JEE Mathematics Quadratic Equations MCQs Set 11

Answer: (a) infinitely many

Explanation:
1. \( D=4(1+3m)^2-4(1+m^2)(1+8m)=-8m(2m-1)^2 \).
2. No real root means D < 0: \( m\gt0 \) and \( m\ne\frac12 \).
3. Every positive integer qualifies, so there are infinitely many.

Teacher's Note:
1. Expand the discriminant and factor it.
2. Count the integers in the solution set.

Answer: (c) 10

Explanation:
1. Put \( s=\sqrt x \). If s ≥ 2: \( s-2+s^2-4s+2=s^2-3s=0 \), so s = 3 (x = 9).
2. If s < 2: \( 2-s+s^2-4s+2=s^2-5s+4=0 \), so s = 1 (x = 1); s = 4 is rejected.
3. Sum = 9 + 1 = 10.

Teacher's Note:
1. Substitute s = √x and split the modulus at s = 2.
2. Discard roots that do not satisfy their case.

Answer: (c) 7

Explanation:
1. Leading coefficient positive: \( 1+2m\gt0\Rightarrow m\gt-\frac12 \).
2. D < 0: \( 4(1+3m)^2-16(1+2m)(1+m)=4(m^2-6m-3)\lt0\Rightarrow3-2\sqrt3\lt m\lt3+2\sqrt3 \), i.e. about −0.46 < m < 6.46.
3. Integers m = 0, 1, …, 6: seven values.

Teacher's Note:
1. Positive for all x needs a > 0 and D < 0.
2. Estimate 2√3 ≈ 3.46 to list the integers.

Answer: (d) −300

Explanation:
1. Let the roots be r and \( r^3 \): \( r^4=\frac{256}{81} \), so \( r=\pm\frac43 \).
2. For r = 4/3 the roots are 4/3 and 64/27 and \( -\frac k{81}=\frac43+\frac{64}{27}=\frac{100}{27} \), so k = −300.
3. (For r = −4/3 we get k = 300, which is not among the options.)

Teacher's Note:
1. Use the product of roots to find r.
2. Then use the sum of roots to find k.

Answer: (d) 2

Explanation:
1. Write as \( x^2+(3-\lambda)x+(2-\lambda)=0 \). Sum of squares \( =(3-\lambda)^2-2(2-\lambda)=\lambda^2-4\lambda+5=(\lambda-2)^2+1 \).
2. Least at \( \lambda=2 \) (the roots are real since \( D=(\lambda-1)^2\ge0 \)).

Teacher's Note:
1. Express the sum of squares as a quadratic in λ.
2. Complete the square; check the roots stay real.

Answer: (a) (4, 5]

Explanation:
1. Distinct real roots: \( m^2-16\gt0 \); since the roots are positive, \( m\gt4 \).
2. Axis \( \frac m2\in(1,5) \) holds. \( f(1)=5-m\ge0 \) and \( f(5)=29-5m\ge0 \) give \( m\le5 \).
3. So \( 4\lt m\le5 \): the interval (4, 5].

Teacher's Note:
1. Apply D > 0, axis inside the interval, and f(1), f(5) ≥ 0.
2. Check the end value m = 5, which gives roots 1 and 4.

Answer: (c) 3

Explanation:
1. \( D=121-24\alpha \) must be a perfect square. Positive integers with D ≥ 0 give α ≤ 5.
2. α = 3: D = 49; α = 4: D = 25; α = 5: D = 1. (α = 1, 2 give 97 and 73, not squares.)
3. So 3 values.

Teacher's Note:
1. Rational roots need a perfect-square discriminant.
2. Try each possible α from 1 to 5.

Answer: (a) 11

Explanation:
1. Let \( f(x)=(c-5)x^2-2cx+(c-4) \): \( f(0)=c-4,\ f(2)=c-24,\ f(3)=4c-49 \).
2. Roots separated as required: \( f(0)f(2)\lt0 \) and \( f(2)f(3)\lt0 \). First: 4 < c < 24. Second: \( (c-24)(4c-49)\lt0\Rightarrow\frac{49}4\lt c\lt24 \).
3. Integers 13 to 23: 11 values.

Teacher's Note:
1. Use sign changes of f at 0, 2 and 3.
2. List the integers in the intersection.

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

Explanation:
1. \( \lambda+\frac1\lambda=1 \) means \( r_1^2+r_2^2=r_1r_2 \), i.e. \( (r_1+r_2)^2=3r_1r_2 \).
2. Sum \( =-\frac{m-4}{3m} \), product \( =\frac2{3m^2} \): \( \frac{(m-4)^2}{9m^2}=\frac2{m^2} \), so \( (m-4)^2=18 \).
3. m = 4 ± 3√2; the least value is \( 4-3\sqrt2 \).

Teacher's Note:
1. Convert the ratio condition into a relation between sum and product.
2. Solve (m − 4)² = 18 and pick the smaller root.

Answer: (a) −6

Explanation:
1. \( \frac{4+3\sqrt3}{2+\sqrt3}=(4+3\sqrt3)(2-\sqrt3)=2\sqrt3-1 \).
2. This zero satisfies \( (x+1)^2=12 \), i.e. \( x^2+2x-11=0 \): a = 2, b = −11.
3. Dividing g by this quadratic leaves remainder \( 2x+1 \), so \( g=2(2\sqrt3-1)+1=4\sqrt3-1 \): c = 4, d = −1.
4. Sum: 2 − 11 + 4 − 1 = −6.

Teacher's Note:
1. Rationalise the zero first, then form its minimal polynomial.
2. Reduce g modulo that polynomial to evaluate it at the zero.

Answer: (c) 5

Explanation:
1. Quarter-discriminant: \( (a+b)^2-(a-b+8)=a^2+(2b-1)a+b^2+b-8 \).
2. This must be positive for every a: \( (2b-1)^2-4(b^2+b-8)=33-8b\lt0\Rightarrow b\gt\frac{33}8 \).
3. The smallest natural number is 5.

Teacher's Note:
1. Treat the discriminant as a quadratic in a.
2. Positive for all a means its own discriminant is negative.

Answer: (a) \( \left(-\frac34,\frac32\right) \)

Explanation:
1. \( 4x^2+6x+4=4\left(x+\frac34\right)^2+\frac74\ge\frac74 \) with equality at x = −3/4.
2. \( 4y^2-12y+25=4\left(y-\frac32\right)^2+16\ge16 \) with equality at y = 3/2.
3. The product is at least \( \frac74\cdot16=28 \), so equality forces (−3/4, 3/2).

Teacher's Note:
1. Find the minimum of each factor separately.
2. The product equals its minimum value only when both factors are at their minima.

Answer: (a) 20

Explanation:
1. Let \( t=\sqrt{x^2+18x+45}\ge0 \). Then \( t^2-15=2t \), so \( t^2-2t-15=0 \) and t = 5 (t = −3 rejected).
2. Hence \( x^2+18x+45=25\Rightarrow x^2+18x+20=0 \), with real roots (D = 244 > 0).
3. Product of roots = 20.

Teacher's Note:
1. Substitute t for the square root to get a quadratic in t.
2. Reject negative t and return to x.

Answer: (b) 4

Explanation:
1. \( \frac{2x^2+2x+3}{x^2+x+1}=2+\frac1{x^2+x+1} \).
2. \( x^2+x+1\ge\frac34 \), so the expression is at most \( 2+\frac43=\frac{10}3\approx3.33 \).
3. The smallest integer p is 4.

Teacher's Note:
1. Rewrite the fraction as 2 + 1/(x² + x + 1).
2. Find the maximum using the minimum of the denominator.

Answer: (a) \( (-2,1) \)

Explanation:
1. The denominator is positive, so multiply through.
2. Left part: \( 4x^2+(a+3)x+1\gt0 \) for all x requires \( (a+3)^2-16\lt0\Rightarrow-7\lt a\lt1 \).
3. Right part: \( x^2+(2-a)x+4\gt0 \) requires \( (2-a)^2-16\lt0\Rightarrow-2\lt a\lt6 \). Intersection: (−2, 1).

Teacher's Note:
1. Clear the positive denominator to get two quadratic inequalities.
2. Intersect the conditions from both.

Answer: (a) 277

Explanation:
1. Put y = f(x): \( (y-1)x^2+(2y-a)x+(3y-b)=0 \). Real x needs \( D\ge0 \), i.e. \( 8y^2-(4b-4a+12)y-(a^2-4b)\le0 \).
2. This must hold exactly for −5 ≤ y ≤ 4, so −5 and 4 are its roots: sum \( -1=\frac{4b-4a+12}8\Rightarrow b=a-5 \) and product \( -20=-\frac{a^2-4b}8\Rightarrow a^2-4b=160 \).
3. Then \( a^2-4a-140=0 \), so a = 14 (natural), b = 9 and \( a^2+b^2=196+81=277 \).

Teacher's Note:
1. Write y = f(x) as a quadratic in x and demand D ≥ 0.
2. The endpoints of the range are the roots of the resulting quadratic in y.

Answer: (a) −6

Explanation:
1. \( x^3-4x^2+x+6=(x+1)(x-2)(x-3) \).
2. Each quadratic is (x + 1) times one of (x − 2), (x − 3). Try \( x^2+ax+b=(x+1)(x-2)=x^2-x-2 \): a = −1, b = −2, then \( x^2+bx+c=x^2-2x+c \) must be \( (x+1)(x-3)=x^2-2x-3 \), so c = −3.
3. (The other assignment gives b = −3 and −1 inconsistently.) Sum = −1 − 2 − 3 = −6.

Teacher's Note:
1. Factorise the LCM and distribute the factors between the two quadratics.
2. Check consistency of b in both quadratics.

Answer: (a) 30

Explanation:
1. \( a+b=2c,\ ab=-5d,\ c+d=2a,\ cd=-5b \). So \( b=2c-a \) and \( d=2a-c \).
2. Substitute in the product relations: \( 2ac-a^2=-10a+5c \) and \( 2ac-c^2=-10c+5a \). Subtract: \( c^2-a^2=15(c-a) \).
3. Since c ≠ a, a + c = 15, hence \( a+b+c+d=2c+2a=30 \).

Teacher's Note:
1. Eliminate b and d using the two sum relations.
2. Subtracting the two product equations factorises nicely, using c ≠ a.

Answer: (a) 146

Explanation:
1. The difference of the two quadratics is \( (a-b)(x-1) \), so both share the root x = 1 and f has the three roots 1, r, s with \( x^2+ax+b=(x-1)(x-r) \) and \( x^2+bx+a=(x-1)(x-s) \).
2. Comparing: \( a=-(1+r)=s \) and \( b=r=-(1+s) \), so r + s = −1. Product of roots: \( 1\\cdot r\\cdot s=-72 \).
3. Then r, s = −9 and 8, and the sum of squares is \( 1+81+64=146 \).

Teacher's Note:
1. Subtract the two quadratics to find a common root.
2. Use the sum and product relations of the cubic's roots.

Answer: (a) \( 2\\sqrt2-2 \)

Explanation:
1. Rationalise: \( y=\\frac{(x^2+2)^2-(x^4+4)}{x\\,(x^2+2+\\sqrt{x^4+4})}=\\frac{4x^2}{x(x^2+2+\\sqrt{x^4+4})}=\\frac{4}{x+\\frac2x+\\sqrt{x^2+\\frac4{x^2}}} \).
2. By AM-GM, \( x+\\frac2x\\ge2\\sqrt2 \) and \( x^2+\\frac4{x^2}\\ge4 \), both with equality at \( x=\\sqrt2 \).
3. So the denominator is at least \( 2\\sqrt2+2 \) and \( y_{max}=\\frac4{2\\sqrt2+2}=2(\\sqrt2-1)=2\\sqrt2-2 \).

Teacher's Note:
1. Rationalise the numerator to turn the expression into a reciprocal of a sum.
2. Apply AM-GM to each term of the new denominator.

Answer: (a) \( 1-\\sqrt2\\ \\text{or}\\ 5+\\sqrt{10} \)

Explanation:
1. The vertex is at \( x=\\frac p2 \). If \( 0\\le p\\le4 \) the least value is \( 2-2p=3 \), giving p = −1/2, not in [0, 4]: rejected.
2. If p < 0 the minimum is at x = 0: \( p^2-2p+2=3\\Rightarrow p=1-\\sqrt2 \) (negative, accepted).
3. If p > 4 the minimum is at x = 2: \( p^2-10p+18=3\\Rightarrow p=5+\\sqrt{10} \) (accepted).

Teacher's Note:
1. Consider three cases according to where the vertex lies relative to [0, 2].
2. Check each solution against its case.

Answer: (a) 86

Explanation:
1. The product of all four roots is −1984 = (−32)(62), so the polynomial is \( (x^2+px-32)(x^2+qx+62) \).
2. Matching \( x^3 \): p + q = −18. Matching x: \( 62p-32q=200 \). Solving gives p = −4, q = −14.
3. Matching \( x^2 \): \( k=62-32+pq=30+56=86 \).

Teacher's Note:
1. Use the total product of roots to find the product of the other pair.
2. Match coefficients of the factorised form.