JEE Mathematics Miscellaneous Problems MCQs Set 03

Mathematics Objective Questions and Answers: Miscellaneous Problems

Access targeted multiple-choice questions for Miscellaneous Problems designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Navigate directly to the 50 objective questions for Miscellaneous Problems using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: If \( (1+x+x^2+x^3)^7=\sum_{r=0}^{21}a_rx^r \), then the value of \( a_0+a_4+a_8+a_{12}+a_{16}+a_{20} \) is
(a) 1024
(b) 2048
(c) 4096
(d) 8192
Show Answer & Explanation

Answer: (c) 4096

Explanation:
1. Let f(x) = (1 + x + x² + x³)⁷. The sum of coefficients of powers that are multiples of 4 is \( \frac14[f(1)+f(i)+f(-1)+f(-i)] \).
2. f(1) = 4⁷ = 16384, while 1 + i + i² + i³ = 0 gives f(i) = f(−1) = f(−i) = 0.
3. Sum = 16384/4 = 4096.

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Question 2: The number of different ways in which 24 identical apples can be distributed among 3 boys and 3 girls such that the total number of apples received by the boys is double that received by the girls is
(a) 6885
(b) 6120
(c) 5148
(d) 7425
Show Answer & Explanation

Answer: (a) 6885

Explanation:
1. Boys get 16 and girls get 8 apples.
2. Boys: \( \binom{18}2=153 \) ways; girls: \( \binom{10}2=45 \) ways.
3. Total = 153 × 45 = 6885.

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Question 3: If \( |\vec a|=3,\ |\vec b|=4,\ |\vec c|=5 \) and \( \vec a\perp(\vec b+\vec c),\ \vec b\perp(\vec c+\vec a),\ \vec c\perp(\vec a+\vec b) \), then \( \sqrt2\,|\vec a+\vec b+\vec c| \) is equal to
(a) 5
(b) 10
(c) 10√2
(d) 50
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. The three conditions give a·b + a·c = 0, b·c + b·a = 0, c·a + c·b = 0, so a·b = b·c = c·a = 0.
2. Then \( |\vec a+\vec b+\vec c|^2=9+16+25=50 \).
3. So \( \sqrt2\cdot\sqrt{50}=10 \).

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Question 4: Three times the least value of the area bounded by the line y = mx + 1 and the parabola \( y=x^2+2x-3 \), m being a parameter, is
(a) 16
(b) 24
(c) 32
(d) 48
Show Answer & Explanation

Answer: (c) 32

Explanation:
1. Intersections satisfy \( x^2+(2-m)x-4=0 \) with discriminant \( D=(2-m)^2+16 \).
2. Area between line and parabola = \( \frac{D^{3/2}}6 \), least when m = 2: \( \frac{64}6=\frac{32}3 \).
3. Three times this is 32.

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Question 5: If f(x) and φ(x) are continuous functions on [0, 4] satisfying f(x) = f(4 − x), φ(x) + φ(4 − x) = 3 and \( \int_0^4f(x)\,dx=2 \), then \( \int_0^4f(x)\varphi(x)\,dx \) is
(a) 2
(b) 3
(c) 6
(d) 1.5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. By the property \( \int_0^4g(x)dx=\int_0^4g(4-x)dx \): \( I=\int_0^4f(x)\varphi(4-x)\,dx \) using f(4 − x) = f(x).
2. Adding: \( 2I=\int_0^4f(x)[\varphi(x)+\varphi(4-x)]dx=3\cdot2=6 \).
3. So I = 3.

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Question 6: The value of \( \int_{10}^{100}(x-[x])\,dx \) is
(a) 90
(b) 45
(c) 50
(d) 4500
Show Answer & Explanation

Answer: (b) 45

Explanation:
1. x − [x] is the fractional part, which has period 1 and mean ½.
2. Over 90 unit intervals the integral is 90 × ½ = 45.

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Question 7: If \( f(x)=\int_a^x\frac{dt}{f(t)} \) (x > 0) and \( \int_a^1\frac{dx}{f(x)}=\sqrt2 \), then f(50) is
(a) 5
(b) 10
(c) 25
(d) 50
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. Differentiating, f′(x) = 1/f(x), so \( f\,f'=1 \) and \( f^2=2x+c \).
2. f(a) = 0 gives c = −2a. Also \( \int_a^1\frac{dx}{f}=f(1)-f(a)=f(1)=\sqrt2 \), so 2 − 2a = 2 and a = 0.
3. Therefore f² = 2x and f(50) = √100 = 10.

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Question 8: Consider the closed figure C made by the lines \( |x|+|y|=\sqrt2 \). Let S be the region inside C such that any point in it is nearer to the side x + y = √2 than to the origin. Then three times the area of S is
(a) 2
(b) 3
(c) 4
(d) 6
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Rotate axes: u = (x + y)/√2, v = (y − x)/√2. Then C becomes the square |u| ≤ 1, |v| ≤ 1, the side is u = 1 and the origin is unchanged.
2. Nearer to the side than the origin: \( 1-u\lt\sqrt{u^2+v^2} \), which simplifies to \( u\gt\frac{1-v^2}{2} \).
3. Area \( =\int_{-1}^{1}\left(1-\frac{1-v^2}2\right)dv=\int_{-1}^1\frac{1+v^2}2dv=\frac43 \), so three times the area is 4.

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Question 9: Let \( 2^{a_1}+3^{a_2}+5^{a_3}+7^{a_4}+9^{a_5} \) be divisible by 4, where a₁, a₂, a₃, a₄, a₅ are digits (0 to 9). The largest possible number of values of (a₁, a₂, a₃, a₄, a₅) is
(a) 25000
(b) 45000
(c) 50000
(d) 40000
Show Answer & Explanation

Answer: (b) 45000

Explanation:
1. Modulo 4: 2^{a₁} is 1, 2, 0 (for a₁ = 0, 1, ≥2); 3^{a₂} is 1 or 3 (even or odd a₂); 5^{a₃} ≡ 1; 7^{a₄} is 1 or 3; 9^{a₅} ≡ 1.
2. Sum ≡ 2^{a₁} + 3^{a₂} + 7^{a₄} + 2 (mod 4). Enumerating all cases for (a₁, a₂, a₄) gives 450 valid triples out of 1000.
3. a₃ and a₅ are free (10 × 10 = 100 choices), so the count is 450 × 100 = 45000.

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Question 10: Let \( f(x)=ax^4+bx^3+cx^2+dx+\lambda \), where \( \frac{f(1)}1=\frac{f(2)}2=\frac{f(3)}3 \) and \( \lim_{x\to\infty}\frac{f(x)}{x^4}=1 \). If f(i) + f(−i) + k = 0, where \( i=\sqrt{-1} \), then k is
(a) 10
(b) 20
(c) −20
(d) 40
Show Answer & Explanation

Answer: (b) 20

Explanation:
1. a = 1, and f(x) − cx has roots 1, 2, 3, so \( f(x)=(x-1)(x-2)(x-3)(x-r)+cx \).
2. The even part of f is \( x^4+(11+6r)x^2+6r \); at x = i this is \( 1-(11+6r)+6r=-10 \).
3. So f(i) + f(−i) = −20 and k = 20.

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Question 11: A plane exists which is perpendicular to the line \( \frac x3=\frac{y-1}5=\frac{z+1}7 \) and passes through the line \( \frac{x+3}{10}=\frac{y-1}{-20}=\frac{z+1}k \). Then k is
(a) 5
(b) 10
(c) 20
(d) −10
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. The normal of the plane is along (3, 5, 7), and it must be perpendicular to the contained line's direction (10, −20, k).
2. 30 − 100 + 7k = 0, so k = 10.

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