JEE Mathematics Miscellaneous Questions MCQs Set 04

Answer: (c) \( 2(\sqrt2 - 1)\,m^2 \)

Explanation:
1. If each corner is a right isosceles triangle with legs \( x \), the octagon side is \( x\sqrt2 \) and \( 2x + x\sqrt2 = 1 \), so \( x = \frac1{2 + \sqrt2} \).
2. Area \( = 1 - 4\cdot\frac{x^2}2 = 1 - \frac{2}{(2+\sqrt2)^2} = 2\sqrt2 - 2 \).

Teacher's Note:
1. Express the cut-off corners in terms of x.

Answer: (d) \( \frac\pi4 \)

Explanation:
1. Let \( x = \cos\theta \) with \( 0 \lt \theta \lt \frac\pi4 \). The second term is \( \cos^{-1}\cos(\theta - \frac\pi4) = \frac\pi4 - \theta \).
2. The sum is \( \theta + \frac\pi4 - \theta = \frac\pi4 \).

Teacher's Note:
1. Substitute x = cos θ.

Answer: (d) \( {}^{2n+1}C_n \)

Explanation:
1. By the hockey-stick identity \( \sum_{r=0}^n\binom{n+r}r = \binom{2n+1}n \) (check \( n = 1 \): \( 1 + 2 = 3 = \binom31 \)).

Teacher's Note:
1. Hockey stick identity.

Answer: (b) 0

Explanation:
1. \( \mathrm{Re}(w) = \frac{|z|^2 - 1}{|z + 1|^2} = 0 \) because \( |z| = 1 \) (so \( w \) is purely imaginary).

Teacher's Note:
1. Multiply by the conjugate.

Answer: (a) \( \frac\pi4 \)

Explanation:
1. \( \tan^{-1}\frac{2r}{2 + r^2 + r^4} = \tan^{-1}(r^2 + r + 1) - \tan^{-1}(r^2 - r + 1) \), since \( 1 + (r^2+r+1)(r^2-r+1) = 2 + r^2 + r^4 \).
2. The series telescopes to \( \frac\pi2 - \tan^{-1}1 = \frac\pi4 \).

Teacher's Note:
1. Write each term as a difference of arctangents.

Answer: (a) \( \frac{3}{20} \)

Explanation:
1. Triples of distinct numbers from 1-6 with sum 12: \( \{1,5,6\}, \{2,4,6\}, \{3,4,5\} \), i.e. \( 3\times3! = 18 \) ordered triples.
2. Total ordered triples \( = 6\cdot5\cdot4 = 120 \). Probability \( = \frac{18}{120} = \frac3{20} \).

Teacher's Note:
1. Count ordered triples of distinct numbers.

Answer: (b) 1

Explanation:
1. The numerator is \( g(\alpha) \) and the denominator \( g(\beta) \) where \( g(x) = x^2 + px - r \).
2. Since \( \alpha^2 + p\alpha = -q \) and likewise for \( \beta \), both equal \( -q - r \). The ratio is 1.

Teacher's Note:
1. Write the products as values of the second polynomial.

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. With roots \( \alpha, 2\alpha \): \( 3\alpha = -\frac{3p-1}{D} \) and \( 2\alpha^2 = \frac2D \) with \( D = p^2 - 5p + 3 \).
2. Eliminating \( \alpha \): \( 9D = (3p - 1)^2 \), so \( -45p + 27 = -6p + 1 \) and \( p = \frac23 \).

Teacher's Note:
1. Use the sum and product of the roots.

Answer: (c) x + 3

Explanation:
1. On \( (3, 4) \), \( [x] = 3 \), so \( f(x) = x - 3 \) and the inverse is \( x + 3 \).

Teacher's Note:
1. Find [x] on the interval.

Answer: (b) 2

Explanation:
1. The terms are \( |x - 2|, |x|, |x + 2| \). On \( [-2, 2] \) the first and third add to 4, so \( 2|x| = 4 \) and \( x = \pm2 \).

Teacher's Note:
1. Use the AP condition 2·middle = sum of ends.

Answer: (a) \( {}^{n+2}C_{r+1} \)

Explanation:
1. Group as \( ({}^nC_{r+1} + {}^nC_r) + ({}^nC_r + {}^nC_{r-1}) = {}^{n+1}C_{r+1} + {}^{n+1}C_r = {}^{n+2}C_{r+1} \).

Teacher's Note:
1. Apply Pascal's rule twice.

Answer: (a) 8

Explanation:
1. \( \frac{n(n-3)}2 \geq n + 10 \) gives \( n^2 - 5n - 20 \geq 0 \), so \( n \geq 7.6 \).
2. The least integer is 8.

Teacher's Note:
1. Number of diagonals is n(n-3)/2.

Answer: (d) 0 or -3

Explanation:
1. The determinant of the direction vectors and the vector joining the two points: \( \begin{vmatrix}1&1&-k\\k&2&1\\-1&1&1\end{vmatrix} = -k^2 - 3k = 0 \).
2. So \( k = 0 \) or \( -3 \).

Teacher's Note:
1. Use the coplanarity determinant.

Answer: (c) \( z \in \left(-\frac12, 2\right) \)

Explanation:
1. \( y = 2x + 1 \), so \( z = \frac{2 - x}{2x + 1} \), which decreases from 2 (as \( x \to 0 \)) towards \( -\frac12 \) (as \( x \to \infty \)).
2. So \( z \in \left(-\frac12, 2\right) \).

Teacher's Note:
1. Express z in terms of x.

Answer: (c) 8

Explanation:
1. Orders \( m\times n \) with \( mn = 24 \) correspond to the divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24, i.e. 8.

Teacher's Note:
1. Count divisors of 24.

Answer: (d) \( 2n\pi + \frac{5\pi}6 \)

Explanation:
1. \( \sin\theta = \frac12 \) gives \( \frac\pi6 \) or \( \frac{5\pi}6 \); \( \cos\theta = -\frac{\sqrt3}2 \) gives \( \frac{5\pi}6 \) or \( \frac{7\pi}6 \).
2. The common solution is \( 2n\pi + \frac{5\pi}6 \).

Teacher's Note:
1. Intersect the two solution sets.

Answer: (b) \( -\frac{e^2}2 + e - \frac32 \)

Explanation:
1. \( f(x) = e^x \) and \( g(x) = x^2 - e^x \).
2. \( \int_0^1x^2e^xdx = e - 2 \) and \( \int_0^1e^{2x}dx = \frac{e^2 - 1}2 \), so the value is \( e - 2 - \frac{e^2 - 1}2 = -\frac{e^2}2 + e - \frac32 \).

Teacher's Note:
1. Solve for f and g first.

Answer: (b) \( (-\infty, 2) \)

Explanation:
1. \( [x + [x]] = 2[x] \), so \( 2[x] \leq 2 \) and \( [x] \leq 1 \).
2. So \( x \lt 2 \).

Teacher's Note:
1. [x + n] = [x] + n for an integer n.

Answer: (c) inclined at \( \cos^{-1}\frac{8 + t}{9\sqrt{1 + t^2}} \)

Explanation:
1. \( \vec f(\frac54) = (1, \frac14, 2) \) with magnitude \( \frac94 \). For \( 0 \lt t \lt 1 \): \( \vec f(t) = (0, t, 1) \) with magnitude \( \sqrt{1 + t^2} \).
2. Dot product \( = \frac t4 + 2 \), so \( \cos\theta = \frac{t + 8}{9\sqrt{1 + t^2}} \).

Teacher's Note:
1. Evaluate both vectors.

Answer: (b) \( \left(0, \frac34\right) \)

Explanation:
1. \( x^2 = \frac k2 y - \frac32 \), so the latus rectum is \( \frac k2 = 2 \), giving \( k = 4 \).
2. The vertex is where \( x = 0 \): \( y = \frac3k = \frac34 \).

Teacher's Note:
1. Write in the form x² = 4a(y - y₀).

Answer: (c) \( n + \frac1n \)

Explanation:
1. Near 0 the numerator behaves like \( [(a - n)n - 1]x\cdot nx \), so the limit is \( n[(a - n)n - 1] \).
2. Setting this to 0: \( a = n + \frac1n \).

Teacher's Note:
1. Use small-angle approximations.

Answer: (a) a > 2

Explanation:
1. The parabola opens upward, so \( 1 \) lies between the roots exactly when \( f(1) \lt 0 \): \( 1 - 2a + 3 \lt 0 \).
2. So \( a > 2 \).

Teacher's Note:
1. Use the sign of f(1).

Answer: (c) 1

Explanation:
1. \( \cos^7x \leq \cos^2x \) and \( \sin^5x \leq \sin^2x \), so the left side is at most 1, with equality only if \( \cos x \in \{0, 1\} \) and \( \sin x \in \{0, 1\} \).
2. In the open interval only \( x = \frac\pi2 \) qualifies.

Teacher's Note:
1. Compare with cos² + sin².

Answer: (c) \( |c| > \sqrt2|b| \)

Explanation:
1. \( \min f = 2c^2 - b^2 \) and \( \max g = b^2 + c^2 \).
2. \( 2c^2 - b^2 > b^2 + c^2 \) gives \( c^2 > 2b^2 \), i.e. \( |c| > \sqrt2|b| \).

Teacher's Note:
1. Find the vertex values.

Answer: (d) \( \left(\frac53, \frac13\right) \)

Explanation:
1. \( PQ \) is vertical, so \( PR \) is horizontal: \( R \) has \( y = 1 \) and lies on \( 2x + y = 3 \), so \( R = (1, 1) \).
2. Centroid \( = \left(\frac{2 + 2 + 1}3, \frac{1 - 1 + 1}3\right) = \left(\frac53, \frac13\right) \).

Teacher's Note:
1. Find R from the right angle.

Answer: (c) 0

Explanation:
1. Substitute \( x = \frac1u \): the integrand changes to \( -\frac{\sin(u - 1/u)}{u} \) with the limits reversed, so \( I = -I \).
2. Hence \( I = 0 \).

Teacher's Note:
1. Try x → 1/x.

Answer: (a) -1

Explanation:
1. \( |\vec a + \vec b + \vec c|^2 = 3 + 2(\cos\alpha + \cos\beta + \cos\gamma) = 1 \).
2. So the sum is \( -1 \).

Teacher's Note:
1. Square the magnitude.

Answer: (d) 5

Explanation:
1. The line \( x = -6 \) is at distance 9 from \( A \). On \( x = 6 \): \( 9 + (y - 4)^2 \leq 16 \) gives \( |y - 4| \leq 2 \) (as \( \sqrt7 \lt 3 \)), so \( y = 2, 3, 4, 5, 6 \).

Teacher's Note:
1. Only the line x = 6 is close enough.

Answer: (c) \( \frac{13\times{}^4C_2\times{}^{12}C_2\times4^2}{{}^{52}C_4} \)

Explanation:
1. Choose the number of the pair (13), the 2 suits (\( {}^4C_2 \)), then two other different numbers (\( {}^{12}C_2 \)) with any suit each (\( 4^2 \)).
2. Divide by \( {}^{52}C_4 \).

Teacher's Note:
1. Pick the pair first, then the two singles.

Answer: (a) \( f'(t) + f'''(t) \)

Explanation:
1. \( \dot x = (f' + f''')\cos t \) and \( \dot y = -(f' + f''')\sin t \).
2. Speed \( = \sqrt{\dot x^2 + \dot y^2} = |f' + f'''| \).

Teacher's Note:
1. Differentiate x and y.