JEE Mathematics Quadratic Equations MCQs Set 20

Answer: (b) infinitely many

Explanation:
1. No real root means \( D\lt0 \).
2. \( D=4\left[(1+3m)^2-(1+m^2)(1+8m)\right]=4(-8m^3+8m^2-2m)=-8m(2m-1)^2 \).
3. \( D\lt0\iff m\gt0 \) and \( m\ne\frac12 \), which contains infinitely many integers. Answer: (b)

Teacher's Note:
1. Factorise the discriminant completely before solving the inequality.
2. The coefficient \( 1+m^2 \) is always positive, so the equation is always quadratic.

Answer: (b) \( 8\sqrt5 \)

Explanation:
1. Sum of roots \( =\frac{3}{m^2+1} \), greatest when \( m^2+1 \) is least, i.e. \( m=0 \).
2. Then the equation is \( x^2-3x+1=0 \): \( \alpha+\beta=3 \), \( \alpha\beta=1 \).
3. \( |\alpha-\beta|=\sqrt{9-4}=\sqrt5 \) and \( \alpha^2+\alpha\beta+\beta^2=(\alpha+\beta)^2-\alpha\beta=8 \).
4. \( |\alpha^3-\beta^3|=|\alpha-\beta|(\alpha^2+\alpha\beta+\beta^2)=8\sqrt5 \). Answer: (b)

Teacher's Note:
1. Maximise the sum of roots first; it fixes the value of m.
2. Use the factorisation of \( \alpha^3-\beta^3 \) together with symmetric functions.

Answer: (a) 1

Explanation:
1. Put \( t=2^x\gt0 \).
2. Case \( t\ge1 \) (\( x\ge0 \)): \( 5+t-1=t^2-2t\Rightarrow t^2-3t-4=0\Rightarrow t=4 \) (reject \( -1 \)), so \( x=2 \).
3. Case \( 0\lt t\lt1 \) (\( x\lt0 \)): \( 5+1-t=t^2-2t\Rightarrow t^2-t-6=0\Rightarrow t=3 \) or \( -2 \), neither lies in \( (0,1) \), so no root.
4. Hence there is exactly one real root. Answer: (a)

Teacher's Note:
1. Split the modulus at \( 2^x=1 \) and check each root against its case.
2. A root that violates the case condition must be rejected.

Answer: (d) 3

Explanation:
1. The roots are rational when \( D=121-24\alpha \) is a perfect square.
2. \( D\ge0\Rightarrow\alpha\le5 \). Testing: \( \alpha=1:97 \), \( \alpha=2:73 \) (not squares); \( \alpha=3:49 \), \( \alpha=4:25 \), \( \alpha=5:1 \) (squares).
3. So \( \alpha=3,4,5 \), three values. Answer: (d)

Teacher's Note:
1. Rational roots need a perfect-square discriminant.
2. The condition \( D\ge0 \) limits the values to test.

Answer: (a) 11

Explanation:
1. Let \( f(x)=(c-5)x^2-2cx+(c-4) \). Then \( f(0)=c-4 \), \( f(2)=c-24 \), \( f(3)=4c-49 \).
2. One root in each interval needs \( f(0)f(2)\lt0 \) and \( f(2)f(3)\lt0 \).
3. \( (c-4)(c-24)\lt0\Rightarrow 4\lt c\lt24 \) and \( (c-24)(4c-49)\lt0\Rightarrow\frac{49}{4}\lt c\lt24 \).
4. Together \( \frac{49}4\lt c\lt24 \), so \( c=13,14,\ldots,23 \): 11 values. Answer: (a)

Teacher's Note:
1. A sign change of f across an interval shows a root inside it.
2. Take the intersection of both conditions and count integers.

Answer: (d) -300

Explanation:
1. Let the roots be \( \alpha \) and \( \alpha^3 \). The product is \( \alpha^4=\frac{256}{81} \), so \( \alpha=\pm\frac43 \).
2. \( k=-81(\alpha+\alpha^3)=-81\alpha(1+\alpha^2)=-81\cdot\left(\pm\frac43\right)\cdot\frac{25}{9}=\mp300 \).
3. So \( k=\pm300 \); only \( -300 \) appears among the options. Answer: (d)

Teacher's Note:
1. Use the product of roots first to find the root.
2. Both signs of the root give valid values of k; check which one is listed.

Answer: (c) \( 4-3\sqrt2 \)

Explanation:
1. \( \lambda+\frac1\lambda=\frac\alpha\beta+\frac\beta\alpha=1\Rightarrow\alpha^2+\beta^2=\alpha\beta\Rightarrow(\alpha+\beta)^2=3\alpha\beta \).
2. \( \alpha+\beta=-\frac{m(m-4)}{3m^2} \) and \( \alpha\beta=\frac2{3m^2} \), so \( \frac{(m-4)^2}{9m^2}=\frac2{m^2} \).
3. \( (m-4)^2=18\Rightarrow m=4\pm3\sqrt2 \). The least value is \( 4-3\sqrt2 \). Answer: (c)

Teacher's Note:
1. Convert the condition on the ratio into a condition on sum and product.
2. Cancel \( m^2 \) only after noting \( m\ne0 \).

Answer: (d) 11

Explanation:
1. The left side is \( \sum_{k=1}^{n}(x+k-1)(x+k)=nx^2+n^2x+\frac{n(n^2-1)}3 \).
2. So \( nx^2+n^2x+\frac{n(n^2-1)}3-10n=0\Rightarrow3x^2+3nx+n^2-31=0 \).
3. Consecutive integer roots mean \( (\alpha-\beta)^2=1 \): \( (\alpha+\beta)^2-4\alpha\beta=n^2-\frac{4(n^2-31)}3=1 \).
4. \( 3n^2-4n^2+124=3\Rightarrow n^2=121\Rightarrow n=11 \). Answer: (d)

Teacher's Note:
1. Sum the series using standard formulas for squares and odd numbers.
2. Two consecutive integers differ by exactly 1.

Answer: (c) 3

Explanation:
1. Since \( \alpha^2=6\alpha+2 \) and \( \beta^2=6\beta+2 \), multiply by \( \alpha^8,\beta^8 \): \( \alpha^{10}=6\alpha^9+2\alpha^8 \) and \( \beta^{10}=6\beta^9+2\beta^8 \).
2. Subtracting: \( a_{10}=6a_9+2a_8 \), so \( a_{10}-2a_8=6a_9 \).
3. Hence \( \frac{a_{10}-2a_8}{2a_9}=\frac{6a_9}{2a_9}=3 \). Answer: (c)

Teacher's Note:
1. Use the equation satisfied by each root to build a recurrence.
2. Subtract the two relations to get a relation among the \( a_n \).

Answer: (d) \( \frac{2\sqrt{13}}9 \)

Explanation:
1. \( \alpha+\beta=-\frac qp \) and \( \alpha\beta=\frac rp \). A.P. gives \( 2q=p+r \).
2. \( \frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=-\frac qr=4\Rightarrow q=-4r \), then \( 2(-4r)=p+r\Rightarrow p=-9r \).
3. So \( \alpha+\beta=-\frac{q}{p}=-\frac49 \) and \( \alpha\beta=\frac rp=-\frac19 \).
4. \( (\alpha-\beta)^2=\frac{16}{81}+\frac49=\frac{52}{81} \), so \( |\alpha-\beta|=\frac{2\sqrt{13}}9 \). Answer: (d)

Teacher's Note:
1. Express p and q in terms of r using the two conditions.
2. Use \( (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta \).

Answer: (a) 1:2:3

Explanation:
1. \( x^2+2x+3=0 \) has discriminant \( 4-12\lt0 \), so its roots are non-real conjugates.
2. The second equation has real coefficients, so a non-real root brings its conjugate too; both roots are common.
3. Hence the equations are proportional: \( \frac a1=\frac b2=\frac c3 \), so \( a:b:c=1:2:3 \). Answer: (a)

Teacher's Note:
1. Non-real roots of a real-coefficient equation occur in conjugate pairs.
2. Two quadratics with both roots in common have proportional coefficients.

Answer: (c) \( \beta\in[1,\infty) \)

Explanation:
1. The coefficients are real, so complex roots are conjugates. Two distinct roots with real part 1 are \( 1+iy \) and \( 1-iy \) with \( y\ne0 \).
2. Product of roots \( =\beta=(1+iy)(1-iy)=1+y^2 \).
3. So \( \beta=1+y^2\gt1 \), hence it is necessary that \( \beta\in[1,\infty) \). Answer: (c)

Teacher's Note:
1. Distinct roots on one vertical line must be conjugates.
2. The product of conjugate roots is the square of the modulus.

Answer: (c) 1

Explanation:
1. The roots are \( \frac{1\pm\sqrt3i}{2}=e^{\pm i\pi/3} \), i.e. \( -\omega \) and \( -\omega^2 \).
2. \( \alpha^{2009}+\beta^{2009}=2\cos\frac{2009\pi}{3} \). Since \( 2009=6\cdot334+5 \), the angle reduces to \( \frac{5\pi}{3} \).
3. \( 2\cos\frac{5\pi}3=2\cdot\frac12=1 \). Answer: (c)

Teacher's Note:
1. Write the roots as \( e^{\pm i\pi/3} \) and use De Moivre's theorem.
2. Reduce the exponent modulo 6, because \( e^{i\pi/3} \) has order 6.

Answer: (c) greater than \( -4ab \)

Explanation:
1. Imaginary roots: \( c^2-4ab\lt0 \), so \( c^2\lt4ab \) and \( -c^2\gt-4ab \); also \( ab\gt0 \).
2. The expression has leading coefficient \( 3b^2\gt0 \), so it has a minimum value \( \frac{4(3b^2)(2c^2)-(6bc)^2}{4(3b^2)}=\frac{-12b^2c^2}{12b^2}=-c^2 \).
3. Therefore the expression is at least \( -c^2\gt-4ab \). Answer: (c) It cannot exceed \( 4ab \) always, since its minimum \( -c^2\le0\lt4ab \).

Teacher's Note:
1. Use the minimum value formula \( \frac{4ac-b^2}{4a} \) for an upward parabola.
2. Compare the minimum with the bound from the discriminant condition.

Answer: (a) \( (-3,3) \)

Explanation:
1. \( \alpha+\beta=-a \), \( \alpha\beta=1 \), so \( |\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4\alpha\beta}=\sqrt{|a^2-4|} \).
2. For \( |a|\ge2 \): \( \sqrt{a^2-4}\lt\sqrt5\Rightarrow a^2\lt9\Rightarrow2\le|a|\lt3 \). For \( |a|\lt2 \) the difference is at most 2, which is less than \( \sqrt5 \).
3. So \( |a|\lt3 \), i.e. \( a\in(-3,3) \). Answer: (a)

Teacher's Note:
1. Express the difference of roots through the discriminant.
2. Square the inequality only when both sides are non-negative.

Answer: (b) \( -1\lt m\lt3 \)

Explanation:
1. \( x^2-2mx+m^2-1=(x-m)^2-1=0 \), so the roots are \( x=m\pm1 \).
2. Both lie in \( (-2,4) \): \( m-1\gt-2 \) and \( m+1\lt4 \), i.e. \( m\gt-1 \) and \( m\lt3 \).
3. So \( -1\lt m\lt3 \). Answer: (b)

Teacher's Note:
1. Factorise the equation to find the roots directly.
2. The smaller root must exceed \( -2 \) and the larger must be below 4.

Answer: (a) 1

Explanation:
1. Let the roots be \( n \) and \( n+1 \). Then \( b=2n+1 \) and \( c=n(n+1) \).
2. \( b^2-4c=(2n+1)^2-4n(n+1)=4n^2+4n+1-4n^2-4n=1 \). Answer: (a)

Teacher's Note:
1. Take the two integers as \( n \) and \( n+1 \).
2. Compute the discriminant directly.

Answer: (c) smaller than \( \alpha \)

Explanation:
1. Let \( f(x)=a_nx^n+\ldots+a_1x \). The second equation is \( f'(x)=0 \).
2. \( f(0)=0 \) and \( f(\alpha)=0 \), and \( f \) is a polynomial, so Rolle's theorem gives a point \( c\in(0,\alpha) \) with \( f'(c)=0 \).
3. Hence \( f'(x)=0 \) has a positive root smaller than \( \alpha \). Answer: (c)

Teacher's Note:
1. Recognise the second polynomial as the derivative of the first.
2. Apply Rolle's theorem on \( [0,\alpha] \).

Answer: (a) \( \frac{49}4 \)

Explanation:
1. \( x=4 \) in \( x^2+px+12=0 \): \( 16+4p+12=0\Rightarrow p=-7 \).
2. The second equation is \( x^2-7x+q=0 \) with equal roots, so \( D=49-4q=0 \).
3. Hence \( q=\frac{49}4 \). Answer: (a)

Teacher's Note:
1. Find p from the given root first.
2. Equal roots mean the discriminant is zero.

Answer: (b) \( x^2-18x+16=0 \)

Explanation:
1. \( \frac{\alpha+\beta}2=9\Rightarrow\alpha+\beta=18 \) and \( \sqrt{\alpha\beta}=4\Rightarrow\alpha\beta=16 \).
2. The equation is \( x^2-(\alpha+\beta)x+\alpha\beta=0 \), i.e. \( x^2-18x+16=0 \). Answer: (b)

Teacher's Note:
1. Convert the two means into sum and product.
2. Use \( x^2-(\text{sum})x+\text{product}=0 \).

Answer: (c) harmonic progression

Explanation:
1. \( \alpha+\beta=-\frac ba \) and \( \alpha\beta=\frac ca \). Then \( \frac1{\alpha^2}+\frac1{\beta^2}=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{b^2-2ac}{c^2} \).
2. The condition gives \( -\frac ba=\frac{b^2-2ac}{c^2}\Rightarrow 2a^2c=ab^2+bc^2 \).
3. Dividing by \( abc \): \( \frac{2a}b=\frac bc+\frac ca \), i.e. \( \frac ca,\frac ab,\frac bc \) are in A.P.
4. Therefore their reciprocals \( \frac ac,\frac ba,\frac cb \) are in H.P. Answer: (c)

Teacher's Note:
1. Write the sum of squares of reciprocals in terms of the sum and product of roots.
2. Reciprocals of an A.P. form an H.P.

Answer: (b) 4

Explanation:
1. For \( x\ge0 \): \( x^2-3x+2=0\Rightarrow x=1,2 \).
2. For \( x\lt0 \): \( x^2+3x+2=0\Rightarrow x=-1,-2 \).
3. All four values satisfy their cases, so there are 4 real solutions. Answer: (b)

Teacher's Note:
1. Split the modulus at \( x=0 \).
2. Check each root against its case.

Answer: (a) \( \frac23 \)

Explanation:
1. Let the roots be \( \alpha \) and \( 2\alpha \). Then \( 3\alpha=-\frac{3a-1}{a^2-5a+3} \) and \( 2\alpha^2=\frac2{a^2-5a+3} \).
2. Eliminating \( \alpha \): \( \alpha=-\frac{3a-1}{3(a^2-5a+3)} \), so \( \frac{(3a-1)^2}{9(a^2-5a+3)^2}=\frac1{a^2-5a+3} \), i.e. \( (3a-1)^2=9(a^2-5a+3) \).
3. \( 9a^2-6a+1=9a^2-45a+27\Rightarrow39a=26\Rightarrow a=\frac23 \). Answer: (a)

Teacher's Note:
1. Take the roots as \( \alpha \) and \( 2\alpha \) and use both sum and product.
2. Eliminate \( \alpha \) by squaring the sum relation.

Answer: (b) \( 3x^2-19x+3=0 \)

Explanation:
1. Both \( \alpha \) and \( \beta \) satisfy \( t^2-5t+3=0 \), and they are distinct, so they are its two roots: \( \alpha+\beta=5 \), \( \alpha\beta=3 \).
2. \( \frac\alpha\beta+\frac\beta\alpha=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{25-6}3=\frac{19}3 \) and the product is 1.
3. The equation is \( x^2-\frac{19}3x+1=0 \), i.e. \( 3x^2-19x+3=0 \). Answer: (b)

Teacher's Note:
1. Spot that \( \alpha,\beta \) are the two roots of one quadratic.
2. Form the new equation from its sum and product.

Answer: (b) 2

Explanation:
1. \( 3^{2x^2-7x+7}=3^2\Rightarrow2x^2-7x+7=2\Rightarrow2x^2-7x+5=0 \).
2. \( D=49-40=9\gt0 \), so there are two distinct real roots (\( x=1 \) and \( x=\frac52 \)). Answer: (b)

Teacher's Note:
1. Equal bases give equal exponents.
2. The discriminant decides the number of real roots.