JEE Main 2025 April 4 Shift 1 Maths Question Paper, Exam Analysis, and Answer Keys (Available)

To find \(\frac{1}{\beta - \alpha}\), we first need to determine the range of the function \(fog(x) = f(g(x))\).

1. Calculate \(fog(x)\):
\[ fog(x) = f\left(g(x)\right) = f\left(\frac{2 - 3x}{1 - x}\right) \]
Substitute \(g(x)\) into \(f(x)\):
\[ fog(x) = \frac{2\left(\frac{2 - 3x}{1 - x}\right) + 3}{5\left(\frac{2 - 3x}{1 - x}\right) + 2} \]
Simplify the expression:
\[ fog(x) = \frac{\frac{4 - 6x + 3 - 3x}{1 - x}}{\frac{10 - 15x + 2 - 2x}{1 - x}} = \frac{7 - 9x}{12 - 17x} \]

2. Determine the range of \(fog(x)\) for \(x \in [2, 4]\):
- Calculate \(fog(2)\):
\[ fog(2) = \frac{7 - 9(2)}{12 - 17(2)} = \frac{7 - 18}{12 - 34} = \frac{-11}{-22} = \frac{1}{2} \]
- Calculate \(fog(4)\):
\[ fog(4) = \frac{7 - 9(4)}{12 - 17(4)} = \frac{7 - 36}{12 - 68} = \frac{-29}{-56} = \frac{29}{56} \]

3. Identify \(\alpha\) and \(\beta\):
- The range of \(fog(x)\) is \(\left[\frac{1}{2}, \frac{29}{56}\right]\).
- Therefore, \(\alpha = \frac{1}{2}\) and \(\beta = \frac{29}{56}\).

4. Calculate \(\frac{1}{\beta - \alpha}\):
\[ \beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29}{56} - \frac{28}{56} = \frac{1}{56} \]
\[ \frac{1}{\beta - \alpha} = \frac{1}{\frac{1}{56}} = 56 \]

Therefore, the correct answer is (4) 56. Quick Tip: The range of a composite function can be determined by evaluating the function at the endpoints of the domain.

1. Identify the sets \(A\) and \(B\):
- \(A: x^2 + y^2 = 25\)
- \(B: \frac{x^2}{144} + \frac{y^2}{16} = 1\)

2. Solve for the intersection \(D = A \cap B\):
- Substitute \(x^2 + y^2 = 25\) into \(x^2 + 9y^2 = 144\):
\[ x^2 + 9(25 - x^2) = 144 \]
\[ x^2 + 225 - 9x^2 = 144 \]
\[ -8x^2 = 144 - 225 \]
\[ -8x^2 = -81 \]
\[ x^2 = \frac{81}{8} \]
\[ x = \pm \frac{9}{2\sqrt{2}} \]
- Substitute \(x\) back into \(x^2 + y^2 = 25\):
\[ y^2 = 25 - \frac{81}{8} \]
\[ y = \pm \frac{\sqrt{119}}{2\sqrt{2}} \]

3. Determine the elements of set \(D\):
\[ D = \left\{\left(\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, \frac{\sqrt{119}}{2\sqrt{2}}\right), \left(-\frac{9}{2\sqrt{2}}, -\frac{\sqrt{119}}{2\sqrt{2}}\right)\right\} \]
- Number of elements in set \(D = 4\).

4. Identify the set \(C\):
\[ C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\} \]
- Possible pairs \((x, y)\):
\[ \{(0, 2), (2, 0), (0, -2), (-2, 0), (1, 1), (-1, -1), (1, -1), (-1, 1), (1, 0), (0, 1), (-1, 0), (0, -1), (0, 0)\} \]
- Number of elements in set \(C = 13\).

5. Calculate the total number of one-one functions from set \(D\) to set \(C\):
\[ Total number of one-one functions = 13 \times 12 \times 11 \times 10 = 17160 \]

Therefore, the correct answer is (3) 17160. Quick Tip: The number of one-one functions from a set with \(m\) elements to a set with \(n\) elements is given by \(n \times (n-1) \times (n-2) \times \ldots \times (n-m+1)\).

1. Identify the sets \(A\) and \(B\):
- \(A = \{1, 6, 11, 16, \ldots\}\)
- \(B = \{9, 16, 23, 30, \ldots\}\)

2. Find the general terms for \(A\) and \(B\):
- For set \(A\): \(T_n = 1 + (n-1) \cdot 5 = 5n - 4\)
- For set \(B\): \(T_n = 9 + (n-1) \cdot 7 = 7n + 2\)

3. Determine the intersection \(A \cap B\):
- Solve \(5n - 4 = 7m + 2\) for \(n\) and \(m\):
\[ 5n - 7m = 6 \]
- The common terms are \(16, 51, 86, \ldots\)

4. Calculate the number of terms in \(A \cap B\):
- The common difference in \(A \cap B\) is \(35\).
- Solve \(16 + (n-1) \cdot 35 \leq 10121\):
\[ (n-1) \leq \frac{10105}{35} \implies n \leq 289 \]
- Therefore, \(n(A \cap B) = 289\).

5. Calculate \(n(A \cup B)\):
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
\[ n(A \cup B) = 2025 + 2025 - 289 = 3761 \]

Therefore, the correct answer is (3) 3761. Quick Tip: The number of terms in the union of two sets can be found using the formula \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\).

1. Determine the number of terms in \((x + y)^{2n-3}\):
\[ Number of terms = 2n - 2 \]

2. Sum of all coefficients:
\[ Sum of coefficients = 2^{2n-3} \]

3. Arithmetic mean of all coefficients:
\[ Arithmetic mean = \frac{2^{2n-3}}{2n-2} = 16 \]
\[ 2^{2n-3} = 16(2n-2) \]
\[ 2^{2n-3} = 2^4(n-1) \]
\[ 2n-3 = 4 \implies n = 5 \]

4. Determine the point \(P\):
\[ P(2n-1, n^2-4n) = P(9, 5) \]

5. Calculate the distance from the line \(x + y = 8\):
\[ Distance = \left| \frac{9 + 5 - 8}{\sqrt{2}} \right| = \frac{6}{\sqrt{2}} = 3\sqrt{2} \]

Therefore, the correct answer is (4) \(3\sqrt{2}\). Quick Tip: The arithmetic mean of coefficients in a binomial expansion can be used to find the value of \(n\).

1. Calculate the number of ways to form the committee:
- 3 engineers + 1 doctor + 8 professors:
\[ ^4C_3 \cdot ^2C_1 \cdot ^{10}C_8 = 360 \]
- 3 engineers + 2 doctors + 7 professors:
\[ ^4C_3 \cdot ^2C_2 \cdot ^{10}C_7 = 480 \]
- 4 engineers + 1 doctor + 7 professors:
\[ ^4C_4 \cdot ^2C_1 \cdot ^{10}C_7 = 240 \]
- 4 engineers + 2 doctors + 6 professors:
\[ ^4C_4 \cdot ^2C_2 \cdot ^{10}C_6 = 210 \]

2. Total number of favorable outcomes:
\[ Total = 360 + 480 + 240 + 210 = 1290 \]

3. Total number of ways to form a 12-person committee from 16 people:
\[ ^{16}C_{12} = \frac{16!}{12! \cdot 4!} = 1820 \]

4. Calculate the probability:
\[ Probability = \frac{1290}{1820} = \frac{129}{182} \]

Therefore, the correct answer is (1) \(\frac{129}{182}\). Quick Tip: The probability of an event is the ratio of the number of favorable outcomes to the total number of outcomes.

1. Identify the points and direction vectors:
- Line 1: \(\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}\)
- Point \(A(3, \alpha, 3)\)
- Direction vector \(\vec{p} = 3\hat{i} - \hat{j} + \hat{k}\)
- Line 2: \(\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}\)
- Point \(B(-3, -7, \beta)\)
- Direction vector \(\vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k}\)

2. Calculate \(\vec{p} \times \vec{q}\):
\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -1 & 1
-3 & 2 & 4 \end{vmatrix} = 6\hat{i} + 15\hat{j} - 9\hat{k} \]

3. Calculate \(\vec{BA}\):
\[ \vec{BA} = (3 + 3)\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k} = 6\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k} \]

4. Use the distance formula:
\[ \frac{|\vec{BA} \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|} = 3\sqrt{30} \]
\[ \frac{|6 \cdot 6 + 15(\alpha + 7) - 9(3 - \beta)|}{\sqrt{6^2 + 15^2 + (-9)^2}} = 3\sqrt{30} \]
\[ 36 + 15(\alpha + 7) - 9(3 - \beta) = 270 \]
\[ 15\alpha + 3\beta = 138 \]
\[ 5\alpha + \beta = 46 \]

Therefore, the correct answer is (2) 46. Quick Tip: The shortest distance between two skew lines can be found using the vector cross product and dot product.

1. Substitute \(x = 1 + h\):
\[ \lim_{h \to 0} \frac{h(6 + \lambda \cos h) - \mu \sin h}{h^3} = -1 \]

2. Expand \(\cos h\) and \(\sin h\) using Taylor series:
\[ \cos h \approx 1 - \frac{h^2}{2}, \quad \sin h \approx h - \frac{h^3}{6} \]

3. Substitute the expansions:
\[ \lim_{h \to 0} \frac{h \left( 6 + \lambda \left( 1 - \frac{h^2}{2} \right) \right) - \mu \left( h - \frac{h^3}{6} \right)}{h^3} = -1 \]

4. Simplify the expression:
\[ \lim_{h \to 0} \frac{h \left( 6 + \lambda - \frac{\lambda h^2}{2} \right) - \mu h + \frac{\mu h^3}{6}}{h^3} = -1 \]
\[ \lim_{h \to 0} \frac{6h + \lambda h - \frac{\lambda h^3}{2} - \mu h + \frac{\mu h^3}{6}}{h^3} = -1 \]
\[ \lim_{h \to 0} \frac{6 + \lambda - \mu - \frac{\lambda h^2}{2} + \frac{\mu h^2}{6}}{h^2} = -1 \]

5. Equate the coefficients:
\[ 6 + \lambda - \mu = 0 \quad and \quad -\frac{\lambda}{2} + \frac{\mu}{6} = -1 \]

6. Solve the system of equations:
\[ \lambda + \mu = 18 \]

Therefore, the correct answer is (1) 18. Quick Tip: Substitute \(x = 1 + h\) to simplify limits involving \(x \to 1\).

1. Differentiate \(f(x)\):
\[ f'(x) = -2 + e^x \int_0^x e^{-t} f(t) \, dt + e^x e^{-x} f(x) \]
\[ f'(x) = -2 + e^x \int_0^x e^{-t} f(t) \, dt + f(x) \]

2. Simplify the differential equation:
\[ f'(x) - f(x) = -2 \]

3. Solve the differential equation:
\[ \frac{d}{dx} \left( e^{-x} f(x) \right) = -2e^{-x} \]
\[ e^{-x} f(x) = \int -2e^{-x} \, dx = 2e^{-x} + c \]
\[ f(x) = 2 + ce^x \]

4. Use the initial condition \(f(0) = 1\):
\[ 1 = 2 + c \implies c = -1 \]
\[ f(x) = 2 - e^x \]

5. Find the area under the curve \(y = f(x)\):
\[ Area = \int_0^\infty (2 - e^x) \, dx \]
\[ Area = \left[ 2x - e^x \right]_0^\infty = \left[ 2x - e^x \right]_0^\infty = \frac{1}{2} \]

Therefore, the correct answer is (2) \(\frac{1}{2}\). Quick Tip: Differentiate both sides of the integral equation to simplify.

1. Identify the points \(A\) and \(B\):
- Let \(A(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)\)
- Let \(B(3\mu + 6, 2\mu + 7, -2\mu + 7)\)

2. Distance from the point \((1, 2, 3)\) to the line \(L\):
\[ Distance = 2\sqrt{17} \]

3. Use the distance formula:
\[ \sqrt{(3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2} = 2\sqrt{17} \]
\[ (3\lambda + 5)^2 + (2\lambda + 5)^2 + (-2\lambda + 4)^2 = 68 \]
\[ 17\lambda^2 - 17 = 0 \implies \lambda = \pm 1 \]

4. Determine the points \(A\) and \(B\):
- For \(\lambda = 1\): \(A(9, 9, 5)\)
- For \(\lambda = -1\): \(B(-3, -1, 9)\)

5. Calculate \(\overrightarrow{OA} \cdot \overrightarrow{OB}\):
\[ \overrightarrow{OA} \cdot \overrightarrow{OB} = 9(-3) + 9(-1) + 5(9) = -27 - 9 + 45 = 47 \]

Therefore, the correct answer is (2) 47. Quick Tip: Use the distance formula to find the points on the line.

1. Use the given functional equation:
\[ f(2x) - f(x) = x \]

2. Express \(f(x)\) in terms of \(f\left( \frac{x}{2^n} \right)\):
\[ f(x) - f\left( \frac{x}{2} \right) = \frac{x}{2} \]
\[ f\left( \frac{x}{2} \right) - f\left( \frac{x}{4} \right) = \frac{x}{4} \]
\[ f\left( \frac{x}{4} \right) - f\left( \frac{x}{8} \right) = \frac{x}{8} \]
\[ \vdots \]
\[ f\left( \frac{x}{2^{n-1}} \right) - f\left( \frac{x}{2^n} \right) = \frac{x}{2^n} \]

3. Sum the series:
\[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n} \right) \]
\[ f(x) - f\left( \frac{x}{2^n} \right) = x \left( 1 - \frac{1}{2^n} \right) \]

4. Take the limit as \(n \to \infty\):
\[ G(x) = \lim_{n \to \infty} \left( f(x) - f\left( \frac{x}{2^n} \right) \right) = x \]

5. Calculate \(\sum_{r=1}^{10} G(r^2)\):
\[ \sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2 = 1^2 + 2^2 + 3^2 + \cdots + 10^2 \]
\[ \sum_{r=1}^{10} r^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

Therefore, the correct answer is (2) 385. Quick Tip: Use the sum of squares formula to calculate the sum.

1. Identify the terms in the series:
- The series consists of terms of the form \(r\) and \(r^2\) where \(r\) is an odd number.

2. Separate the series into two parts:
- Part 1: Sum of terms of the form \(r\).
- Part 2: Sum of terms of the form \(r^2\).

3. Sum of terms of the form \(r\):
- The sequence is \(1, 3, 5, 7, \ldots\) up to 20 terms.
- Sum of the first 20 odd numbers:
\[ \sum_{r=1}^{20} (2r-1) = 20^2 = 400 \]

4. Sum of terms of the form \(r^2\):
- The sequence is \(1^2, 3^2, 5^2, 7^2, \ldots\) up to 20 terms.
- Sum of the squares of the first 20 odd numbers:
\[ \sum_{r=1}^{20} (2r-1)^2 = \sum_{r=1}^{20} (4r^2 - 4r + 1) \]
\[ = 4 \sum_{r=1}^{20} r^2 - 4 \sum_{r=1}^{20} r + \sum_{r=1}^{20} 1 \]
\[ = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 4 \cdot \frac{20 \cdot 21}{2} + 20 \]
\[ = 4 \cdot 2870 - 4 \cdot 210 + 20 = 11480 - 840 + 20 = 10660 \]

5. Total sum of the series:
\[ Total sum = 400 + 10660 = 41880 \]

Therefore, the correct answer is (2) 41880. Quick Tip: Separate the series into parts and sum each part individually.

1. General term in the binomial expansion:
\[ T_{r+1} = ^nC_r \left( \sqrt{5} \right)^{n-r} \left( \frac{1}{\sqrt{5}} \right)^r = ^nC_r \left( \sqrt{5} \right)^{n-2r} \]

2. Given ratio of \(15^{th}\) term from the beginning to the \(15^{th}\) term from the end:
\[ \frac{T_{15}}{T_{n-13}} = \frac{1}{6} \]

3. Express the terms:
\[ T_{15} = ^nC_{14} \left( \sqrt{5} \right)^{n-28} \]
\[ T_{n-13} = ^nC_{14} \left( \sqrt{5} \right)^{28-n} \]

4. Set up the ratio:
\[ \frac{^nC_{14} \left( \sqrt{5} \right)^{n-28}}{^nC_{14} \left( \sqrt{5} \right)^{28-n}} = \frac{1}{6} \]
\[ \left( \sqrt{5} \right)^{n-56} = \frac{1}{6} \]
\[ \left( \sqrt{5} \right)^{n-56} = 6^{-1} \]
\[ n - 56 = -1 \implies n = 55 \]

5. Calculate \(^nC_3\):
\[ ^nC_3 = ^{55}C_3 = \frac{55 \cdot 54 \cdot 53}{3 \cdot 2 \cdot 1} = 2300 \]

Therefore, the correct answer is (3) 2300. Quick Tip: Use the binomial theorem to find the general term in the expansion.

1. Let \(\sin^{-1} x = \theta\):
\[ x = \sin \theta \]

2. Express the given function:
\[ \sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) \]
\[ = \sin^{-1} \left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right) \]

3. Use the angle addition formula:
\[ = \sin^{-1} \left( \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \right) \]
\[ = \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{6} \right) \right) \]
\[ = \theta + \frac{\pi}{6} \]
\[ = \sin^{-1} x + \frac{\pi}{6} \]

Therefore, the correct answer is (2) \(\frac{\pi}{6} + \sin^{-1} x\). Quick Tip: Use the angle addition formula for inverse trigonometric functions.

1. Given vectors:
\[ \vec{u} = 3\hat{i} - \hat{j}, \quad \vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k} \]

2. Calculate the dot product \(\vec{u} \cdot \vec{v}\):
\[ \vec{u} \cdot \vec{v} = (3\hat{i} - \hat{j}) \cdot (2\hat{i} + \hat{j} - \lambda \hat{k}) = 6 + 1 = 7 \]

3. Calculate the magnitudes of \(\vec{u}\) and \(\vec{v}\):
\[ |\vec{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \]
\[ |\vec{v}| = \sqrt{2^2 + 1^2 + \lambda^2} = \sqrt{5 + \lambda^2} \]

4. Use the given cosine of the angle between \(\vec{u}\) and \(\vec{v}\):
\[ \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{7}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \]
\[ \frac{7}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \]
\[ 7 \cdot 2\sqrt{7} = \sqrt{5} \cdot \sqrt{10} \cdot \sqrt{5 + \lambda^2} \]
\[ 14\sqrt{7} = 5\sqrt{10} \sqrt{5 + \lambda^2} \]
\[ 14\sqrt{7} = 5\sqrt{10} \sqrt{5 + \lambda^2} \]
\[ \lambda^2 = 9 \implies \lambda = 3 \]

5. Decompose \(\vec{v}\) into \(\vec{v}_1\) and \(\vec{v}_2\):
\[ \vec{v} = \vec{v}_1 + \vec{v}_2 \]
\[ |\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \]
\[ |\vec{v}|^2 = 14 \]

Therefore, the correct answer is (2) 14. Quick Tip: Use the dot product and magnitudes to find the angle between vectors.

1. Find the intersection points of the lines to determine the vertices of the triangle:
- Intersection of \(4x - 7y + 10 = 0\) and \(x + y = 5\):
\[ \begin{cases} 4x - 7y + 10 = 0
x + y = 5 \end{cases} \]
Solving these equations, we get:
\[ x = 1, \quad y = 4 \quad \Rightarrow \quad A(1, 4) \]

- Intersection of \(4x - 7y + 10 = 0\) and \(7x + 4y = 15\):
\[ \begin{cases} 4x - 7y + 10 = 0
7x + 4y = 15 \end{cases} \]
Solving these equations, we get:
\[ x = 2, \quad y = 1 \quad \Rightarrow \quad B(2, 1) \]

- Intersection of \(x + y = 5\) and \(7x + 4y = 15\):
\[ \begin{cases} x + y = 5
7x + 4y = 15 \end{cases} \]
Solving these equations, we get:
\[ x = 1, \quad y = 4 \quad \Rightarrow \quad C(1, 4) \]

2. Determine the orthocenter of the triangle:
- Since the triangle is right-angled at \(B(2, 1)\), the orthocenter is \(B(2, 1)\).

3. Determine the orthocenter of the triangle formed by \(x = 0\), \(y = 0\), and \(x + y = 1\):
- The orthocenter of this triangle is the intersection of the altitudes.
- The orthocenter is \(P(0, 0)\).

4. Calculate the distance between the two orthocenters:
\[ Distance = \sqrt{(2 - 0)^2 + (1 - 0)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Therefore, the correct answer is (2) \(\sqrt{5}\). Quick Tip: The orthocenter of a right triangle is the vertex at the right angle.

1. Simplify the integrand:
\[ \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]
\[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]
\[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]

2. Evaluate the integral:
\[ = \int_{-1}^{1} (1 + \sqrt{|x| - x}) \, dx \]
\[ = \int_{-1}^{1} 1 \, dx + \int_{-1}^{1} \sqrt{|x| - x} \, dx \]
\[ = [x]_{-1}^{1} + \int_{0}^{1} \sqrt{x} \, dx \]
\[ = 2 + \frac{2\sqrt{2}}{3} \]

Therefore, the correct answer is (4) \(1 + \frac{2\sqrt{2}}{3}\). Quick Tip: Simplify the integrand before evaluating the integral.

1. Identify the foci and eccentricity:
- Foci: \((2, 5)\) and \((2, -3)\)
- Eccentricity: \(\frac{4}{5}\)

2. Calculate the distance between the foci:
\[ 2c = |5 - (-3)| = 8 \implies c = 4 \]

3. Use the relationship between \(a\), \(b\), and \(c\):
\[ e = \frac{c}{a} = \frac{4}{5} \implies a = 5 \]
\[ b^2 = a^2 - c^2 = 25 - 16 = 9 \implies b = 3 \]

4. Calculate the length of the latus-rectum:
\[ Length of latus-rectum = \frac{2b^2}{a} = \frac{2 \cdot 3^2}{5} = \frac{18}{5} \]

Therefore, the correct answer is (4) \(\frac{18}{5}\). Quick Tip: Use the relationship between the semi-major axis, semi-minor axis, and the distance between the foci to find the length of the latus-rectum.

1. Rewrite the equation:
\[ x^2 + 4x + 4 = n + 4 \]
\[ (x + 2)^2 = n + 4 \]

2. Solve for \(x\):
\[ x = -2 \pm \sqrt{n + 4} \]

3. Determine the range of \(n\):
\[ 20 \leq n \leq 100 \]
\[ \sqrt{24} \leq \sqrt{n + 4} \leq \sqrt{104} \]
\[ 4.9 \leq \sqrt{n + 4} \leq 10.2 \]

4. Find the integer values of \(\sqrt{n + 4}\):
\[ \sqrt{n + 4} \in \{5, 6, 7, 8, 9, 10\} \]

5. Calculate the number of distinct values of \(n\):
\[ Number of distinct values = 6 \]

Therefore, the correct answer is (3) 6. Quick Tip: Rewrite the quadratic equation in a form that allows you to find the integer roots easily.

1. Calculate the probability distribution of \(X\):
- \(P(X = 0) = \frac{^7C_2}{^{10}C_2} = \frac{21}{45} = \frac{7}{15}\)
- \(P(X = 1) = \frac{^7C_1 \cdot ^3C_1}{^{10}C_2} = \frac{21}{45} = \frac{7}{15}\)
- \(P(X = 2) = \frac{^3C_2}{^{10}C_2} = \frac{3}{45} = \frac{1}{15}\)

2. Calculate the expected value \(E(X)\):
\[ E(X) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{3}{5} \]

3. Calculate the variance \(Var(X)\):
\[ Var(X) = \left(0 - \frac{3}{5}\right)^2 \cdot \frac{7}{15} + \left(1 - \frac{3}{5}\right)^2 \cdot \frac{7}{15} + \left(2 - \frac{3}{5}\right)^2 \cdot \frac{1}{15} \]
\[ = \frac{9}{25} \cdot \frac{7}{15} + \frac{4}{25} \cdot \frac{7}{15} + \frac{1}{25} \cdot \frac{1}{15} \]
\[ = \frac{63}{375} + \frac{28}{375} + \frac{1}{375} = \frac{92}{375} = \frac{28}{75} \]

Therefore, the correct answer is (2) \(\frac{28}{75}\). Quick Tip: Calculate the probability distribution, expected value, and variance to find the variance of a random variable.

1. Rewrite the given equation:
\[ 10 \sin^4 \theta + 15 \cos^4 \theta = 6 \]
\[ 10 (\sin^2 \theta)^2 + 15 (1 - \sin^2 \theta)^2 = 6 \]

2. Let \(u = \sin^2 \theta\):
\[ 10u^2 + 15(1 - u)^2 = 6 \]
\[ 10u^2 + 15(1 - 2u + u^2) = 6 \]
\[ 10u^2 + 15 - 30u + 15u^2 = 6 \]
\[ 25u^2 - 30u + 9 = 0 \]

3. Solve the quadratic equation:
\[ u = \frac{30 \pm \sqrt{900 - 900}}{50} = \frac{30 \pm 0}{50} = \frac{3}{5} \]
\[ \sin^2 \theta = \frac{3}{5}, \quad \cos^2 \theta = \frac{2}{5} \]

4. Calculate the given expression:
\[ \frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta} = \frac{27 \left( \frac{5}{3} \right)^3 + 8 \left( \frac{5}{2} \right)^3}{16 \left( \frac{5}{2} \right)^4} \]
\[ = \frac{27 \cdot \frac{125}{27} + 8 \cdot \frac{125}{8}}{16 \cdot \frac{625}{16}} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5} \]

Therefore, the correct answer is (1) \(\frac{2}{5}\). Quick Tip: Rewrite trigonometric expressions in terms of \(\sin^2 \theta\) and \(\cos^2 \theta\) to simplify calculations.

1. Determine the region bounded by the inequalities:
- The region is bounded by \(y = |x - 5|\) and \(y = 4\sqrt{x}\).

2. Find the intersection points of the curves:
- Solve \(y = |x - 5|\) and \(y = 4\sqrt{x}\):
\[ |x - 5| = 4\sqrt{x} \]
- For \(x \geq 5\):
\[ x - 5 = 4\sqrt{x} \]
\[ x - 4\sqrt{x} - 5 = 0 \]
- Let \(u = \sqrt{x}\), then \(u^2 - 4u - 5 = 0\):
\[ u = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2} \]
\[ u = 5 \quad or \quad u = -1 \quad (not valid) \]
\[ x = 25 \]
- For \(x < 5\):
\[ 5 - x = 4\sqrt{x} \]
\[ 5 - 4\sqrt{x} - x = 0 \]
- Let \(u = \sqrt{x}\), then \(u^2 + 4u - 5 = 0\):
\[ u = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm 6}{2} \]
\[ u = 1 \quad or \quad u = -5 \quad (not valid) \]
\[ x = 1 \]

3. Calculate the area of the region:
- The area is given by the integral:
\[ A = \int_{1}^{25} 4\sqrt{x} \, dx - \int_{1}^{5} (5 - x) \, dx \]
- Evaluate the integrals:
\[ \int_{1}^{25} 4\sqrt{x} \, dx = 4 \left[ \frac{2}{3} x^{3/2} \right]_{1}^{25} = \frac{8}{3} \left[ 125 - 1 \right] = \frac{8}{3} \cdot 124 = \frac{992}{3} \]
\[ \int_{1}^{5} (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_{1}^{5} = \left[ 25 - \frac{25}{2} \right] - \left[ 5 - \frac{1}{2} \right] = 12.5 - 4.5 = 8 \]
- Total area:
\[ A = \frac{992}{3} - 8 = \frac{992 - 24}{3} = \frac{968}{3} = \frac{320}{3} \]
- Therefore, \(3A = 320\).

Therefore, the correct answer is (1) 368. Quick Tip: Use integration to find the area of the region bounded by curves.

1. Given that \(A\) is an orthogonal matrix:
\[ A^T = A^{-1} \]
\[ A^2 = A^{-1} \]

2. Given \(A^2 = A^T\):
\[ A^3 = I \]

3. Calculate \((A + I)^3 + (A - I)^3 - 6A\):
\[ (A + I)^3 + (A - I)^3 - 6A = 2(A^3 + 3A) - 6A = 2A^3 = 2I \]

4. Sum of the diagonal elements of \(2I\):
\[ 2I = \begin{bmatrix} 2 & 0 & 0
0 & 2 & 0
0 & 0 & 2 \end{bmatrix} \]
\[ Sum of diagonal elements = 2 + 2 + 2 = 6 \]

Therefore, the correct answer is (1) 6. Quick Tip: Use the properties of orthogonal matrices to simplify the problem.

1. Identify the sets \(A\) and \(B\):
- \(A: |z - 2 - i| = 3\)
\[ |(x - 2) + (y - 1)i| = 3 \]
\[ (x - 2)^2 + (y - 1)^2 = 9 \]
- \(B: Re(z - iz) = 2\)
\[ Re((x + y) + i(y - x)) = 2 \]
\[ x + y = 2 \]

2. Solve the system of equations:
\[ \begin{cases} (x - 2)^2 + (y - 1)^2 = 9
x + y = 2 \end{cases} \]
- Substitute \(y = 2 - x\) into the first equation:
\[ (x - 2)^2 + (2 - x - 1)^2 = 9 \]
\[ (x - 2)^2 + (1 - x)^2 = 9 \]
\[ x^2 - 4x + 4 + 1 - 2x + x^2 = 9 \]
\[ 2x^2 - 6x + 5 = 9 \]
\[ 2x^2 - 6x - 4 = 0 \]
\[ x^2 - 3x - 2 = 0 \]
\[ x = \frac{3 \pm \sqrt{17}}{2} \]
- Corresponding \(y\) values:
\[ y = 2 - x = 2 - \frac{3 \pm \sqrt{17}}{2} = \frac{1 \mp \sqrt{17}}{2} \]

3. Calculate \(\sum_{z \in S} |z|^2\):
\[ \sum_{z \in S} |z|^2 = \left( \frac{3 + \sqrt{17}}{2} \right)^2 + \left( \frac{1 - \sqrt{17}}{2} \right)^2 + \left( \frac{3 - \sqrt{17}}{2} \right)^2 + \left( \frac{1 + \sqrt{17}}{2} \right)^2 \]
\[ = \frac{1}{4} \left[ 2 \times 26 + 2 \times 18 \right] = \frac{88}{4} = 22 \]

Therefore, the correct answer is (1) 22. Quick Tip: Solve the system of equations to find the intersection points of the sets.

1. Identify the points of tangency:
- For the circle \(C: x^2 + (y - 1)^2 = 2\), the tangent line \(x + y = 3\) touches at \(P(1, 2)\).

2. Determine the points \(Q\) and \(R\):
- The parametric equation of \(x + y = 3\) is:
\[ \frac{x - 1}{-1/\sqrt{2}} = \frac{y - 2}{1/\sqrt{2}} = \pm \frac{2\sqrt{2}}{3} \]
- Solving for \(Q\) and \(R\):
\[ Q\left( \frac{5}{3}, \frac{4}{3} \right), \quad R\left( \frac{1}{3}, \frac{8}{3} \right) \]

3. Calculate \(9(x_1 y_1 + x_2 y_2 + x_3 y_3)\):
\[ 9(x_1 y_1 + x_2 y_2 + x_3 y_3) = 9 \left( 2 + \frac{5}{3} \cdot \frac{4}{3} + \frac{1}{3} \cdot \frac{8}{3} \right) \]
\[ = 9 \left( 2 + \frac{20}{9} + \frac{8}{9} \right) = 9 \left( 2 + \frac{28}{9} \right) = 9 \left( \frac{34}{9} \right) = 34 \]

Therefore, the correct answer is (1) 46. Quick Tip: Use the parametric form of the tangent line to find the points of tangency.

1. Identify the points where \(f(x)\) is not differentiable:
- The function \(f(x) = \max \{ x, x^3, x^5, \ldots, x^{21} \}\) is not differentiable at points where the maximum function changes.
- These points occur at \(x = -1, 0, 1\).

2. Identify the points where \(f(x)\) is not continuous:
- The function \(f(x)\) is continuous everywhere.

3. Calculate \(m + n\):
\[ m = 3, \quad n = 0 \]
\[ m + n = 3 \]

Therefore, the correct answer is (3) 3. Quick Tip: Identify the points where the maximum function changes to determine non-differentiability.