JEE Mathematics Quadratic Equations MCQs Set 05

Answer: (c) 16

Explanation:
1. \( \log_2(xy)\ge6\Rightarrow xy\ge64 \) (x, y > 0).
2. By AM-GM, \( x+y\ge2\sqrt{xy}\ge2\cdot8=16 \).

Teacher's Note:
1. Combine the logarithms to get a lower bound on xy.
2. Then use AM-GM.

Answer: (c) 11

Explanation:
1. \( f(x,n)=\sum(\log_xk-1)=\log_x(n!)-n \).
2. Equating f(x, 10) and f(x, 11): \( \log_x11-1=0 \), so \( \log_x11=1 \) and x = 11.

Teacher's Note:
1. Split log_x(k/x) into log_x k − 1.
2. f(x, 11) − f(x, 10) is the term k = 11.

Answer: (b) 16

Explanation:
1. The even prime is 2: \( 4-2\lambda+12=0\Rightarrow\lambda=8 \).
2. Equal roots of \( x^2+8x+\mu=0 \): \( 64-4\mu=0\Rightarrow\mu=16 \).

Teacher's Note:
1. The only even prime is 2.
2. Equal roots mean zero discriminant.

Answer: (c) \( b^2-4ac \)

Explanation:
1. Let the roots be p and q. Then \( a+b+c=a(1-(p+q)+pq)=a(1-p)(1-q) \).
2. \( 1-p=\frac{-1}{\alpha-1},\ 1-q=\frac{-1}{\alpha} \), so \( (1-p)(1-q)=\frac1{\alpha(\alpha-1)} \).
3. Also \( p-q=\frac{\alpha^2-(\alpha^2-1)}{\alpha(\alpha-1)}=\frac1{\alpha(\alpha-1)} \). Hence \( (a+b+c)^2=a^2(p-q)^2=b^2-4ac \).

Teacher's Note:
1. a + b + c is a times f(1), which factors as (1 − p)(1 − q).
2. a²(p − q)² is the discriminant.

Answer: (d) \( (-\infty,-1) \)

Explanation:
1. Let f(x) be the quadratic. Conditions: D ≥ 0, f(2) > 0, and axis \( \frac{5p}2<2 \).
2. \( D=400p^2-16(25p^2+15p-66)=1056-240p\ge0\Rightarrow p\le4.4 \).
3. \( f(2)=25p^2-25p-50>0\Rightarrow(p-2)(p+1)>0 \) gives p < −1 or p > 2; the axis condition gives p < 4/5.
4. Intersection: \( p<-1 \).

Teacher's Note:
1. Both roots less than k needs D ≥ 0, f(k) > 0 and axis < k.
2. Intersect the three conditions.

Answer: (a) \( 0\lt m\le1 \)

Explanation:
1. Let \( u=a+b,\ v=c+d \) with u + v = 2 and u, v > 0.
2. By AM-GM, \( uv\le\left(\frac{u+v}{2}\right)^2=1 \), and uv > 0. So 0 < m ≤ 1.

Teacher's Note:
1. Treat a + b and c + d as two positive numbers with a fixed sum.
2. AM-GM gives the upper bound.

Answer: (a) \( \left|\frac rp-7\right|\ge4\sqrt3 \)

Explanation:
1. AP: \( q=\frac{p+r}2 \). Real roots need \( q^2\ge4pr\Rightarrow(p+r)^2\ge16pr \), i.e. \( p^2-14pr+r^2\ge0 \).
2. Divide by p² with \( t=\frac rp \): \( t^2-14t+1\ge0\Rightarrow(t-7)^2\ge48 \), i.e. \( \left|\frac rp-7\right|\ge4\sqrt3 \).

Teacher's Note:
1. Use the discriminant condition with q = (p + r)/2.
2. Complete the square in t = r/p.

Answer: (b) 1

Explanation:
1. Let the roots be 2k and 3k. Sum: \( 5k=\frac5a \) and product: \( 6k^2=\frac6a \).
2. So k = 1/a and \( k^2=\frac1a \), giving \( \frac1{a^2}=\frac1a \) and a = 1.

Teacher's Note:
1. Write the roots as 2k and 3k.
2. Eliminate k between the sum and product.

Answer: (c) \( x\in[1,2) \)

Explanation:
1. Since \( x-[x]=\{x\} \) and \( x-\{x\}=[x] \), the equation is \( [x]+\{x\}-1=[x]\{x\} \).
2. Rearrange: \( ([x]-1)(\{x\}-1)=0 \). As {x} < 1, we need [x] = 1.
3. Hence \( x\in[1,2) \).

Teacher's Note:
1. Replace x − [x] by {x} and x − {x} by [x].
2. {x} is never equal to 1.

Answer: (b) 4

Explanation:
1. Write x = n + f (n integer, 0 ≤ f < 1): \( [x]-2x=-n-2f \), so \( |n+2f|=4 \).
2. n + 2f = 4: (n = 4, f = 0) or (n = 3, f = 1/2): x = 4, 3.5.
3. n + 2f = −4: (n = −4, f = 0) or (n = −5, f = 1/2): x = −4, −4.5. Total 4 solutions.

Teacher's Note:
1. Split x into integer and fractional parts.
2. Since 0 ≤ 2f < 2, n is determined up to two values for each target.

Answer: (c) −3

Explanation:
1. Write both sides as powers of 2: \( 4(x^2+3x-1)=3(x^2+3x+2) \).
2. This gives \( x^2+3x-10=0 \), whose roots sum to −3.

Teacher's Note:
1. Express 16 and 8 as powers of 2.
2. The sum of roots of x² + 3x − 10 is −3.

Answer: (b) c = 1

Explanation:
1. The equation is \( ax^3+bx+(c-1)=0 \) which has no x² term, so \( \alpha+\beta+\gamma=0 \).
2. Since α + β = 0, γ = 0, so \( c-1=0 \), i.e. c = 1.

Teacher's Note:
1. Sum of the roots of a cubic without an x² term is zero.
2. γ = 0 is a root, so the constant term vanishes.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Put \( t=2^x\in(0,1] \): \( t^2-(a-3)t+(a-4)=0 \) has roots t = 1 and \( t=a-4 \).
2. Non-positive x means t ≤ 1, so \( 03. R is also true (a negative leading coefficient and D ≤ 0), but it is not used to obtain A.

Teacher's Note:
1. Substitute t = 2^x; x ≤ 0 means 0 < t ≤ 1.
2. Note that t = 1 is always a root.