Mathematics Objective Questions and Answers: Miscellaneous Problems
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Multiple Choice Questions
(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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(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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(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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(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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(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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(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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(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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(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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(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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(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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(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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Yes, our JEE Mathematics Miscellaneous Problems MCQs Set 03 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.
By solving our JEE Mathematics Miscellaneous Problems MCQs Set 03, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.
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