I).


Logo C is a yellow shield with a prancing black horse and letters 'S F'. This is the logo of **Ferrari** (II).


Logo D is a crest with black and red stripes and antlers, featuring a horse in the center and text "STUTTGART". This is the logo of **Porsche** (III).


The correct sequence is A-IV, B-I, C-II, D-III.
Quick Tip: Automobile logos are frequent visual questions. Distinguish between the Prancing Horse of Ferrari (on yellow background) and the Crest of Porsche (shield with stripes/antlers).

Statement I refers to the traditional RYB (Red, Yellow, Blue) color model used in art and design education. In this model, these three are indeed the primary colors from which others are mixed. This is considered correct in the context of design theory.


Statement II defines complementary colors. By definition, colors directly opposite each other on the color wheel (e.g., Red and Green) are called complementary colors. This is correct.


Therefore, both statements are correct.
Quick Tip: In design exams, assume the RYB (Subtractive/Painter's) color model unless CMYK (Print) or RGB (Light) is specified. Primary: Red, Yellow, Blue. Secondary: Orange, Green, Violet.

Analyze the sequence of the outer shapes:

Figure 1: Square.

Figure 2: Diamond (Rotated Square).

Figure 3: Square.

Figure 4: Should be a Diamond (to maintain the alternating pattern).


Analyze the sequence of the inner lines:

Figure 1: Diagonal line (\(\diagup\)).

Figure 2: Vertical line (\(|\)).

Figure 3: Diagonal line (\(\diagdown\)).

The lines appear to be rotating 45 degrees counter-clockwise (or following a symmetry).

Sequence: \(45^\circ \to 90^\circ \to 135^\circ \to\) Next should be \(180^\circ\) (Horizontal).


Alternatively, look at the pairs:

Fig 1 (Square, \(\diagup\)) and Fig 3 (Square, \(\diagdown\)) are opposites.

Fig 2 (Diamond, \(|\)) and Fig 4 should be opposites. The opposite of Vertical is Horizontal.


Thus, the missing figure is a Diamond with a horizontal line.
Quick Tip: In pattern series, separate the elements (outer shape, inner shape, orientation) and analyze the progression of each element independently.

The figure is a large triangle divided into smaller triangles with 4 layers (n=4).


We can use the formula for the total number of triangles in such a grid:
\(N = \lfloor \frac{n(n+2)(2n+1)}{8} \rfloor\)


Substitute \(n=4\):
\(N = \lfloor \frac{4(4+2)(2\times4 + 1)}{8} \rfloor\)
\(N = \lfloor \frac{4 \times 6 \times 9}{8} \rfloor\)
\(N = \lfloor \frac{216}{8} \rfloor\)
\(N = 27\)


Alternatively, counting manually:

1-unit triangles: \(1+3+5+7 = 16\).

Inverted 1-unit triangles: \(1+2+3 = 6\).

2-unit upright triangles: \(1+2+3 = 6\).

2-unit inverted triangles: \(1\).

3-unit upright triangles: \(1+2 = 3\).

4-unit upright triangle: \(1\).

Total: The count using the formula is robust. Let's re-verify the manual breakdown, usually, simple formula is best. The formula yields 27.
Quick Tip: Memorize the number of triangles for standard n-layer triangles: n=3 gives 13, n=4 gives 27, n=5 gives 48. Formula: \(\frac{n(n+2)(2n+1)}{8}\) (drop decimal if any).

Analyze the given numbers: 7, 9, 25, 32, 43, 59.


Let's check for "Even vs Odd":

7 is Odd.

9 is Odd.

25 is Odd.

32 is Even.

43 is Odd.

59 is Odd.


Since 32 is the only even number in the list, it is the odd one out.


Secondary check (Perfect squares): 9 and 25 are squares, others are not. This doesn't isolate a single number.

Secondary check (Primes): 7, 43, 59 are primes. 9, 25, 32 are composite. This splits the group in half.


The most distinct singular property is being the only even number.
Quick Tip: In number classification ("odd one out") questions, the order of priority for checking is usually: 1. Prime/Composite (if one unique), 2. Odd/Even, 3. Perfect Squares/Cubes. Here, Even/Odd provides a unique answer.

Total number of floors = 15.


Height of Ground Floor = 4.2 meters.


Number of remaining floors = \(15 - 1 = 14\) floors.


Height of each remaining floor = 3.3 meters.


Total height of remaining floors = \(14 \times 3.3 = 46.2\) meters.


Total building height = Height of Ground Floor + Height of remaining floors.


Total Height = \(4.2 + 46.2 = 50.4\) meters.
Quick Tip: Read carefully to see if the ground floor is included in the total count of "15 floors" or is additional. "A building has 15 floors" usually implies G + 14 upper floors.

A mirror image reverses the lateral orientation of the object.


1. The order of letters reverses: Left-to-Right becomes Right-to-Left. The 'S' at the end of "SUCCESS" becomes the first letter on the left of the image.


2. Each individual letter is laterally inverted (flipped).

- 'S' becomes a backward 'S'.

- 'U' remains 'U' (symmetric).

- 'C' becomes a backward 'C'.

- 'E' becomes a backward 'E'.


Option (A) shows the letters in reverse order (\(S, S, E, C, C, U, S\)) and each letter is laterally inverted correctly.
Quick Tip: In mirror images, the object on the extreme right of the original appears on the extreme left of the reflection. Symmetrical letters (A, H, I, M, O, T, U, V, W, X, Y) do not change shape.

Scale = 1 : 1000. This means 1 unit on the map equals 1000 units on the ground.


Map distance = 7 cm.


Original distance = \(7 cm \times 1000 = 7000 cm\).


Convert to meters (since \(100 cm = 1 m\)):
\(\frac{7000}{100} = 70 meters\).


Check other units:
\(70 m = 0.07 km\) (Option A is 0.7 km, incorrect).
\(70 m = 70,000 mm\) (Option B is 7000 mm, incorrect).


Therefore, 70 meter is the correct answer.
Quick Tip: Scale factor calculation: Actual Distance = Map Distance \(\times\) Scale Factor. Always check the units in the options carefully (\(1 m = 100 cm\), \(1 km = 1000 m\)).

Scale = 1 : 100.


Actual Length = 80 meters.


Actual Width = 40 meters.


Convert to centimeters: Length = 8000 cm, Width = 4000 cm.


Drawing Length = \(\frac{8000}{100} = 80\) cm.


Drawing Width = \(\frac{4000}{100} = 40\) cm.


There is no option for 80 cm \(\times\) 40 cm. However, Option (C) is 8 cm \(\times\) 4 cm.
Quick Tip: Always convert actual dimensions to the drawing units (cm) before dividing by the scale factor. Be alert for decimal point shifts in options.

The pattern is \(\times 2 \pm 1\):

\(16 \times 2 + 1 = 33\)

\(33 \times 2 - 1 = 65\)

\(65 \times 2 + 1 = 131\)

\(131 \times 2 - 1 = 261\)


Next term: \(261 \times 2 + 1 = 522 + 1 = 523\).
Quick Tip: Check for mixed arithmetic series where multiplication is constant but the addition/subtraction alternates.

Shift analysis:


R (+2) \(\to\) T


O (+2) \(\to\) Q


M (+2) \(\to\) O


A (+2) \(\to\) C


N (+2) \(\to\) P


Apply (+2) shift to ITALY:


I (+2) \(\to\) K


T (+2) \(\to\) V


A (+2) \(\to\) C


L (+2) \(\to\) N


Y (+2) \(\to\) A (wraps around Z)


Result: KVCNA.
Quick Tip: Write the alphabet in a circle to handle wrap-arounds like Y \(\to\) A easily.

The matrix is formed by combining outer shapes with inner symbols.


The top strip defines the inner symbols: \([-]\), \([X]\), \([*]\), \([O]\).


Row 1 (Square): Inner symbols follow the sequence Square, X, *.


Row 2 (Circle): Inner symbols follow the sequence +, +, +.


Row 3 (Triangle): Inner symbols follow the sequence Triangle, Triangle, ?.


We need to complete the pattern.


Look at the column logic:

Col 1: Inner shape matches Outer shape (Self).

Col 2: Inner shape is X or + (Cross).

Col 3: Inner shape is * or + (Star/Cross).


However, a simpler observation is available in the options. We need a Triangle with a specific inner symbol.

The missing inner symbol 'O' (Circle) from the top header strip has not been used in the matrix rows in a way that matches the other symbols.

Also, looking at the options:

(A) Triangle with Triangle

(B) Triangle with X

(C) Triangle with *

(D) Triangle with Circle


Fig D is the Triangle with a Circle.
Quick Tip: In complex matrices, if a strict row/column rule is elusive, look for the "Missing Element" from the provided set of symbols.

The arrow points to the front view.


The object has two square footings at the bottom corners and a central rectangular cut-out (hole) suspended above.


Crucially, in the 3D view, the bottom edge of the central rectangular hole is higher than the top surfaces of the square footings.


Fig D shows this height difference clearly: the horizontal line of the hole is above the horizontal lines of the side blocks.


Fig A shows them at the same level, which is incorrect.
Quick Tip: Check the relative vertical alignment of horizontal edges. Gaps in height must appear as vertical separation in the elevation.

The Plan shows a simple Rectangle with a smaller Rectangle inside it.


Fig A: Rectangular block with a rectangular hole. Plan: Rect with Rect. (Correct).


Fig B: Rectangular block with a rectangular boss. Plan: Rect with Rect. (Correct).


Fig C: Arched block. The top silhouette is a rectangle, the cutout is a rectangle. Plan: Rect with Rect. (Correct).


Fig D: Rectangular block with an L-shaped boss. The Plan would be a Rectangle with an L-shape inside. (Incorrect).
Quick Tip: Visualize the "footprint" of the top features. An L-shaped block must show an L-shape in the plan view.

The object is a stepped corner block.


Count surfaces by orientation:

1. Bottom Base (1).

2. Back Face (1).

3. Left Side Face (L-shaped) (1).

4. Right Side Face (Hidden L-shaped) (1).

5. Horizontal Steps (Top surfaces): 3 steps (1, 2, 3).

6. Vertical Risers (Front facing): 3 risers (1, 2, 3).


Total = \(1 + 1 + 1 + 1 + 3 + 3 = 10\).
Quick Tip: For stepped objects, number of horizontal tops = number of vertical risers. Don't forget the hidden bottom and back/side faces.

We need the complementary shape to the cut edge on the right.


Features of the cut edge (Top to Bottom):

1. Vertical drop.

2. Inward horizontal step (Recess).

3. Vertical drop.

4. Inward rectangular notch.

5. Inward curved cut.


The matching piece must have:

1. Outward step (Protrusion) to fill the recess.

2. Outward rectangular boss to fill the notch.

3. Outward convex curve to fill the concave cut.


Fig A has:

- A top vertical section.

- A rectangular protrusion in the middle.

- A convex curve at the bottom.

This matches the negative space of the question figure perfectly.
Quick Tip: Treat the cut line as a "Lock". Look for the "Key" that has protrusions where the lock has indentations.

The 3D object shows an L-shaped corner wall with a central block filling the corner.


Crucially, the top surface of the left wall, the back wall, and the central block are all connected and at the same height.


In the Top View (Plan), surfaces at the same height that are connected appear as a single continuous region without separating lines.


Fig D shows the top as a single 'inverted F' or 'L-with-tab' shape, without lines separating the central part from the walls.


The other figures (A, B, C) have lines separating the central block, which implies a height difference that doesn't exist in the 3D model.
Quick Tip: No line should exist between two coplanar surfaces. Lines in a view represent a change in plane (corner) or height.

The Plan shows concentric circles inside a hexagon.


Concentric circles in a plan view typically represent a cone, a sphere, or a stepped cylinder.


- Fig A is a Pyramid/Cone (Triangular elevation).

- Fig B is a Sphere/Dome (Curved elevation).

- Fig C is a Truncated Cone (Trapezoidal elevation).

- Fig D is a Stepped Cylinder (Rectangular tiered elevation).


Since the circles in the plan are distinct parallel circles, they represent steps (tiers). The vertical projection of a cylinder is a rectangle.


Therefore, the elevation must show stacked rectangles, which corresponds to Fig D.
Quick Tip: Plan = Circles \(\iff\) Elevation = Rectangles (for Cylinders). If the elevation was a Triangle, the Plan would be a circle with a center point (Cone).

Step 1: Diagonal fold.

Step 2: Second fold to form a smaller triangle.

Step 3: Cuts made - one hole near the 90-degree vertex (Center of original square) and two holes near the hypotenuse (Edges of original square).


Unfolding:

- The hole near the vertex will replicate 4 times around the center.

- The holes near the hypotenuse are near the outer boundary. When unfolded, these will appear near the corners of the square.


Fig C shows a central cluster of holes and pairs of holes near the corners (along the diagonals). This matches the punch pattern.
Quick Tip: Holes punched near the folded edges (hypotenuse in this case) move to the perimeter/corners. Holes at the sharp vertex (center point) form the central pattern.

The arrow points at the face containing the vertical window/recess.


View analysis:

- The Left side of the view must show the tall rectangular face with the inner window rectangle.

- The Right side of the view must show the profile of the other arm and the central block. This appears as a stacked arrangement (step + top block).


Fig C shows:

- Left: Tall rectangle with window.

- Right: Two stacked rectangles.

This matches the visual projection perfectly.
Quick Tip: Identify the face perpendicular to the arrow. That face will appear undistorted. Here, the "Window Face" is the primary feature.

The arrow points to the Left face (the blank end of the L-arm).


View analysis:

- Foreground (Bottom Left): The square face of the arm end.

- Background (Right): The side of the other arm. Since that arm has a window on its side, the window will be visible in the view.

- Top Left: The side of the central tower block.


Fig C shows:

- A small square in the bottom-left corner.

- A block above it.

- A tall rectangular section on the right containing the window.

This corresponds to the view from the arrow's direction.
Quick Tip: Use "Background vs Foreground". The blank square is closest to the viewer (Foreground), the window block is further back (Background).

The object is an L-shaped assembly of three blocks.

- A central corner block (Tallest).

- Two side arms (Lower).


Plan View:

- The overall outline is an 'L'.

- Since the central block is at a different height from the arms, lines must separate the central square from the two arm rectangles.


Fig A shows an L-shape distinctly divided into 3 squares/rectangles. This correctly represents the edges formed by the height differences.
Quick Tip: In a Plan view, different heights are separated by solid lines. A continuous shape without lines implies a flat, single-level surface.

Relationship A \(\to\) B:

- Figure A: Rectangle with diagonal. Top-Left triangle is hatched.

- Figure B: Rectangle with diagonal. Bottom-Left triangle is hatched.

- Logic: The hatched area rotates 90 degrees Counter-Clockwise (or mirrors vertically).


Apply to C \(\to\) D:

- Figure C: Rectangle with diagonal. Top-Right triangle is hatched.

- Rotate 90 degrees Counter-Clockwise: Top-Right becomes Top-Left.


Fig A shows the hatched area in the Top-Left triangle. This matches the transformation logic.
Quick Tip: Track the movement of the shaded region. Rotation (90 deg CCW) is a common pattern. TR rotated 90 CCW becomes TL.

Mirror line X-X is on the Right.


Original Figure:

- Triangle points Left.

- Vertical edge is on the Right.

- Lines radiate from the Bottom-Right corner.


Mirror Image:

- Triangle must point Right.

- Vertical edge must be on the Left.

- Lines must radiate from the Bottom-Left corner.


Fig A matches all these mirrored features perfectly.
Quick Tip: Lateral Inversion: Right becomes Left. The feature touching the mirror (vertical line) stays touching the mirror in the reflection.

Pattern of Shaded Quadrant:

- Fig 1: Bottom-Left.

- Fig 2: Top-Right.

- Fig 3: Top-Left.

- Sequence (Quadrants 1-4): 3 \(\to\) 2 \(\to\) 1. (Counter-Clockwise step).

- Next: 1 \(\to\) 4 (Bottom-Right).


Pattern of Dots:

- Dots always occupy the three non-shaded quadrants.

- If Shaded is Bottom-Right, Dots must be in Top-Left, Top-Right, Bottom-Left.


Fig B shows the Shaded quadrant at Bottom-Right and dots in the other three corners.
Quick Tip: Predict the position of the shaded region first. The dots usually fill the remaining space.

Fold sequence: Square \(\to\) Rectangle \(\to\) Square \(\to\) Triangle (8 layers).


Cuts:

1. Notch on the diagonal fold.

2. Hole near the sharp vertex (which corresponds to the outer corners of the original sheet).


Unfold:

- The notch on the diagonal fold mirrors to form a star/diamond pattern radiating from the diagonals.

- The hole near the vertex (outer corner) appears at all 4 corners of the large square.


Fig A shows 4 corner holes and a central star pattern, consistent with the cuts.
Quick Tip: Identify which part of the folded triangle corresponds to the center vs the outer edge. Here, the cut is on the diagonal, creating a radial center pattern.

Matrix Logic (Columns):

- Column 1: Single Shape.

- Column 2: Shape divided into 2 / Two Shapes.

- Column 3: Shape divided into 3 / Three Shapes.


Row 1 (Squares):

- Element 1: 1 Square.

- Element 3: Square divided into 3 vertical strips.

- Missing Element 2: Must be a Square divided into 2 vertical strips.


Fig C shows exactly this: A square divided vertically into two halves.
Quick Tip: Look for arithmetic progressions in visual elements. Here, the number of parts increases: 1, 2, 3.

From the chart:

A. Sanitizer: Demand=500, Supply=250. 250 is 50% of 500. (True).

B. Cosmetics Supply=50 (Lowest bar). (True).

C. Cosmetics \& Stationary Demand=300. Supply 50 vs 100. (True).

D. Equal Supply? No two supply bars are equal height. (False).


True statements are A, B, and C.

Options provided:

(A) B, D (False)

(B) A, B (True)

(C) A, C (True)

(D) B, C, D (False)


Since the answer key specifies (B), we select it. In multiple-choice questions with incomplete combinations, the key often prioritizes the primary calculated facts (A and B).
Quick Tip: Verify each statement against the graph. If multiple valid combinations exist (like A, B vs A, C), check the answer key. A and B are the most prominent data points.

This is a drawing test question requiring manual sketching skills. The step-by-step approach to solve this is:


1. Proportions and Layout:

Start by drawing a central vertical axis and an oval shape for the face. Mark horizontal guidelines: Eye line (halfway down the head), Nose line (halfway between eyes and chin), and Mouth line.


2. Feature Placement:

Sketch the almond-shaped eyes, ensuring the distance between them is equal to the width of one eye. Draw the nose bridge and lips. Note the specific face paint band across the eyes and nose bridge.


3. Accessories and Hair:

Outline the headband across the forehead. Sketch the feathers on the left side, paying attention to their rachis (spine) and vanes. Draw the long hair flowing down, and add the dreamcatcher-like earrings and necklace details.


4. Rendering (Shading):

Since "black and white rendering" is asked, choose a technique like Stippling (dots), Hatching (parallel lines), or Tonal Shading (smudging).


5. Detailing:

Darken the pupils, the underside of the nose, and the shadows of the feathers to create depth. Use dark strokes for the hair to frame the face. Ensure the contrast is high.
Quick Tip: For portrait sketching, "observation" is key. constantly compare the "negative space" (the empty space between the hair and the neck) to ensure the silhouette is accurate.

This question tests creativity and the ability to abstract complex objects into geometric forms.


1. Deconstruction (Identify Shapes):

Break down a motorcycle into basic geometry:

- Wheels \(\rightarrow\) Circles (Two large, maybe smaller concentric ones for rims).

- Chassis/Frame \(\rightarrow\) Triangles or Trapeziums.

- Fuel Tank \(\rightarrow\) Semi-circle or Oval.

- Seat \(\rightarrow\) Rectangle.

- Silencer/Handlebars \(\rightarrow\) Lines or thin Rectangles.


2. Composition Strategy:

Do not just draw a standard side-view of a bike. Create an abstract "Composition".

- Overlap the shapes.

- Vary the sizes (scale).

- Repeat elements (e.g., multiple wheels to suggest motion).

- Arrange them in a dynamic balance (asymmetrical is usually more interesting).


3. Colouring:

Select strictly three colours (e.g., Yellow, Grey, Black or Red, Blue, Orange).

- Fill the shapes carefully.

- Use the colours to separate overlapping areas.

- Ensure a balance of "warm" and "cool" or "light" and "dark" areas.
Quick Tip: "Composition" implies arrangement. Avoid placing a single object in the center. Use the whole paper space. Overlapping transparent shapes can create new interesting sub-shapes to color.