JEE Mathematics Previous Year Questions MCQs Set 03

Answer: (a) -2, -32

Explanation:
1. Let the common ratio be r: \( \alpha(1+r)=1 \) and \( \alpha r^2(1+r)=4 \), so \( r^2=4 \).
2. For r = 2, p = 2/9 is not an integer. For r = -2: \( \alpha=-1,\beta=2,\gamma=-4,\delta=8 \).
3. p = \( \alpha\beta=-2 \) and q = \( \gamma\delta=-32 \).

Teacher's Note:
1. Use the sums of roots to find r.
2. Keep only integer p and q.

Answer: (b) \( \tfrac15(n-4) \)

Explanation:
1. \( T_5+T_6=0 \) gives \( \binom n4a^{n-4}b^4=\binom n5a^{n-5}b^5 \).
2. So \( a/b=\binom n5/\binom n4=(n-4)/5 \).

Teacher's Note:
1. Write the 5th and 6th terms.
2. Divide and simplify the ratio of binomial coefficients.

Answer: (a) 14

Explanation:
1. Total functions = \( 2^4=16 \).
2. Remove the two constant functions, which are not onto: 16 - 2 = 14.

Teacher's Note:
1. Count all functions and subtract those missing an element of F.
2. Inclusion-exclusion.

Answer: (b) 7

Explanation:
1. \( T_n=\binom n3 \), so \( T_{n+1}-T_n=\binom n2 \).
2. \( \binom n2=21 \) gives n = 7.

Teacher's Note:
1. Use Pascal's identity.
2. Solve \( n(n-1)=42 \).

Answer: (c) 11

Explanation:
1. First sum = \( n[4+(2n-1)3]=n(6n+1) \).
2. Second sum = \( \tfrac n2[114+2(n-1)]=n(n+56) \).
3. Equating: 6n + 1 = n + 56, so n = 11.

Teacher's Note:
1. Use \( S=\tfrac m2[2a+(m-1)d] \).
2. Cancel the common factor n.

Answer: (d) 4k

Explanation:
1. A right angle at the origin means \( z_2/z_1=\pm i \).
2. This ratio is itself an nth root of unity, so \( i^n=1 \) and n is a multiple of 4.

Teacher's Note:
1. Ratio of two nth roots is an nth root.
2. i is a 4th root of unity.

Answer: (d) -1

Explanation:
1. \( f(f(x))=\dfrac{\alpha^2x}{(\alpha+1)x+1} \) after simplification.
2. Setting this equal to x for all x forces \( \alpha+1=0 \) and \( \alpha^2=1 \), so \( \alpha=-1 \).

Teacher's Note:
1. Compose f with itself.
2. Compare coefficients.

Answer: (d) {-1, 0, 1}

Explanation:
1. The graphs of y = x and y = x³ cross at x = -1, 0, 1.
2. At each crossing the function switches branches with different slopes, so it is not differentiable at -1, 0 and 1.

Teacher's Note:
1. Find where x = x³.
2. Compare slopes on either side.

Answer: (c) -3

Explanation:
1. f'(1) = 2 + b, tangent: \( y-1=(2+b)(x-1) \).
2. Intercepts: \( x=\tfrac{1+b}{2+b} \), \( y=-(1+b) \); area = \( \tfrac12\left|\tfrac{(1+b)^2}{2+b}\right|=2 \).
3. b = -3 gives intercepts 2 and 2 and area 2, in the first quadrant.

Teacher's Note:
1. Write the tangent equation and its intercepts.
2. Check each option in the area formula.

Answer: (b) 1

Explanation:
1. \( g(x)=1+\{x\}\ge1\gt0 \) for every x.
2. So f of a positive number is 1.

Teacher's Note:
1. Fractional part is non-negative.
2. Check the sign of g(x).

Answer: (d) \( (-3,+\infty)-\{-1,-2\} \)

Explanation:
1. Need x + 3 \gt 0, so x \gt -3.
2. The denominator \( (x+1)(x+2)\ne0 \) removes -1 and -2.

Teacher's Note:
1. Apply the log and denominator restrictions.
2. Intersect.

Answer: (c) \( \pi/2 \)

Explanation:
1. Using \( \int_{-c}^{c}\dfrac{g(x)}{1+a^x}dx=\int_0^c g(x)dx \) for even g, the integral is \( \int_0^\pi\cos^2x\,dx \).
2. This equals \( \pi/2 \).

Teacher's Note:
1. Add the integral to the same integral with x replaced by -x.
2. \( \tfrac1{1+a^x}+\tfrac1{1+a^{-x}}=1 \).

Answer: (a) 2

Explanation:
1. Substituting: \( x=\dfrac{5}{3+4m} \).
2. 3 + 4m must be a divisor of 5 with m an integer: 3 + 4m = -1 or -5, giving m = -1 or -2.
3. So there are 2 values.

Teacher's Note:
1. Solve for x.
2. List divisors of 5.

Answer: (d) \( x=\tfrac32 \)

Explanation:
1. \( (y+2)^2=-4x+2=-4\left(x-\tfrac12\right) \).
2. Vertex at x = 1/2, a = 1, opening left, so the directrix is x = 1/2 + 1 = 3/2.

Teacher's Note:
1. Complete the square in y.
2. Directrix is a distance a from the vertex on the opposite side.

Answer: (b) 1

Explanation:
1. \( \log_4(x-1)=\tfrac12\log_2(x-1) \), so \( x-1=(x-3)^2 \).
2. \( x^2-7x+10=0 \) gives x = 2 or 5; x = 2 violates x \gt 3, so only x = 5.

Teacher's Note:
1. Change both logs to base 2.
2. Check the domain x \gt 3.

Answer: (d) \( \dfrac{1}{|m-n|} \)

Explanation:
1. Vertices: (0,0), \( \left(\tfrac1{n-m},\tfrac n{n-m}\right) \), (0,1) and \( \left(\tfrac1{n-m},\tfrac{n}{n-m}+1\right) \).
2. Base along the y-axis is 1 and the horizontal distance is \( 1/|m-n| \), so the area is \( 1/|m-n| \).

Teacher's Note:
1. Find the intersection points.
2. Area = base x horizontal distance.

Answer: (c) \( \tan\beta+2\tan\gamma \)

Explanation:
1. From \( \alpha=\tfrac\pi2-\beta \): \( \tan\alpha=\cot\beta \). From \( \alpha=\beta+\gamma \): \( 2\beta+\gamma=\tfrac\pi2 \), so \( \tan\gamma=\cot2\beta=\dfrac{1-t^2}{2t} \) with \( t=\tan\beta \).
2. Then \( \tan\beta+2\tan\gamma=t+\dfrac{1-t^2}{t}=\dfrac1t=\cot\beta=\tan\alpha \).

Teacher's Note:
1. Test with specific angles.
2. Use \( \tan(\pi/2-\beta)=\cot\beta \).

Answer: (c) (-6, 11)

Explanation:
1. The characteristic polynomial is \( \lambda^3-6\lambda^2+11\lambda-6 \) and det A = 6.
2. By Cayley-Hamilton, \( A^3-6A^2+11A-6I=0 \), so \( 6A^{-1}=A^2-6A+11I \).
3. Hence c = -6, d = 11.

Teacher's Note:
1. Use the Cayley-Hamilton theorem.
2. Multiply by \( A^{-1} \).

Answer: (a) 5

Explanation:
1. \( F'(x)=f(x/2)f'(x/2)+g(x/2)g'(x/2) \) and \( g'=f''=-f \).
2. Since \( g=f' \), \( F'(x)=f f'-f' f=0 \), so F is constant and F(10) = 5.

Teacher's Note:
1. Differentiate F.
2. A zero derivative means constant.

Answer: (b) 5051

Explanation:
1. Integrating by parts, \( I_{101}=\tfrac{5050}{5051}I_{100} \).
2. So the quantity equals 5051.

Teacher's Note:
1. Relate the two integrals by parts.
2. Ratio = 5051/5050.