JEE Mathematics Quadratic Equations MCQs Set 16

Answer: (c) \( \alpha<0<\beta<|\alpha| \)

Explanation:
1. Product \( c<0 \): the roots have opposite signs, so \( \alpha<0<\beta \).
2. Sum \( \alpha+\beta=-b<0 \), so \( \beta<-\alpha=|\alpha| \).

Teacher's Note:
1. Signs of roots follow from product and sum.
2. Sum negative with one positive root means the negative root is larger in size.

Answer: (d) \( \left(-\frac34,-\frac12\right) \)

Explanation:
1. \( f'(x)=2+6x+12x^2>0 \) (discriminant \( 36-96<0 \)), so \( f \) has exactly one real root \( s \).
2. \( f(-\frac34)=-\frac12<0 \) and \( f(-\frac12)=\frac14>0 \), so \( s\in(-\frac34,-\frac12) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \lambda<\frac43 \)

Explanation:
1. Real roots need \( (a+b+c)^2\ge3\lambda(ab+bc+ca) \), i.e. \( \lambda\le\frac{(a+b+c)^2}{3(ab+bc+ca)} \).
2. In a triangle \( a^2+b^2+c^2<2(ab+bc+ca) \), so \( (a+b+c)^2<4(ab+bc+ca) \) and the right side is less than \( \frac43 \).
3. Hence real roots force \( \lambda<\frac43 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 4

Explanation:
1. At integer \( x \) the left side takes values \( 6 \) (at \( x=0,\pm1 \)), \( 8 \) (at \( \pm2 \)), \( 12 \) (at \( \pm3 \)), \( 16 \) (at \( \pm4 \)), \( 20 \) (at \( \pm5 \)).
2. These equal \( 4k \) for \( k=2,3,4,5 \) (\( 6 \) is not a multiple of 4, and \( k=1 \) would need 4, below the minimum 6): 4 values.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( -2,\ -32 \)

Explanation:
1. Let the GP be \( \alpha,\alpha r,\alpha r^2,\alpha r^3 \). Then \( \alpha(1+r)=1 \) and \( \alpha r^2(1+r)=4 \), so \( r^2=4 \).
2. For \( r=2 \): \( \alpha=\frac13 \), giving non-integral \( p \). For \( r=-2 \): \( \alpha=-1 \), terms \( -1,2,-4,8 \).
3. So \( p=\alpha\beta=-2 \) and \( q=\gamma\delta=-32 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) one root in \( (-\infty,a) \) and the other in \( (b,+\infty) \)

Explanation:
1. Let \( f(x)=(x-a)(x-b)-1 \). Then \( f(a)=-1<0 \) and \( f(b)=-1<0 \), while \( f(x)\to+\infty \) at both ends.
2. So there is a root in \( (-\infty,a) \) and another in \( (b,\infty) \).

Teacher's Note:
1. Evaluate the quadratic at a and b.
2. A negative value between positive ends puts a on the root side.

Answer: (b) \( (-\infty,-\sqrt2)\cup(\sqrt2,\infty) \)

Explanation:
1. For \( x\ge-2 \): \( x^2-(x+2)+x=x^2-2>0 \) gives \( x>\sqrt2 \) or \( -2\le x<-\sqrt2 \).
2. For \( x<-2 \): \( x^2+(x+2)+x=x^2+2x+2>0 \) always (discriminant negative).
3. Union: \( (-\infty,-\sqrt2)\cup(\sqrt2,\infty) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) six

Explanation:
1. Write the equation as \( (x-2y-1)^2+2(y-6)^2=102 \).
2. Positive integer solutions need \( 2(y-6)^2\le102 \): the equation has solutions \( (1,5),(5,7),(21,5),(25,7),(25,13),(29,13) \).
3. So there are six solutions.

Teacher's Note:
1. Complete the square in x and then in y.
2. Check each candidate pair in the original equation.

Answer: (d) \( (-\infty,-2)\cup(2,\infty) \)

Explanation:
1. \( x^2-x+2 \) has discriminant \( 1-8<0 \), so it is always positive.
2. Hence \( f(x)>0 \) iff \( x^2>4 \), i.e. \( x<-2 \) or \( x>2 \).

Teacher's Note:
1. A quadratic with negative discriminant and positive leading coefficient is always positive.
2. Reduce the inequality to the remaining factor.

Answer: (a) \( 0,\ -1 \)

Explanation:
1. Put \( x=1-p \): \( (1-p)^2+p(1-p)+1-p=(1-p)\cdot2=0 \), so \( p=1 \).
2. The equation \( x^2+x+1=1 \) becomes \( x(x+1)=0 \), so \( x=0,-1 \).

Teacher's Note:
1. Use the factor theorem.
2. After finding p, substitute and solve the quadratic.

Answer: (d) 4

Explanation:
1. With \( c=4-a-2b \): \( ab+bc+ca=4(a+b)-(a+b)^2-b^2 \).
2. Let \( s=a+b \): \( 4s-s^2-b^2\le4s-s^2\le4 \), with equality at \( s=2,\ b=0 \) (i.e. \( a=2,b=0,c=2 \)).
3. The maximum is 4.

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (d) \( (-5,2) \)

Explanation:
1. The inequality \( x^2+2ax+10-3a>0 \) for all x needs discriminant \( 4a^2-4(10-3a)<0 \).
2. So \( a^2+3a-10<0 \), i.e. \( (a+5)(a-2)<0 \) and \( -5<a<2 \).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (b) 1210

Explanation:
1. From the sums of roots: \( a+b=10c \) and \( c+d=10a \). Since \( a \) and \( c \) are roots of their equations, \( a^2-10ac-11d=0 \) and \( c^2-10ac-11b=0 \).
2. Subtract: \( a^2-c^2=11(d-b) \). Also \( d-b=(10a-c)-(10c-a)=11(a-c) \), so \( (a-c)(a+c)=121(a-c) \) and \( a+c=121 \) (as \( a\ne c \)).
3. Then \( b+d=9(a+c)=1089 \) and \( a+b+c+d=121+1089=1210 \).

Teacher's Note:
1. Use sum of roots and the fact that each root satisfies its own equation.
2. Subtract the two equations to get a + c.

Answer: (a) \( \frac29(2p-q)(2q-p) \)

Explanation:
1. \( \alpha+\beta=p \) and \( \frac\alpha2+2\beta=q \). Solving: \( \beta=\frac{2q-p}{3} \) and \( \alpha=\frac{2(2p-q)}{3} \).
2. Then \( r=\alpha\beta=\frac29(2p-q)(2q-p) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 1

Explanation:
1. \( \log_4(x-1)=\frac12\log_2(x-1) \), so \( x-1=(x-3)^2 \), i.e. \( x^2-7x+10=0 \) and \( x=2 \) or \( 5 \).
2. The log of \( x-3 \) needs \( x>3 \), so only \( x=5 \) is valid.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 2

Explanation:
1. Discriminant \( 64(k-1)>0 \), so \( k>1 \).
2. Both roots \( \ge4 \): \( f(4)=16(k-1)(k-2)\ge0 \) and the vertex \( 4k\ge4 \), so \( k\ge2 \).
3. Smallest \( k=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \)

Explanation:
1. Let \( m=\alpha\beta \). From \( \alpha^3+\beta^3=-p^3+3pm=q \): \( m=\frac{p^3+q}{3p} \).
2. Sum of roots \( =\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{p^2}{m}-2=\frac{3p^3}{p^3+q}-2=\frac{p^3-2q}{p^3+q} \) and product 1.
3. So \( (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 3

Explanation:
1. Let the roots be \( \alpha,\alpha^2 \): product \( \alpha^3=1 \).
2. As \( p>0 \) the roots are non-real, so \( \alpha=\omega \) (non-real cube root of 1) and the sum is \( \omega+\omega^2=-1 \).
3. Then \( -\frac p3=-1 \) gives \( p=3 \).

Teacher's Note:
1. Use sum and product of roots.
2. Cube roots of unity satisfy 1 + w + w squared = 0.

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

Explanation:
1. \( \frac{\alpha^2+\beta^2}{\alpha\beta}=\frac52 \) gives \( 2\alpha^2-5\alpha\beta+2\beta^2=0 \), i.e. \( (2\alpha-\beta)(\alpha-2\beta)=0 \).
2. So one root is twice the other; with difference 1 the roots are 1 and 2.
3. The equation is \( x^2-3x+2=0 \).

Teacher's Note:
1. Factorise the homogeneous quadratic in alpha and beta.
2. Sum 3 and product 2.