JEE Mathematics Miscellaneous Problems MCQs Set 01

Download JEE MCQs for JEE Mathematics: Miscellaneous Problems

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Chapter-wise Objective Questions: Miscellaneous Problems

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Multiple Choice Questions

Question 1: Every divisor of \( 2^{10}\cdot3^8\cdot5^7 \) has the form \( 2^m3^n5^p \) with \( 0\le m\le10,\ 0\le n\le8,\ 0\le p\le7 \), and the sum of all divisors is the product of the sums of geometric series. A proper divisor means a divisor other than 1 and the number itself. The number of proper divisors of 16200 is
(a) 60
(b) 24
(c) 58
(d) 22
Show Answer & Explanation

Answer: (c) 58

Explanation:
1. \( 16200=2^3\cdot3^4\cdot5^2 \).
2. Number of divisors \( =(3+1)(4+1)(2+1)=60 \).
3. Excluding 1 and 16200 itself: 60 − 2 = 58.

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Question 2: Every divisor of \( 2^{10}\cdot3^8\cdot5^7 \) has the form \( 2^m3^n5^p \) with \( 0\le m\le10,\ 0\le n\le8,\ 0\le p\le7 \), and the sum of all divisors is the product of the sums of geometric series. A proper divisor means a divisor other than 1 and the number itself. The sum of the proper divisors of 360 is
(a) 1170
(b) 810
(c) 1169
(d) 809
Show Answer & Explanation

Answer: (d) 809

Explanation:
1. \( 360=2^3\cdot3^2\cdot5 \).
2. Sum of all divisors \( =(1+2+4+8)(1+3+9)(1+5)=15\times13\times6=1170 \).
3. Excluding 1 and 360: 1170 − 1 − 360 = 809.

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Question 3: Every divisor of \( 2^{10}\cdot3^8\cdot5^7 \) has the form \( 2^m3^n5^p \) with \( 0\le m\le10,\ 0\le n\le8,\ 0\le p\le7 \), and the sum of all divisors is the product of the sums of geometric series. A proper divisor means a divisor other than 1 and the number itself. The sum of the odd divisors (other than 1) of 10800 is
(a) 1239
(b) 1240
(c) 1238
(d) 2479
Show Answer & Explanation

Answer: (a) 1239

Explanation:
1. \( 10800=2^4\cdot3^3\cdot5^2 \).
2. Odd divisors come from the factors of 3 and 5 only: \( (1+3+9+27)(1+5+25)=40\times31=1240 \).
3. Excluding 1: 1239.

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Question 4: The number of positive integral solutions of \( x_1+x_2+\dots+x_m=n \) equals the coefficient of \( x^n \) in \( (x+x^2+x^3+\dots)^m \), which is \( {}^{n-1}C_{m-1} \). The number of positive integral solutions of \( x_1+x_2+x_3+x_4=10 \) is
(a) \( {}^9C_6 \)
(b) \( {}^{10}C_4 \)
(c) \( {}^9C_4 \)
(d) \( {}^{10}C_5 \)
Show Answer & Explanation

Answer: (a) \( {}^9C_6 \)

Explanation:
1. Number \( ={}^{10-1}C_{4-1}={}^9C_3=84 \).
2. Since \( {}^9C_3={}^9C_6 \), the answer is option (a).

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Question 5: The number of positive integral solutions of \( x_1+x_2+\dots+x_m=n \) equals the coefficient of \( x^n \) in \( (x+x^2+x^3+\dots)^m \), which is \( {}^{n-1}C_{m-1} \). The number of negative integral solutions of \( x_1+x_2+x_3+x_4+x_5+10=0 \) is
(a) \( {}^{10}C_4 \)
(b) \( {}^{10}C_5 \)
(c) \( {}^9C_5 \)
(d) \( {}^9C_6 \)
Show Answer & Explanation

Answer: (c) \( {}^9C_5 \)

Explanation:
1. Put \( x_i=-y_i \) with \( y_i\ge1 \): \( y_1+\dots+y_5=10 \).
2. Number \( ={}^{9}C_4=126={}^9C_5 \).

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Question 6: The number of positive integral solutions of \( x_1+x_2+\dots+x_m=n \) equals the coefficient of \( x^n \) in \( (x+x^2+x^3+\dots)^m \), which is \( {}^{n-1}C_{m-1} \). The number of ways in which 10 identical things can be distributed among 3 persons so that each gets at least one thing is
(a) \( {}^8C_3 \)
(b) \( {}^{10}C_4 \)
(c) \( {}^9C_3 \)
(d) \( {}^9C_2 \)
Show Answer & Explanation

Answer: (d) \( {}^9C_2 \)

Explanation:
1. This is the number of positive solutions of \( x_1+x_2+x_3=10 \): \( {}^9C_2=36 \).

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Question 7: The rank of a matrix is r if some minor of order r is non-zero and every minor of order r + 1 is zero. The characteristic equation of a square matrix A is \( |A-xI|=0 \), whose roots are the eigenvalues. Let \( A=\begin{bmatrix}8&-6&2\\-6&7&-4\\2&-4&3\end{bmatrix} \). The rank of A is
(a) 3
(b) 2
(c) 1
(d) none of these
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \( |A|=8(21-16)+6(-18+8)+2(24-14)=40-60+20=0 \), so the rank is less than 3.
2. A 2 × 2 minor \( \begin{vmatrix}8&-6\\-6&7\end{vmatrix}=20\neq0 \), so the rank is 2.

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Question 8: The rank of a matrix is r if some minor of order r is non-zero and every minor of order r + 1 is zero. The characteristic equation of a square matrix A is \( |A-xI|=0 \), whose roots are the eigenvalues. Let \( A=\begin{bmatrix}8&-6&2\\-6&7&-4\\2&-4&3\end{bmatrix} \). The characteristic equation of A is
(a) \( x^3-18x^2+45x=0 \)
(b) \( x^3-6x^2+9x-4=0 \)
(c) \( x^3+18x^2-45x=0 \)
(d) \( x^3+6x^2-9x+4=0 \)
Show Answer & Explanation

Answer: (a) \( x^3-18x^2+45x=0 \)

Explanation:
1. Trace \( =8+7+3=18 \). Sum of principal 2×2 minors \( =(56-36)+(24-4)+(21-16)=45 \). Determinant = 0.
2. Characteristic equation: \( x^3-18x^2+45x-0=0 \).

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Question 9: Gregory's series gives \( \theta=\tan\theta-\frac13\tan^3\theta+\frac15\tan^5\theta-\dots \) for \( |\theta|<\frac\pi4 \). The sum of the series \( 1-\frac13+\frac15-\frac17+\dots \) to infinity is
(a) \( \frac\pi2 \)
(b) \( \frac\pi4 \)
(c) 1
(d) \( \frac\pi8 \)
Show Answer & Explanation

Answer: (b) \( \frac\pi4 \)

Explanation:
1. Put tan θ = 1, i.e. \( \theta=\frac\pi4 \), in Gregory's series.
2. The series equals \( \frac\pi4 \).

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Question 10: Gregory's series gives \( \theta=\tan\theta-\frac13\tan^3\theta+\frac15\tan^5\theta-\dots \) for \( |\theta|<\frac\pi4 \). The sum of the series \( \frac{1}{1\cdot3}+\frac{1}{5\cdot7}+\frac{1}{9\cdot11}+\dots \) to infinity is
(a) \( \frac\pi2 \)
(b) \( \frac\pi4 \)
(c) 1
(d) \( \frac\pi8 \)
Show Answer & Explanation

Answer: (d) \( \frac\pi8 \)

Explanation:
1. Each term \( \frac{1}{(4k-3)(4k-1)}=\frac12\left(\frac{1}{4k-3}-\frac{1}{4k-1}\right) \).
2. The sum is \( \frac12\left(1-\frac13+\frac15-\frac17+\dots\right)=\frac12\cdot\frac\pi4=\frac\pi8 \).

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Question 11: Gregory's series gives \( \theta=\tan\theta-\frac13\tan^3\theta+\frac15\tan^5\theta-\dots \) for \( |\theta|<\frac\pi4 \). The sum of the series \( 1-\frac{1}{3\cdot3}+\frac{1}{5\cdot3^2}-\frac{1}{7\cdot3^3}+\dots \) to infinity is
(a) \( \frac\pi2 \)
(b) \( \frac{\pi}{\sqrt3} \)
(c) \( \frac{\pi}{2\sqrt3} \)
(d) \( \frac{\sqrt3\pi}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{\pi}{2\sqrt3} \)

Explanation:
1. Take \( \tan\theta=\frac{1}{\sqrt3} \): \( \theta=\frac\pi6=\frac{1}{\sqrt3}\left(1-\frac{1}{3\cdot3}+\frac{1}{5\cdot3^2}-\dots\right) \).
2. So the series equals \( \sqrt3\cdot\frac\pi6=\frac{\pi}{2\sqrt3} \).

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Question 12: A second-degree curve inscribing the triangle ABC, where the sides AB, BC and CA have equations \( L_1=0,\ L_2=0,\ L_3=0 \), has the equation \( L_1L_2+\lambda L_2L_3+\mu L_3L_1=0 \). Let the equations of AB, BC and CA be y = 2x, y + 2x = 0 and x = 1 respectively. The circumcentre of the triangle ABC is
(a) \( \left(\frac54,\frac58\right) \)
(b) \( \left(\frac52,0\right) \)
(c) \( \left(0,\frac54\right) \)
(d) \( \left(\frac54,-\frac58\right) \)
Show Answer & Explanation

Answer: (b) \( \left(\frac52,0\right) \)

Explanation:
1. Vertices: A = (0, 0) (intersection of y = 2x and y = −2x), B = (1, 2), C = (1, −2).
2. By symmetry the circumcentre is (h, 0) with \( h^2=(1-h)^2+4 \), so \( 2h=5 \) and \( h=\frac52 \).

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Question 13: A second-degree curve inscribing the triangle ABC, where the sides AB, BC and CA have equations \( L_1=0,\ L_2=0,\ L_3=0 \), has the equation \( L_1L_2+\lambda L_2L_3+\mu L_3L_1=0 \). Let the equations of AB, BC and CA be y = 2x, y + 2x = 0 and x = 1 respectively. The parabola circumscribing the triangle ABC and passing through the point (4, 4) has the focus
(a) (0, 1)
(b) (4, 0)
(c) (1, 0)
(d) (0, 4)
Show Answer & Explanation

Answer: (c) (1, 0)

Explanation:
1. The curve is \( (y-2x)(y+2x)+\lambda(y+2x)(x-1)+\mu(x-1)(y-2x)=0 \). It is a parabola when the quadratic terms form a perfect square.
2. Imposing that and passing through (4, 4) gives two possibilities; one is a pair of parallel lines \( (2x-y)^2-4(2x-y)=0 \) (degenerate), the other \( \lambda=1,\ \mu=-1 \).
3. This gives \( y^2=4x \), a genuine parabola with focus (1, 0).

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Question 14: A second-degree curve inscribing the triangle ABC, where the sides AB, BC and CA have equations \( L_1=0,\ L_2=0,\ L_3=0 \), has the equation \( L_1L_2+\lambda L_2L_3+\mu L_3L_1=0 \). Let the equations of AB, BC and CA be y = 2x, y + 2x = 0 and x = 1 respectively. The parabola circumscribing the triangle ABC and passing through the point (4, 4) has the latus rectum
(a) 4
(b) 1
(c) 16
(d) 8
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. As above, the parabola is \( y^2=4x \).
2. Latus rectum = 4a = 4.

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Question 15: The director circle of a circle is the locus of points from which tangents to the circle are at right angles. \( C_1,C_2,C_3,\dots \) is a sequence of circles such that \( C_{r+1} \) is the director circle of \( C_r \). Let the equation of \( C_1 \) be \( x^2+y^2=4 \). The length of a chord of \( C_{11} \) which touches \( C_{10} \) is
(a) \( 64\sqrt2 \)
(b) 64
(c) \( 32\sqrt2 \)
(d) 32
Show Answer & Explanation

Answer: (a) \( 64\sqrt2 \)

Explanation:
1. The director circle of \( x^2+y^2=a^2 \) is \( x^2+y^2=2a^2 \), so the squared radii are 4, 8, 16, …: \( r_n^2=2^{n+1} \).
2. So \( r_{10}^2=2^{11}=2048 \) and \( r_{11}^2=2^{12}=4096 \).
3. A chord of C11 touching C10 has length \( 2\sqrt{4096-2048}=2\sqrt{2048}=64\sqrt2 \).

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Question 16: The director circle of a circle is the locus of points from which tangents to the circle are at right angles. \( C_1,C_2,C_3,\dots \) is a sequence of circles such that \( C_{r+1} \) is the director circle of \( C_r \). Let the equation of \( C_1 \) be \( x^2+y^2=4 \). The distance of the chord of contact of tangents from a point on \( C_{11} \) with respect to \( C_{10} \) from the centre of \( C_{10} \) is
(a) \( 64\sqrt2 \)
(b) \( 32\sqrt2 \)
(c) 64
(d) 32
Show Answer & Explanation

Answer: (d) 32

Explanation:
1. From the previous result \( r_{10}^2=2048 \) and \( r_{11}=64 \).
2. The chord of contact from a point at distance d from the centre of a circle of radius r lies at distance \( \frac{r^2}{d} \) from the centre.
3. Distance \( =\frac{2048}{64}=32 \).

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Question 17: The director circle of a circle is the locus of points from which tangents to the circle are at right angles. \( C_1,C_2,C_3,\dots \) is a sequence of circles such that \( C_{r+1} \) is the director circle of \( C_r \). Let the equation of \( C_1 \) be \( x^2+y^2=4 \). The locus of the poles of the tangents of \( C_9 \) with respect to \( C_{10} \) is
(a) \( x^2+y^2=2\times32^2 \)
(b) \( x^2+y^2=32^2 \)
(c) \( x^2+y^2=64^2 \)
(d) \( x^2+y^2=2\times16^2 \)
Show Answer & Explanation

Answer: (c) \( x^2+y^2=64^2 \)

Explanation:
1. \( r_9^2=2^{10}=1024 \), so a tangent of \( C_9 \) is at distance 32 from the origin.
2. Its pole with respect to \( C_{10} \) (\( r^2=2048 \)) is at distance \( \frac{2048}{32}=64 \) from the centre.
3. Locus: \( x^2+y^2=64^2 \).

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Question 18: The equation \( (\lambda^2-4\lambda+3)+(\lambda^2+\lambda-2)x+6x^2-5x^3=0 \) will have two roots equal to zero when
(a) 3
(b) −2
(c) 0
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Two roots zero means the constant term and the coefficient of x both vanish.
2. Constant: \( \lambda^2-4\lambda+3=0 \) gives λ = 1 or 3. Coefficient of x: \( \lambda^2+\lambda-2=0 \) gives λ = 1 or −2.
3. Common value: λ = 1.

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Question 19: The value of \( (\lambda,\mu) \) for which \( (\lambda-\mu)x^3-(\lambda-2)x^2-3x+7=0 \) has two infinite roots is
(a) (0, 0)
(b) (2, 2)
(c) (1, 1)
(d) (2, 0)
Show Answer & Explanation

Answer: (b) (2, 2)

Explanation:
1. Put \( x=\frac1y \): \( 7y^3-3y^2-(\lambda-2)y+(\lambda-\mu)=0 \). Infinite roots in x correspond to zero roots in y.
2. Two zero roots need the coefficient of y and the constant to vanish: \( \lambda=2 \) and \( \lambda=\mu \), so \( (\lambda,\mu)=(2,2) \).

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