JEE Mathematics Quadratic Equations MCQs Set 18

Answer: (a) no solution

Explanation:
1. \( \pi^x>0 \) for every real x.
2. \( -2x^2+6x-9=-2\left(x-\dfrac32\right)^2-\dfrac92\le-\dfrac92<0 \).
3. A positive value can never equal a negative one, so there is no solution.

Teacher's Note:
1. Compare the sign of both sides before solving anything.
2. Completing the square shows the quadratic is always negative.

Answer: (c) real and distinct

Explanation:
1. For \( 2x^2+bx+c=0 \): \( b^2-8c>0 \).
2. For the second equation \( D'=(b-4c)^2-8c(2c-b+1)=b^2-8bc+16c^2-16c^2+8bc-8c=b^2-8c \).
3. So \( D'>0 \) and the roots are real and distinct.

Teacher's Note:
1. Expand the discriminant of the second equation carefully; many terms cancel.
2. It turns out to equal the discriminant of the first equation.

Answer: (d) 3

Explanation:
1. 2 lies between the roots, so \( f(2)<0 \): \( 16-32+\lambda<0\Rightarrow\lambda<16 \).
2. Also \( f(1)>0\Rightarrow\lambda>12 \) and \( f(3)>0\Rightarrow\lambda>12 \).
3. So \( 12<\lambda<16 \) and the integers are 13, 14, 15, i.e. 3 values.

Teacher's Note:
1. Place the numbers 1, 2, 3 relative to the roots using the sign of f.
2. Count integers strictly between the end points.

Answer: (b) \( \left[-2,\dfrac25\right) \)

Explanation:
1. Write \( (\lambda^2+\lambda-2)x^2+(\lambda+2)x-1<0 \) for all x. Note \( \lambda^2+\lambda-2=(\lambda+2)(\lambda-1) \).
2. If \( \lambda=-2 \), the expression is \( -1<0 \), which is true.
3. Otherwise we need \( \lambda<1 \) and \( D=(\lambda+2)^2+4(\lambda+2)(\lambda-1)<0 \), i.e. \( (\lambda+2)(5\lambda-2)<0 \), giving \( -2<\lambda<\dfrac25 \).
4. Combined with \( \lambda=-2 \), the answer is \( \left[-2,\dfrac25\right) \).

Teacher's Note:
1. Do not forget the case in which the quadratic term vanishes.
2. Here that case gives a true statement, so the end point is included.

Answer: (b) have negative real parts

Explanation:
1. If the roots are real, their sum \( -\dfrac ba<0 \) and product \( \dfrac ca>0 \) make both negative.
2. If the roots are complex, the real part of each is \( -\dfrac{b}{2a}<0 \).
3. In both cases the real parts are negative.

Teacher's Note:
1. Treat the real and the complex case separately.
2. Option (a) is wrong because the roots can be complex.

Answer: (b) \( a,\ b,\ c \)

Explanation:
1. Since \( \alpha,\beta,\gamma \) are the roots, \( (x-a)(x-b)(x-c)-d\equiv(x-\alpha)(x-\beta)(x-\gamma) \).
2. Hence \( (x-\alpha)(x-\beta)(x-\gamma)+d\equiv(x-a)(x-b)(x-c) \), whose roots are a, b, c.

Teacher's Note:
1. Both cubics have the same leading coefficient, so the identity holds exactly.
2. Shifting the constant term moves between the two equations.

Answer: (c) \( -(q+r) \)

Explanation:
1. \( (\alpha-\gamma)(\alpha-\delta)=\alpha^2+p\alpha-r \) because \( \gamma,\delta \) are roots of \( x^2+px-r \).
2. Since \( \alpha^2+p\alpha=-q \), the value is \( -q-r=-(q+r) \).

Teacher's Note:
1. The product of differences with all roots of a polynomial equals the polynomial evaluated at that point.
2. Use the equation satisfied by the first root to simplify.

Answer: (d) \( c^2>2b^2 \)

Explanation:
1. \( f(x)=(x+b)^2+2c^2-b^2 \), so \( \min f=2c^2-b^2 \).
2. \( g(x)=-(x+c)^2+b^2+c^2 \), so \( \max g=b^2+c^2 \).
3. \( 2c^2-b^2>b^2+c^2\Rightarrow c^2>2b^2 \).

Teacher's Note:
1. Complete the square for both functions to read off the extreme values.
2. Compare the minimum of one with the maximum of the other.

Answer: (b) \( [a^2-m^2,\ a^2) \)

Explanation:
1. \( |\alpha-\beta|=\sqrt{4a^2-4b}=2\sqrt{a^2-b} \).
2. Distinct roots: \( a^2-b>0 \). Difference at most \( 2m \): \( \sqrt{a^2-b}\le m\Rightarrow b\ge a^2-m^2 \).
3. So \( a^2-m^2\le b<a^2 \).

Teacher's Note:
1. Express the difference of roots through the discriminant.
2. Notice which ends of the interval are included.

Answer: (a) at least one of the equations has real roots

Explanation:
1. \( D_1+D_2=p^2+r^2-4(q+s)=p^2+r^2-2pr=(p-r)^2\ge0 \).
2. The sum of the discriminants is non-negative, so at least one of them is non-negative.

Teacher's Note:
1. Add the two discriminants and look for a perfect square.
2. A non-negative sum means they cannot both be negative.

Answer: (c) \( \left[-\dfrac12,1\right] \)

Explanation:
1. \( (a+b+c)^2=1+2(ab+bc+ca)\ge0\Rightarrow ab+bc+ca\ge-\dfrac12 \).
2. \( (a-b)^2+(b-c)^2+(c-a)^2\ge0\Rightarrow 2-2(ab+bc+ca)\ge0\Rightarrow ab+bc+ca\le1 \).
3. So the interval is \( \left[-\dfrac12,1\right] \).

Teacher's Note:
1. Both bounds come from two different non-negative squares.
2. Each bound is attained, so both ends are closed.

Answer: (b) \( \dfrac{a-b}{b-c},\ 1 \)

Explanation:
1. The sum of the coefficients is \( (b-c)+(c-a)+(a-b)=0 \), so \( x=1 \) is a root.
2. The product of the roots is \( \dfrac{a-b}{b-c} \), so the other root is \( \dfrac{a-b}{b-c} \).

Teacher's Note:
1. When the coefficients add to zero, x = 1 is always a root.
2. Use the product of roots to find the other one quickly.

Answer: (b) \( \left(2,\dfrac94\right) \)

Explanation:
1. The equation is \( x^2-x+(p-2)=0 \).
2. Distinct roots: \( 1-4(p-2)>0\Rightarrow p<\dfrac94 \). Positive roots: product \( p-2>0 \) and sum \( 1>0 \).
3. Hence \( 2<p<\dfrac94 \).

Teacher's Note:
1. Rewrite in standard form first, then apply the conditions for positive distinct roots.
2. Sum is already positive, so only the product and discriminant matter.

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

Explanation:
1. Since \( x^4+x^2+1=(x^2+x+1)(x^2-x+1) \), factor out \( x^2+x+1 \) (never zero for real x).
2. The remaining equation is \( (a-1)(x^2+x+1)-(a+1)(x^2-x+1)=0 \), i.e. \( x^2-ax+1=0 \).
3. Real and distinct roots: \( a^2-4>0 \), so \( |a|>2 \).

Teacher's Note:
1. Factorise the quartic to find the common factor.
2. The factor with no real roots can be discarded.

Answer: (c) \( -7 \)

Explanation:
1. Write each term as \( \dfrac{1+\alpha}{1-\alpha}=-1+\dfrac{2}{1-\alpha} \). The required sum is \( -3+2\sum\dfrac{1}{1-\alpha} \).
2. Let \( f(x)=x^3-x-1 \). Then \( \sum\dfrac{1}{1-\alpha}=\dfrac{f'(1)}{f(1)}=\dfrac{2}{-1}=-2 \).
3. The sum is \( -3+2(-2)=-7 \).

Teacher's Note:
1. The sum of \( 1/(k-\alpha) \) over the roots is f'(k)/f(k).
2. Splitting off a constant from each fraction makes the calculation easier.

Answer: (a) both C1 and C2 are satisfied

Explanation:
1. Real roots need \( D\ge0 \) (C1).
2. Positive roots need sum \( -\dfrac ba>0 \) and product \( \dfrac ca>0 \), which is exactly C2.
3. So both conditions are needed together.

Teacher's Note:
1. List separately the requirements for realness, positive sum and positive product.
2. Each condition alone is not enough.

Answer: (d) \( c\in(-6,0) \)

Explanation:
1. Put \( y=\dfrac{x^2-x+c}{x^2+x+2c} \): \( (y-1)x^2+(y+1)x+(2cy-c)=0 \). For every y we need \( D\ge0 \), i.e. \( (1-8c)y^2+(12c+2)y+(1-4c)\ge0 \) for all y.
2. This needs \( 1-8c>0 \) and \( (12c+2)^2-4(1-8c)(1-4c)\le0 \Rightarrow 16c(c+6)\le0 \), so \( -6\le c\le0 \).
3. At \( c=0 \) and \( c=-6 \) numerator and denominator share a factor and the value \( y=1 \) is never taken, so those two end points are excluded.
4. Hence \( c\in(-6,0) \).

Teacher's Note:
1. Check the end points separately, since a common factor can remove a value from the range.
2. A quadratic in y that is non-negative for all y needs a non-positive discriminant.

Answer: (a) irrational

Explanation:
1. \( D=b^2-4ac=20d^2 \).
2. \( D>0 \), so the roots are real and distinct. \( \sqrt{20d^2}=2d\sqrt5 \) is irrational, so the roots are irrational.

Teacher's Note:
1. For integer coefficients, rational roots need a perfect square discriminant.
2. 20 is not a perfect square, so the discriminant is not either.

Answer: (b) \( a>\dfrac{11}{9} \)

Explanation:
1. We need \( f(3)>0 \): \( 9-18a+2-2a+9a^2=9a^2-20a+11>0\Rightarrow(9a-11)(a-1)>0 \).
2. Also \( D=36a^2-4(9a^2-2a+2)=8a-8\ge0\Rightarrow a\ge1 \), and the vertex \( 3a>3\Rightarrow a>1 \).
3. From these, \( a>\dfrac{11}{9} \).

Teacher's Note:
1. Use all three conditions: discriminant, f(3) and the vertex.
2. Intersect the results to find the final range.

Answer: (a) 0

Explanation:
1. The first is \( (6k+2)x^2+rx+(3k-1)=0 \) and the second is \( (12k+4)x^2+px+(6k-2)=0 \).
2. The second equals twice the first, so \( p=2r \) and \( 2r-p=0 \).

Teacher's Note:
1. Both roots common means the equations are proportional.
2. Compare the coefficients of x² and x only.

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

Explanation:
1. Product \( c<0 \): the roots have opposite signs, so \( \alpha<0<\beta \).
2. Sum \( -b<0 \): the negative root has the larger magnitude, so \( \beta<|\alpha| \).

Teacher's Note:
1. The product tells the signs and the sum tells which root is larger in size.
2. Draw a quick number line to visualise the roots.

Answer: (a) \( \dfrac{3b}{2a} \)

Explanation:
1. \( \sum\alpha=0,\ \sum\alpha\beta=a,\ \alpha\beta\gamma=-b \).
2. \( \sum\alpha^2=0-2a=-2a \) and, since \( \sum\alpha=0 \), \( \sum\alpha^3=3\alpha\beta\gamma=-3b \).
3. The ratio is \( \dfrac{-3b}{-2a}=\dfrac{3b}{2a} \).

Teacher's Note:
1. When the sum of the roots is zero, the sum of cubes equals three times the product.
2. Check signs carefully in the ratio.