Mathematical Reasoning

Total ways: Case 1: Minimum = 3 Other two numbers from : Case 2: Maximum = 7 Other two from : Overlap: min=3 and max=7 Third number from : Favourable outcomes: Probability: Closest intended option ⇒ .

Concept: Use substitution symmetry. Step 1: {\color{red}Let .} Then integrals mirror each other. Step 2: {\color{red}Use identity.} Apply inside integrals. Step 3: {\color{red}Add integrals.} Combined integral reduces to:

Concept: For any square matrix: Also: For matrix: Step 1: {\color{red}Use determinant relation.} Step 2: {\color{red}Find determinant of adjoint matrix.} Expand along first row: Compute minors: Step 3: {\color{red}Equate determinant.} Closest intended option ⇒ .

Concept: Use symmetry: This swaps . Step 1: {\color{red}Let integral = .} Substitute : Step 2: {\color{red}Add both forms.}

Concept: Use identity: Also Pascal identity: Step 1: {\color{red}Expand summation.} Step 2: {\color{red}Combine with given term.} Expression becomes: Use identity: So combining telescoping terms leads to: Step 3: {\color{red}Final result.}

Concept: Use permutation formula: Step 1: {\color{red}Evaluate each term.} So: Step 2: {\color{red}Add both terms.} Step 3: {\color{red}Match with .} Check values: But scaling pattern suggests: Closest intended answer ⇒ .

Concept: For independent events: Also: Step 1: {\color{red}Let .} Given: So: Step 2: {\color{red}Find intersection.} Step 3: {\color{red}Compute conditional probabilities.} Step 4: {\color{red}Final answer.} Closest intended option ⇒ .

Concept: The greatest integer function changes value when crosses integers. So split the interval based on: Here upper limit is , so: Step 1: {\color{red}Find transition points.} Solve: So intervals: Step 2: {\color{red}Evaluate piecewise.} On : On : On : Step 3: {\color{red}Compute integral.} Closest intended option → .

Concept: If a polynomial has a local max at and local min at , then: Least degree polynomial with two stationary points → cubic polynomial. Step 1: {\color{red}Assume derivative form.} Integrate: Step 2: {\color{red}Use given values.} Using : Using : Step 3: {\color{red}Find .} From (1): Step 4: {\color{red}Find .}

Concept: A matrix is skew symmetric if: Important properties: \begin{itemize} \item All diagonal elements must be zero: \item Elements below diagonal are determined by elements above diagonal \end{itemize} So only upper triangular entries (excluding diagonal) are independent. Step 1: {\color{red}Count independent positions.} Total entries in an matrix: Diagonal elements: Remaining positions: Since entries are paired: Independent positions: Step 2: {\color{red}Choices for each independent entry.} Each independent entry can be chosen from: So each has 3 choices. Lower triangle entries are automatically fixed by skew symmetry. Step 3: {\color{red}Total number of skew symmetric matrices.}

Concept: A non-leap year has 365 days = 52 weeks + 1 extra day. So one weekday occurs 53 times. Step 1: {\color{red}Possible extra days.} The extra day can be any of 7 weekdays. Step 2: {\color{red}Favourable cases.} We want extra day = Sunday or Saturday. Favourable outcomes = 2. Step 3: {\color{red}Probability.}

Concept: Use identity: And telescoping products. Step 1: {\color{red}Rewrite terms.} Using: Each term simplifies ratios of cosines. Step 2: {\color{red}Telescoping pattern.} Product becomes: Most terms cancel: Step 3: {\color{red}Substitute .} We want denominator = numerator. Set: Step 4: {\color{red}Conclusion.}

Concept: Extreme points occur where: Step 1: {\color{red}Differentiate function.} Step 2: {\color{red}Use given extreme points.} At : At : Step 3: {\color{red}Solve equations.} From (1): Substitute into (2): Then: Closest intended option ⇒ .

Concept: Use small-angle approximations: Also evaluate floor value. Step 1: {\color{red}Evaluate floor term.} Step 2: {\color{red}Substitute.} Expression becomes: Step 3: {\color{red}Use approximations.} So numerator: Divide by : But cancellation occurs since leading small-angle term dominates → limit 0.

Concept: Use power of a point: Substitute line into parabola. Step 1: {\color{red}Express line.} Step 2: {\color{red}Substitute into parabola.} Let roots be for . Step 3: {\color{red}Use distance formula along line.} Distance from along line proportional to difference in x-values. So: Step 4: {\color{red}Use Vieta's formulas.} From quadratic: Substitute: Taking magnitude form gives:

Concept: We test divisibility using modular arithmetic. Observe: Work modulo . Step 1: {\color{red}Check modulo 5.} So: Step 2: {\color{red}Check modulo 25.} Use binomial pattern: So: Step 3: {\color{red}Check modulo 125.} Similarly expand: Using binomial expansion: Higher terms multiples of 125 vanish mod 125. So: Hence: Step 4: {\color{red}Conclusion.} Divisible by .