JEE Mathematics Miscellaneous Questions MCQs Set 05

Answer: (b) \( f(x) = \begin{cases}\frac12 - x, & x \lt \frac12\\ \left(\frac12 - x\right)^2, & x \geq \frac12\end{cases} \)

Explanation:
1. In (a), \( f \) is continuous at \( \frac12 \) but the left derivative is \( -1 \) and the right derivative is \( 0 \), so it is not differentiable at \( \frac12 \).
2. The functions in (b), (c), (d) are continuous on \( [0,1] \) and differentiable on \( (0,1) \).

Teacher's Note:
1. Check differentiability at the break point.

Answer: (d) 2

Explanation:
1. The centres are \( (k, 0) \) and \( (2, 4) \); the radii are \( |k| \) and \( 2 \).
2. External touching: \( \sqrt{(k-2)^2 + 16} = |k| + 2 \), giving \( 16 - 4k = 4|k| \), so \( k = 2 \).

Teacher's Note:
1. Distance between centres equals the sum of radii.

Answer: (a) \( \frac{3\pi}4 \)

Explanation:
1. \( \tan(A+B) = \frac{5}{1 - 6} = -1 \) and the sum exceeds \( \frac\pi2 \), so the sum is \( \pi - \frac\pi4 = \frac{3\pi}4 \).

Teacher's Note:
1. Use the addition formula, then check the quadrant.

Answer: (a) 3

Explanation:
1. With \( b = 2c \): \( \tan\frac{B-C}2 = \frac{b - c}{b + c}\cot\frac A2 = \frac13\cot\frac A2 \).
2. So \( \cot\frac A2\cot\frac{B-C}2 = \cot\frac A2\cdot3\tan\frac A2 = 3 \).

Teacher's Note:
1. Use Napier's analogy.

Answer: (c) 3

Explanation:
1. \( \vec n \) is perpendicular to both \( \hat{i} + \hat{j} \) and \( \hat{i} - \hat{j} \), so \( \vec n = \pm\hat{k} \).
2. \( |\vec w\cdot\vec n| = 3 \).

Teacher's Note:
1. n is along the z-axis.

Answer: (d) \( \frac{32}{3} \)

Explanation:
1. The determinant is upper triangular: \( x\cdot1\cdot2\cdot3 = 6x \).
2. Arrangements: \( 4 + 12 + 24 + 24 = 64 \), so \( x = \frac{64}{6} = \frac{32}3 \).

Teacher's Note:
1. Triangular determinant is the product of the diagonal.
2. Count arrangements for 1, 2, 3, 4 things.

Answer: (c) \( \frac{1}{4\sqrt x} \)

Explanation:
1. \( f(x)f'(x) = \frac12\frac{d}{dx}f(x)^2 = \frac12\cdot\frac1{2\sqrt x} = \frac1{4\sqrt x} \).

Teacher's Note:
1. ff' is half the derivative of f².

Answer: (b) 6

Explanation:
1. The greatest coefficient is the middle one, \( {}^{2n}C_n \), so \( n = 6 \).

Teacher's Note:
1. The middle term is the greatest.

Answer: (a) \( \mathrm{Re}(z^5) = \frac12 \)

Explanation:
1. \( z = \frac12 + i\frac{\sqrt3}2 = e^{i\pi/3} \), so \( z^5 = e^{5i\pi/3} = \frac12 - i\frac{\sqrt3}2 \).
2. So \( \mathrm{Re}(z^5) = \frac12 \).

Teacher's Note:
1. Convert to polar form.

Answer: (d) 1

Explanation:
1. \( R = \frac4{\sqrt3} \), \( r = \frac2{\sqrt3} \), \( h = 2\sqrt3 \).
2. \( \frac{R + r}{h} = \frac{6/\sqrt3}{2\sqrt3} = 1 \).

Teacher's Note:
1. Use the standard equilateral formulas.

Answer: (c) 1

Explanation:
1. Rearrange: \( (a-b)(a+b) = b - c \), \( (b-c)(b+c) = c - a \), \( (c-a)(c+a) = a - b \).
2. Multiply: \( (a+b)(b+c)(c+a)\cdot(a-b)(b-c)(c-a) = (b-c)(c-a)(a-b) \), so the product is 1.

Teacher's Note:
1. Multiply the three relations.

Answer: (c) 0.2

Explanation:
1. Composites up to 25: 15. Those divisible by 5: 10, 15, 20, 25, giving 4 numbers.
2. Non-composites (1 and 9 primes): only 2 is even, giving 1. Probability \( = \frac5{25} = 0.2 \).

Teacher's Note:
1. Count the composites divisible by 5 and even non-composites.

Answer: (a) \( \frac\pi2 \)

Explanation:
1. Shift the origin to \( (-1, 0) \): the curve becomes \( 3X^2 + 5XY - 3Y^2 + 2X + 3Y = 0 \) and the line \( 3X - 2Y = 1 \).
2. Homogenise: \( 9X^2 + 10XY - 9Y^2 = 0 \). The coefficients of \( X^2 \) and \( Y^2 \) add to 0, so the lines are perpendicular.

Teacher's Note:
1. Shift the origin and homogenise.

Answer: (b) 2

Explanation:
1. \( |z| - \frac2{|z|} \leq \left|z - \frac2z\right| = 1 \), so \( |z|^2 - |z| - 2 \leq 0 \).
2. Hence \( |z| \leq 2 \).

Teacher's Note:
1. Use the triangle inequality.

Answer: (c) \( \frac{-4}{3} \)

Explanation:
1. \( b^2 \leq 4a \), so \( 3a - 2b \geq \frac{3b^2}4 - 2b = \frac34\left(b - \frac43\right)^2 - \frac43 \).
2. The least value is \( -\frac43 \) (at \( b = \frac43 \), \( a = \frac49 \)).

Teacher's Note:
1. Replace a by its lower bound b²/4.

Answer: (b) GP

Explanation:
1. \( y_r = \frac{x_r^2}8 \). If \( x_2^2 = x_1x_3 \), then \( y_2^2 = \frac{x_2^4}{64} = \frac{x_1^2x_3^2}{64} = y_1y_3 \).

Teacher's Note:
1. Square the GP condition.

Answer: (b) one-one but not onto

Explanation:
1. \( f'(x) = \frac1{(x+1)^2} > 0 \), so \( f \) is one-one. Its range is \( [0, 1) \), which is smaller than the codomain.

Teacher's Note:
1. Find the range.

Answer: (b) -4

Explanation:
1. Perpendicular tangents meet on the directrix \( x = -4 \), so \( \lambda = -4 \).

Teacher's Note:
1. Locus of perpendicular tangents.

Answer: (a) \( \left[-\frac14, \frac12\right] \)

Explanation:
1. Need \( -1 \leq 2x \leq 1 \) and \( \sin^{-1}2x \geq -\frac\pi6 \), i.e. \( 2x \geq -\frac12 \).
2. So \( x \in \left[-\frac14, \frac12\right] \).

Teacher's Note:
1. Both the arcsine domain and the root condition matter.

Answer: (b) \( \frac{\sin B}{2\sin C} \)

Explanation:
1. \( BD = DC \) and the sine rule in the two small triangles give \( \frac{\sin\frac{2A}3}{\sin B} = \frac{\sin\frac A3}{\sin C} \).
2. So \( 2\cos\frac A3 = \frac{\sin B}{\sin C} \).

Teacher's Note:
1. Equate BD and DC with the sine rule.

Answer: (c) rational numbers

Explanation:
1. The third vertex lies on two altitudes, which have rational equations; solving two linear equations with rational coefficients gives rational coordinates.

Teacher's Note:
1. Intersect two altitudes.

Answer: (b) {-1, 1}

Explanation:
1. The focus is \( (4, 0) \), so the chord is \( y = m(x - 4) \). The distance from \( (6, 0) \) is \( \frac{2|m|}{\sqrt{1 + m^2}} = \sqrt2 \).
2. \( 4m^2 = 2 + 2m^2 \), so \( m = \pm1 \).

Teacher's Note:
1. Distance from the circle's centre equals its radius.

Answer: (b) 3

Explanation:
1. The numerator is about \( f'(2)(2h) \) and the denominator about \( f'(1)h \).
2. The limit is \( \frac{6\cdot2}{4} = 3 \).

Teacher's Note:
1. First-order expansion in h.

Answer: (c) the exterior of the circle \( (x + 32)^2 + (y - 54)^2 = 3600 \)

Explanation:
1. \( G_1 \) is cheaper when \( 40d_1 \lt 60d_2 \), i.e. \( 4d_1^2 \lt 9d_2^2 \): \( 4[(x-40)^2 + y^2] \lt 9[x^2 + (y-30)^2] \).
2. This simplifies to \( x^2 + y^2 + 64x - 108y + 340 > 0 \), i.e. \( (x + 32)^2 + (y - 54)^2 > 3600 \): the exterior.

Teacher's Note:
1. Compare the cost functions.

Answer: (b) \( \frac{13}{20} \)

Explanation:
1. \( P \) lies on the line \( y = x + 2 \); \( A \) and \( B \) are on the same side, so reflect \( B \) to \( B' = (0, 3) \).
2. Line \( AB' \): \( y = 3 - \frac73x \); it meets \( y = x + 2 \) at \( x = \frac3{10} \). Then \( 2k - 1 = \frac3{10} \) gives \( k = \frac{13}{20} \).

Teacher's Note:
1. Reflect B across the line.

Answer: (c) 2

Explanation:
1. The foot of the altitude splits \( BC \) into pieces \( \sqrt{c^2 - 4} = 2\sqrt3 \) and \( \sqrt{b^2 - 4} = \sqrt5 \).
2. \( BC = 2\sqrt3 + \sqrt5 \) or \( 2\sqrt3 - \sqrt5 \), so there are 2 triangles.

Teacher's Note:
1. The foot may lie inside or outside BC.

Answer: (b) \( 4a + c^2 > 2b \)

Explanation:
1. \( f \) has no real zero and \( f(0) = c^2 \geq 0 \), so \( f(x) > 0 \) for all \( x \).
2. \( f(-2) = 4a + 2b + c^2 > 0 \) and \( f(2) = 4a - 2b + c^2 > 0 \), giving \( 4a + c^2 > 2b \). Also \( f(1) > 0 \) and \( f(3) > 0 \) contradict (a) and (c).

Teacher's Note:
1. f keeps the sign of f(0).