Solution:
We are given two equations: \[ x^2 + x + a = 0 \quad \text{and} \quad x^2 + ax + 1 = 0, \] with a common root \( \alpha \). Substituting \( \alpha \) into both equations: 1. From \( x^2 + x + a = 0 \): \( \alpha^2 + \alpha + a = 0 \), 2. From \( x^2 + ax + 1 = 0 \): \( \alpha^2 + a\alpha + 1 = 0 \). Subtracting these equations: \[ (a - 1)(\alpha - 1) = 0. \] This gives two cases: - If \( a = 1 \), the first equation has no real roots since \( \Delta < 0 \). - If \( \alpha = 1 \), substituting into the first equation gives \( a = -2 \). For \( a = -2 \), \( x = 1 \) is a common root. Thus, \( a = -2 \) is the only solution in the interval \( [-3, -1] \). Therefore, the correct answer is **Option C**.

Solution:
Centre and radius of a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) are \( (-g, -f) \) and \( \sqrt{g^2 + f^2 - c} \), respectively.
For \( C_1 \): Centre = \( (2, 2) \), Radius = \( \sqrt{2} \).
For \( C_2 \): Centre = \( (5, 5) \), Radius = \( \sqrt{50 - k} \).
Distance between the centres = \( \sqrt{(5 - 2)^2 + (5 - 2)^2} = \sqrt{18} \).

According to the properties of circles:
If two circles have exactly two common tangents, then \[ |\text{Difference of radii}| < \text{Distance between centres} < |\text{Sum of radii}| \] Substituting: \[ \sqrt{50 - k} - \sqrt{2} < \sqrt{18} < \sqrt{50 - k} + \sqrt{2}. \]
Analyzing options: - **Option A**: For \( k = 17 \), \( |\sqrt{7} - \sqrt{2}| < \sqrt{18} < |\sqrt{7} + \sqrt{2}| \). Does not satisfy. - **Option B**: For \( k = 25 \), \( |\sqrt{5} - \sqrt{2}| < \sqrt{18} < |\sqrt{5} + \sqrt{2}| \). Satisfies. - **Option C**: For \( k = 49 \), \( |\sqrt{1} - \sqrt{2}| < \sqrt{18} < |\sqrt{1} + \sqrt{2}| \). Does not satisfy.

Therefore, the correct answer is **Option B**.

Solution:
The given function is: \[ f(x) = \begin{cases} 2x - 1 & \text{if } x < -1 \\ x^2 + 1 & \text{if } -1 \leq x \leq 1 \\ x + 1 & \text{if } x > 1 \end{cases} \]
**At \( x = -1 \):**
\[ \lim_{x \to -1^-} (2x - 1) = 2(-1) - 1 = -3 \] \[ \lim_{x \to -1^+} (x^2 + 1) = (-1)^2 + 1 = 2 \] Since the left-hand limit (\(-3\)) and the right-hand limit (\(2\)) are not equal, \( f(x) \) is **discontinuous at \( x = -1 \)**.

**At \( x = 1 \):**
\[ \lim_{x \to 1^-} (x^2 + 1) = (1)^2 + 1 = 2 \] \[ \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2 \] Since the left-hand limit and the right-hand limit are equal, \( f(x) \) is **continuous at \( x = 1 \)**.

Therefore, \( f(x) \) is continuous everywhere except at \( x = -1 \).
The correct answer is **Option C**.

Solution:
The given series is \( 3, 7, 13, 21, 31, 43, \ldots \), and the differences between consecutive terms are: \[ 4, 6, 8, 10, \ldots \] which form an arithmetic progression with a common difference of 2.

Let the \( n \)-th term of the series be \( T_n = an^2 + bn + c \). Using the first three terms: - For \( n=1, T_1=3 \): \( a(1)^2 + b(1) + c = 3 \), so \( a + b + c = 3 \). - For \( n=2, T_2=7 \): \( a(2)^2 + b(2) + c = 7 \), so \( 4a + 2b + c = 7 \). - For \( n=3, T_3=13 \): \( a(3)^2 + b(3) + c = 13 \), so \( 9a + 3b + c = 13 \).

Solving these equations gives: \[ a = 1, \, b = 1, \, c = 1 \] Hence, the general term is: \[ T_n = n^2 + n + 1 \] The sum of the first \( n \) terms is: \[ S_n = \sum_{n=1}^N T_n = \sum_{n=1}^N (n^2 + n + 1) \] Using the sum of squares and the sum of integers, we get: \[ S_n = \frac{N^3 + 3N^2 + 5N}{3} \] Substituting \( N = 50 \): \[ S_{50} = \frac{50^3 + 3(50)^2 + 5(50)}{3} = 50 \times 885 \] Therefore, the final answer is **D: \( 50 \times 885 \)**.

Solution:
The expression for \( A_n \) is: \[ A_n = \frac{\sum_{k=1}^n k(k+1)(k+2)}{n \cdot \sum_{k=1}^n k(k+1)} \] Numerator:
Expanding \( \sum_{k=1}^n k(k+1)(k+2) \): \[ \sum_{k=1}^n k(k+1)(k+2) = \sum_{k=1}^n \left(k^3 + 3k^2 + 2k\right) \] Using summation formulas: - \( \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2 \sim \frac{n^4}{4} \) (dominant term), - \( \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \sim \frac{n^3}{3} \) (dominant term), - \( \sum_{k=1}^n k = \frac{n(n+1)}{2} \sim \frac{n^2}{2} \). Thus: \[ \sum_{k=1}^n k(k+1)(k+2) \sim \frac{n^4}{4}. \] Denominator:
Expanding \( \sum_{k=1}^n k(k+1) \): \[ \sum_{k=1}^n k(k+1) = \sum_{k=1}^n \left(k^2 + k\right) \] Using summation formulas: - \( \sum_{k=1}^n k^2 \sim \frac{n^3}{3} \), - \( \sum_{k=1}^n k \sim \frac{n^2}{2} \). Thus: \[ \sum_{k=1}^n k(k+1) \sim \frac{n^3}{3}. \] Multiplying by \( n \): \[ n \cdot \sum_{k=1}^n k(k+1) \sim \frac{n^4}{3}. \] Final Simplification:
For large \( n \): \[ A_n = \frac{\frac{n^4}{4}}{\frac{n^4}{3}} = \frac{3}{4}. \] Therefore: \[ \lim_{n \to \infty} A_n = \frac{3}{4}. \] The correct answer is **A**.

Solution:
Step 1: Differentiate \( f(x) \) to Analyze Monotonicity
To determine if the function is one-to-one, we examine its derivative: \[ f'(x) = 3x^2 - 6x + 6 = 3\left((x-1)^2 + 1\right) \] Since \( (x-1)^2 + 1 > 0 \) for all \( x \in \mathbb{R} \), we conclude: \[ f'(x) > 0 \quad \forall x \in \mathbb{R} \] This means that \( f(x) \) is strictly increasing on \( \mathbb{R} \), so the function is one-to-one (injective).
Step 2: Analyze the Range of \( f(x) \)
Since \( f(x) \) is a cubic polynomial and is strictly increasing, its range is: \[ f(x) \in (-\infty, \infty) \] Thus, \( f(x) \) is onto (surjective) because every real number is covered in the range.
Step 3: Conclusion
- The function \( f(x) \) is one-to-one because it is strictly increasing.
- The function \( f(x) \) is onto because its range covers all real numbers.
Therefore, the correct answer is one-to-one and onto.

Solution:
The word 'PROBABILITY' contains the following letters: \[ \text{P, R, O, B, A, B, I, L, I, T, Y} \] Removing the vowels (O, A, I, I), we are left with the consonants: \[ \text{P, R, B, B, L, T, Y} \] The total number of permutations of these letters, accounting for the repetition of B (which appears twice), is given by: \[ \frac{7!}{2!} = \frac{5040}{2} = 2520 \] Hence, the number of distinct words that can be formed is \( \mathbf{2520} \).
Final Answer: A.

Solution:
The area enclosed between the two curves \( y_1 = f(x) \) and \( y_2 = g(x) \) is calculated as: \[ A = \left| \int_{x_1}^{x_2} \left( f(x) - g(x) \right) dx \right| \] Here, \( f(x) = 6 \), \( g(x) = 2x^2 \), and the limits of integration are found by solving \( 6 = 2x^2 \): \[ x^2 = 3 \quad \Rightarrow \quad x = \pm \sqrt{3} \] Substituting these values, the area becomes: \[ A = \left| \int_{-\sqrt{3}}^{\sqrt{3}} \left(6 - 2x^2 \right) dx \right| \] Compute the integral: \[ A = \left| \left[ 6x - \frac{2x^3}{3} \right]_{-\sqrt{3}}^{\sqrt{3}} \right| \] Evaluate the definite integral: \[ A = \left| \left[ 6\sqrt{3} - \frac{2(\sqrt{3})^3}{3} \right] - \left[ -6\sqrt{3} + \frac{2(-\sqrt{3})^3}{3} \right] \right| \] Simplify: \[ A = |[-4\sqrt{3}] - [4\sqrt{3}]| = 8\sqrt{3} \] Final Answer: \( \mathbf{8\sqrt{3}} \).
Correct Option: D.

Solution:
We know that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Using this property: \[ \lim_{x \to 0} \frac{\sin(x^2)}{x \sin x} = \lim_{x \to 0} \left( \frac{\sin(x^2)}{x^2} \cdot \frac{x}{\sin x} \right) \] - For the first term: \( \lim_{x \to 0} \frac{\sin(x^2)}{x^2} = 1 \) because \( \sin(x^2) \approx x^2 \) as \( x \to 0 \). - For the second term: \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \) because \( \sin x \approx x \) as \( x \to 0 \). Combining these: \[ \lim_{x \to 0} \frac{\sin(x^2)}{x \sin x} = 1 \cdot 1 = 1 \] Final Answer: \( \mathbf{1} \).
Correct Option: B.

Solution:
Using the formula for binomial coefficients: \[ (1 + x)^n = \binom{n}{0} \cdot 1^n + \binom{n}{1} \cdot x + \binom{n}{2} \cdot x^2 + \ldots + \binom{n}{n} \cdot x^n \] On integrating: \[ \frac{(1 + x)^{n+1}}{n+1} = \frac{\binom{n}{0}}{1} + \frac{\binom{n}{1} \cdot x}{2} + \ldots + \frac{\binom{n}{n} \cdot x^n}{n+1} \] Substituting \( x = 1 \): \[ \frac{2^{31}}{31} = \frac{\binom{30}{1}}{2} + \frac{\binom{30}{3}}{4} + \ldots + \frac{\binom{30}{29}}{30} + \frac{\binom{30}{30}}{31} \] Substituting \( x = -1 \): \[ 0 = \frac{\binom{30}{0}}{1} - \frac{\binom{30}{1}}{2} + \frac{\binom{30}{2}}{3} - \frac{\binom{30}{3}}{4} + \ldots - \frac{\binom{30}{30}}{31} \] Adding the two equations: \[ \frac{2^{31}}{31} = 2 \left( \frac{\binom{30}{1}}{2} + \frac{\binom{30}{3}}{4} + \ldots + \frac{\binom{30}{29}}{30} \right) \] Simplifying: \[ \frac{\binom{30}{1}}{2} + \frac{\binom{30}{3}}{4} + \ldots + \frac{\binom{30}{29}}{30} = \frac{2^{30} - 1}{31} \] Final Answer: \( \frac{2^{30} - 1}{31} \).
Correct Option: D.

Solution:
Join \( BD \). The area of quadrilateral ABCD is the sum of areas of triangles ABD and BCD.
In \( \triangle ABD \), applying Pythagoras' Theorem: \[ AB^2 + AD^2 = BD^2 \quad \Rightarrow \quad 24^2 + 18^2 = BD^2 \quad \Rightarrow \quad BD = 30 \] Also, in \( \triangle BCD \), applying Pythagoras' Theorem: \[ BD^2 + BC^2 = CD^2 \quad \Rightarrow \quad 30^2 + 40^2 = 50^2 \] This confirms that \( \triangle BCD \) is a right-angled triangle.
Now, calculate the areas: - Area of \( \triangle ABD \) is: \[ \text{Area of } \triangle ABD = \frac{1}{2} \times AB \times AD = \frac{1}{2} \times 24 \times 18 = 216 \] - Area of \( \triangle BCD \) is: \[ \text{Area of } \triangle BCD = \frac{1}{2} \times BD \times BC = \frac{1}{2} \times 30 \times 40 = 600 \] Therefore, the total area of quadrilateral ABCD is: \[ \text{Area of ABCD} = 216 + 600 = 816 \] Final Answer: 816 \( \text{cm}^2 \).
Correct Option: C.

Solution:
The expression is: \[ \cot\left(\frac{\pi}{40}\right) \cot\left(\frac{2\pi}{40}\right) \ldots \cot\left(\frac{19\pi}{40}\right) \] Now, we know that: \[ \cot A = \tan\left(\frac{\pi}{2} - A\right) \] Thus, \[ \cot \frac{\pi}{20} = \tan\left(\frac{\pi}{2} - \frac{\pi}{40}\right) = \tan \frac{19\pi}{40} \] Also, \[ (\cot A)(\tan A) = 1 \] Hence, all the terms other than \( \cot \frac{10\pi}{40} \) pair up and become 1. Additionally, \[ \cot \frac{10\pi}{40} = \cot \frac{\pi}{4} = 1 \] Therefore, the value of the expression is: \[ 1 \] Final Answer: 1.
Correct Option: A.

Solution:
Consider the function \( f(x) = |2 - |x - 1|| \). We analyze it piecewise as follows: \[ |x - 1| = \begin{cases} x - 1 & \text{if } x \geq 1 \\ 1 - x & \text{if } x < 1 \end{cases} \] Hence, \( f(x) \) will be defined piecewise based on the value of \( x \): \[ f(x) = \begin{cases} |x + 1| & \text{if } x < 1 \\ |3 - x| & \text{if } x \geq 1 \end{cases} \] The derivative \( f'(x) \) will depend on the piecewise definition, and the values for \( f'(x) \) at specific points are: \[ f^{\prime}(-2) = -1, \quad f^{\prime}(0) = 1, \quad f^{\prime}(2) = 1, \quad f^{\prime}(4) = -1 \] Therefore, \[ f^{\prime}(-2) + f^{\prime}(0) + f^{\prime}(2) + f^{\prime}(4) = -1 + 1 + 1 - 1 = 0 \] Hence, the correct answer is: \[ \boxed{0} \] Final Answer: B.
Correct Option: B.

Solution:
Given matrix \( P \): \[ P = \begin{bmatrix} a & b & 0 \\ -1 & 2 & 1 \\ 2 & -3 & -2 \end{bmatrix} \] The determinant \( \det(P) = -2 \). We calculate the minor \( M_{22} \) of \( P \) by eliminating the second row and second column: \[ M_{22} = \det \left[ \begin{array}{cc} a & 0 \\ 2 & -2 \end{array} \right] \] The determinant of this \( 2 \times 2 \) matrix is: \[ \det \left[ \begin{array}{cc} 2 & 0 \\ 2 & -2 \end{array} \right] = 2(-2) - 0(2) = -4 \] Therefore, the minor \( M_{22} \) is \( -4 \). Final Answer: A.
Correct Option: A.

Solution:
Consider the equation \( x^3 - 1 = 0 \). This can be factored as: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) = 0 \] The roots are \( 1, \alpha, \beta \), where \( \alpha \) and \( \beta \) are the roots of \( x^2 + x + 1 = 0 \). Since \( \alpha^3 = 1 \) and \( \beta^3 = 1 \), it follows that: \[ \alpha^{2017} = (\alpha^3)^{672} \cdot \alpha = \alpha, \quad \beta^{2017} = (\beta^3)^{672} \cdot \beta = \beta \] Thus: \[ \alpha^{2017} + \beta^{2017} = \alpha + \beta \] From the properties of the roots of the equation \( x^3 - 1 = 0 \), we know that: \[ 1 + \alpha + \beta = 0 \Rightarrow \alpha + \beta = -1 \] Hence, the value of \( \alpha^{2017} + \beta^{2017} = -1 \). Final Answer: C.
Correct Option: C.

Solution:
Assume \(x = a + 1\), \(y = b + 1\), and \(z = c + 1\), where \(a\), \(b\), and \(c\) are non-negative integers. Substituting in the equation: \[ (a+1) + (b+1) + (c+1) = 10 \quad \Rightarrow \quad a + b + c = 7 \] The number of non-negative integer solutions for \(a\), \(b\), and \(c\) is given by the formula for the number of solutions to the equation: \[ \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} = \frac{9 \times 8}{2} = 36 \] Therefore, the number of different solutions is \(36\). The correct answer is A.
Final Answer: A.
Correct Option: A.

Solution:
We start with the equation: \[ x^2 - y^2 = 2y + 1 \] Rearranging terms, we get: \[ x^2 = y^2 + 2y + 1 \] Which can be rewritten as: \[ x^2 - (y + 1)^2 = 0 \] Factoring the difference of squares, we get: \[ (x - y - 1)(x + y + 1) = 0 \] This represents two straight lines: \[ x - y - 1 = 0 \quad \text{and} \quad x + y + 1 = 0 \] Therefore, the equation represents a pair of straight lines. The correct answer is D.
Final Answer: D

Solution:
Let the number of students who failed in exactly 1 subject be \( x \),
the number of students who failed in exactly 2 subjects be \( y \),
and the number of students who failed in all 3 subjects be \( z \).
We have the following system of equations:

            x + 2y + 3z = 50 + 45 + 40 = 135
            x + y + z = 100 - 1 = 99
            

Subtracting the second equation from the first:

            y + 2z = 36
            

It is given that \( y = 32 \), so:

            2z = 36 - 32 = 4
            z = 2
            

Thus, the number of students failing in all three subjects is 2.
Final Answer: C

Solution:
To make a regular hexagon, three equilateral triangles of side \( \frac{a}{3} \) will be cut from the corners to form a regular hexagon of side \( a \).
Hence, this hexagon can be further divided into 6 equilateral triangles of side \( \frac{a}{3} \).
The area of the hexagon is:
\[ 6 \cdot \frac{\sqrt{3}}{4} \cdot \left( \frac{a}{3} \right)^2 = \frac{\sqrt{3} a^2}{6} \]
Hence, the answer is \( A \).

Solution:
For the function to be differentiable at \( x = 0 \), the derivative must satisfy: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] Now using the value of \( f(x) \), we get: \[ f'(0) = \lim_{h \to 0} \frac{a_0 + a_1|h| + a_2|h|^2 + a_3|h|^3 - a_0}{h} \] Simplifying further: \[ f'(0) = \lim_{h \to 0} \frac{a_1|h| + a_2|h|^2 + a_3|h|^3}{h} \] For \( h < 0 \), \( f'(0) = -a_1 + a_2 h - a_3 h^2 \).
For \( h > 0 \), \( f'(0) = a_1 + a_2 h + a_3 h^2 \).
For the two expressions to match as \( h \to 0 \), it must be that \( a_1 = 0 \). Hence, the correct answer is \( C \).

Solution:
The number of ways to select a non-empty set from \( S = \{1, 2, \dots, 100\} \) is: \[ \binom{100}{1} + \binom{100}{2} + \dots + \binom{100}{100} = 2^{100} - 1 \] Similarly, the number of ways to select a non-empty set from the odd numbers in \( S \) (i.e., \( P = \{1, 3, 5, \dots, 99\} \)) is: \[ 2^{50} - 1 \] Hence, the required number of sets is: \[ 2^{100} - 1 - (2^{50} - 1) = 2^{50}(2^{50} - 1) \] Therefore, the correct answer is \( A \).