JEE Mathematics Quadratic Equations MCQs Set 04

Answer: (a) 7

Explanation:
1. For each root \( r \): \( r-2=\frac{1}{r} \), so \( r^2-2r-1=0 \).
2. Both distinct roots satisfy this quadratic, so it is the given equation: \( b=-2,\ c=-1 \).
3. \( b^2+c^2+bc=4+1+2=7 \).

Teacher's Note:
1. When every root of a quadratic satisfies a second quadratic, compare the two equations directly.
2. Two distinct roots fix a monic quadratic completely, so the coefficients can be read off.

Answer: (a) \( \frac{1}{3} \)

Explanation:
1. \( 1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \).
2. Since \( a^2+b^2+c^2\ge ab+bc+ca \), we get \( 1\ge 3(ab+bc+ca) \).
3. So \( ab+bc+ca\le \frac13 \), with equality when \( a=b=c=\frac13 \).

Teacher's Note:
1. Link the symmetric sum to the square of the total using the identity for (a+b+c)².
2. The bound is attained only when all three numbers are equal.

Answer: (b) \( p\ge -\frac14 \)

Explanation:
1. Put \( u=\sqrt[3]{x} \). Then \( \sqrt[3]{u^3+3p+1}=1+u \).
2. Cubing: \( u^3+3p+1=1+3u+3u^2+u^3 \), so \( p=u^2+u \).
3. \( u^2+u=\left(u+\frac12\right)^2-\frac14\ge -\frac14 \), and every such \( p \) gives a real \( u \).

Teacher's Note:
1. Substituting the cube root removes the radicals and leaves a simple quadratic in u.
2. A real solution exists exactly when p lies in the range of u² + u.

Answer: (c) 12

Explanation:
1. The product of the five roots is \( 420=2^2\cdot3\cdot5\cdot7 \).
2. Each root is an integer greater than 1, and 420 has exactly five prime factors counted with repetition, so the roots are 2, 2, 3, 5, 7.
3. \( f(4)=(4-2)^2(4-3)(4-5)(4-7)=4\cdot1\cdot(-1)\cdot(-3)=12 \).

Teacher's Note:
1. Prime factorisation of the constant term fixes integer roots when each is at least 2.
2. Write f(x) in factored form and substitute directly.

Answer: (b) \( \frac{49}{8} \)

Explanation:
1. Vertical distance \( d(x)=(-2x^2+4x+3)-(x-2)=-2x^2+3x+5 \).
2. The curves meet where \( 2x^2-3x-5=0 \), i.e. \( x=-1 \) and \( x=\frac52 \).
3. The maximum of \( d \) is at \( x=\frac34 \) (inside this interval): \( d=-\frac{9}{8}+\frac94+5=\frac{49}{8} \).

Teacher's Note:
1. Subtract the two functions to get one quadratic and maximise it at its vertex.
2. Check the vertex lies between the intersection points before accepting it.

Answer: (c) 16

Explanation:
1. \( x^4-7x^2+9=(x^2-x-3)(x^2+x-3) \) and \( x-\frac3x+1=\frac{x^2+x-3}{x} \).
2. So \( f(x)=x(x^2-x-3) \) for \( x\ne0 \) and \( x^2+x-3\ne0 \) (the factor \( x^2+x-3 \) cancels, its roots are not zeros of \( f \)).
3. Zeros: roots of \( x^2-x-3=0 \), \( x=\frac{1\pm\sqrt{13}}{2} \) (x = 0 is not in the domain).
4. So a = 1, b = 13, c = 2 and a + b + c = 16.

Teacher's Note:
1. Factorise the numerator as a difference of squares trick: x⁴ − 7x² + 9 = (x² − 3)² − x².
2. A cancelled factor never gives a zero of the function.

Answer: (c) 128

Explanation:
1. From \( x^3=8-4x \): \( x^4=8x-4x^2 \).
2. \( x^5=8x^2-4x^3=8x^2-32+16x \) and \( x^6=16x^2-64x+64 \).
3. \( x^7=16x^3-64x^2+64x=128-64x-64x^2+64x=128-64x^2 \).
4. Hence \( x^7+64x^2=128 \).

Teacher's Note:
1. Reduce higher powers repeatedly using the given cubic relation.
2. The x² and x terms cancel, which is a good sign the reduction is correct.

Answer: (a) 0

Explanation:
1. \( \sqrt{x^2-6x+9}=|x-3| \). Let \( t=(x-3)^2 \); then \( x^2-6x+6=t-3 \).
2. The equation becomes \( \sqrt t+\sqrt{t-3}=1 \), which needs \( t\ge3 \).
3. But then \( \sqrt t\ge\sqrt3\gt 1 \), so the left side exceeds 1 for every admissible \( x \). No real solution.

Teacher's Note:
1. Complete the square under the first radical to spot the substitution.
2. Check the domain before solving; here it already rules out every value.

Answer: (c) 3

Explanation:
1. If the roots are \( r,s \), the new equation has roots \( r^2,s^2 \) and must be the same equation, so \( \{r^2,s^2\}=\{r,s\} \).
2. For real roots this forces \( r,s\in\{0,1\} \) (the other possibility \( r^3=r \), \( r^4=r \) gives only \( r=0,1 \) among reals).
3. The possible equations are \( x^2=0 \), \( x(x-1)=0 \) and \( (x-1)^2=0 \): 3 equations.

Teacher's Note:
1. Real numbers equal to their own squares are only 0 and 1.
2. List the unordered pairs of roots from {0, 1}.

Answer: (b) \( \pi \)

Explanation:
1. \( f(x)=g(x) \) gives \( x^2+2(b-a)x+(1-2ab)=0 \).
2. No intersection means the discriminant is negative: \( (b-a)^2-(1-2ab)\lt 0 \).
3. This simplifies to \( a^2+b^2\lt 1 \), the interior of the unit circle, with area \( \pi \).

Teacher's Note:
1. Non-intersecting graphs mean the equation f(x) = g(x) has no real root.
2. The inequality describes a region whose area follows from a standard shape.

Answer: (a) −11

Explanation:
1. \( c=P(0)=2 \). The product of the zeroes is \( -c=-2 \).
2. Mean of zeroes \( =-\frac a3=-2 \), so \( a=6 \).
3. Sum of coefficients \( 1+a+b+c=-2 \), so \( b=-2-1-6-2=-11 \).

Teacher's Note:
1. Use the relations between roots and coefficients for a cubic.
2. The y-intercept is the constant term, here the value of c.

Answer: (b) 1

Explanation:
1. The equation is \( kx^2-2x-2=0 \) with \( x\ne\frac1k \) (when \( k\ne0 \)).
2. \( k=0 \): \( x+2=-x \Rightarrow x=-1 \), exactly one solution.
3. \( k\ne0 \): discriminant \( 4+8k \). For \( k\gt -\frac12 \) there are two roots; \( x=\frac1k \) is a root only when \( k=-\frac12 \). For \( k=-\frac12 \) the only root \( x=-2=\frac1k \) is rejected, so no solution; for \( k\lt -\frac12 \) no real roots.
4. So exactly one solution occurs only for k = 0: one value.

Teacher's Note:
1. Always exclude the value that makes the denominator zero.
2. The double-root case k = −1/2 looks tempting but its root is not allowed.

Answer: (c) real and unequal roots

Explanation:
1. \( a+b+c=0 \), so \( b=-(a+c) \).
2. Discriminant \( =9b^2-32ac=9(a+c)^2-32ac=9a^2-14ac+9c^2 \).
3. This form has discriminant \( 196-324\lt 0 \) and positive leading coefficient, so it is always positive (a, c not both zero).
4. Hence the roots are real and unequal; they need not be rational.

Teacher's Note:
1. Substitute x = 1 to relate the coefficients first.
2. A positive-definite quadratic form in a and c is positive for all non-zero (a, c).

Answer: (a) \( (-\infty,4) \)

Explanation:
1. Discriminant \( =(4-2k)^2-4(k^2-3k-1)=20-4k\ge0\Rightarrow k\le5 \).
2. \( f(3)=k^2-9k+20=(k-4)(k-5)\gt 0\Rightarrow k\lt 4 \) or \( k\gt 5 \).
3. Vertex \( x=k-2\lt 3\Rightarrow k\lt 5 \).
4. Intersection of the three conditions: \( k\lt 4 \).

Teacher's Note:
1. For both roots less than a number, use discriminant, sign of f at that number, and the position of the vertex.
2. Combine the conditions on a number line.

Answer: (b) two unequal real roots

Explanation:
1. A.P. gives \( 2b=a+c \).
2. Discriminant \( =16b^2-12ac=4(a+c)^2-12ac=4(a^2-ac+c^2) \).
3. \( a^2-ac+c^2=\left(a-\frac c2\right)^2+\frac{3c^2}{4}\gt 0 \) since \( a\ne0 \). So two unequal real roots.
4. The sign of the sum or product of roots depends on a, b, c, so (c) and (d) are not always true.

Teacher's Note:
1. Replace b using the A.P. condition before computing the discriminant.
2. Complete the square to show the discriminant never vanishes.

Answer: (d) 74

Explanation:
1. Let the terms be \( a,\ a+d,\ a+2d,\ a+3d \). Then \( p+q=2a+d=2 \) and \( r+s=2a+5d=18 \).
2. Subtracting: \( 4d=16\Rightarrow d=4 \), \( a=-1 \). The terms are \( -1,3,7,11 \).
3. \( A=pq=-3 \), \( B=rs=77 \), so \( A+B=74 \).

Teacher's Note:
1. Write the four terms with one first term and one common difference.
2. Use sum of roots to get two equations in a and d.

Answer: (a) one

Explanation:
1. \( \sin\frac{9\pi}{4}=\frac{\sqrt2}{2} \), \( \cos\frac{11\pi}{3}=\frac12 \), and \( \log_{1/2}A=-\log_2A \).
2. The equation becomes \( \log_2\big[(3-x)(5-x)\sqrt2\big]=\log_2\big[\sqrt2\,(x+7)\big] \), because \( \frac12=\log_2\sqrt2 \).
3. So \( (3-x)(5-x)=x+7 \Rightarrow x^2-9x+8=0 \Rightarrow x=1 \) or \( 8 \).
4. Domain: \( -7\lt x\lt 3 \). Only \( x=1 \) is valid, so one integer.

Teacher's Note:
1. Convert every logarithm to base 2 and write 1/2 as log₂√2.
2. Always apply the domain restrictions at the end.

Answer: (a) 193

Explanation:
1. \( \log_24=2 \), \( \log_{\sqrt2}8=6 \), \( \log_39^{k-1}=2(k-1) \). Common ratio 3 gives \( 2(k-1)=18 \), so \( k=10 \).
2. \( x^2-x\gt 6\Rightarrow x\lt -2 \) or \( x\gt 3 \); \( |x|\lt 100 \).
3. Integers from −99 to −3: 97; integers from 4 to 99: 96. Total 193.

Teacher's Note:
1. Evaluate each logarithm to a number first, then use the constant ratio.
2. Count integers in each of the two intervals separately.

Answer: (c) \( -\frac12\le k\le4 \)

Explanation:
1. Let \( t=\cos x\in[-1,1] \) and \( g(t)=kt^2-kt+1 \). We need \( g(t)\ge0 \) on \( [-1,1] \).
2. \( k=0 \): \( g=1 \), fine. \( k\gt 0 \): minimum at \( t=\frac12 \) is \( 1-\frac k4\ge0\Rightarrow k\le4 \).
3. \( k\lt 0 \): minimum is at the end \( t=-1 \): \( g(-1)=2k+1\ge0\Rightarrow k\ge-\frac12 \).
4. Combining: \( -\frac12\le k\le4 \).

Teacher's Note:
1. Reduce to a quadratic in t restricted to [−1, 1].
2. The minimum of a quadratic on an interval is at the vertex if inside, otherwise at an end point.

Answer: (c) −3

Explanation:
1. \( x^2-4x \) is decreasing on \( [0,1] \) (vertex at \( x=2 \)), so its minimum is at \( x=1 \): \( -3 \).
2. \( y^2+6y \) is increasing on \( [0,1] \) (vertex at \( y=-3 \)), so its minimum is at \( y=0 \): \( 0 \).
3. The minimum of f is −3 + 0 = −3.

Teacher's Note:
1. Minimise each variable separately since the function is a sum of a function of x and a function of y.
2. Compare the vertex with the allowed interval.

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

Explanation:
1. The parabola \( f(x)=x^2-3x+\lambda \) opens upward.
2. 1 lies strictly between the roots exactly when \( f(1)\lt 0 \): \( 1-3+\lambda\lt 0 \).
3. So \( \lambda\lt 2 \) (the discriminant condition \( \lambda\lt \frac94 \) is then automatic).

Teacher's Note:
1. A number lies between the roots of an upward parabola exactly when f at that number is negative.
2. No separate discriminant condition is needed once f(1) is negative.

Answer: (b) 4

Explanation:
1. Expand: \( \frac{1+x^2+x^6+x^8}{x^4}=\left(x^4+\frac1{x^4}\right)+\left(x^2+\frac1{x^2}\right) \).
2. Each bracket is at least 2 by AM-GM, with equality at \( x^2=1 \).
3. Minimum value = 2 + 2 = 4.

Teacher's Note:
1. Split the fraction so that reciprocal pairs appear.
2. Both AM-GM equalities hold at x = ±1, so the bound is attained.

Answer: (b) 3

Explanation:
1. \( pq=\frac{(p+q)^2-(p^2+q^2)}{2}=\frac{n^4-n^3}{2} \). So \( p,q \) are roots of \( t^2-n^2t+\frac{n^4-n^3}{2}=0 \).
2. Discriminant \( =n^4-2n^4+2n^3=n^3(2-n) \), which must be non-negative, so \( n\le2 \).
3. n = 0: p = q = 0. n = 1: p, q = 1, 0. n = 2: p = q = 2. All give integer p, q.
4. So there are 3 values of n.

Teacher's Note:
1. Build a quadratic in t whose roots are p and q using sum and product.
2. Integer roots need a non-negative discriminant, which bounds n.

Answer: (c) 5

Explanation:
1. The condition says the graph is symmetric about \( x=\frac74 \): \( -\frac{b}{2a}=\frac74 \), so \( b=-\frac{7a}{2} \).
2. \( f(x)=7x+a \) becomes \( ax^2+(b-7)x=0 \), with roots \( 0 \) and \( \frac{7-b}{a} \).
3. Only one solution means \( b=7 \), so \( a=-2 \) and \( a+b=5 \).

Teacher's Note:
1. Equal function values at x + c and c − x mean the axis of symmetry is x = c.
2. One solution of a quadratic with no constant term means the second root is also 0.

Answer: (d) 61

Explanation:
1. If the roots are positive integers \( m,n \): \( a=-(m+n) \), \( b=mn \), so \( a^2+b^2=(m+n)^2+(mn)^2 \).
2. 19, 37 and 50 cannot be written as (m + n)² + (mn)² with positive integers m, n: for example 37 would need m + n = 6 with mn = 1, and 50 would need m + n = 5 with mn = 5 or m + n = 7 with mn = 1, none of which exist.
3. For 61: m + n = 5 and mn = 6 gives m, n = 2, 3, and 25 + 36 = 61.

Teacher's Note:
1. Express a and b through the sum and product of the integer roots.
2. Test each option by splitting it into a perfect square for (m + n)² and one for (mn)².

Answer: (c) 4

Explanation:
1. Odd polynomial: \( P(-x)=-P(x) \). So \( P(3)=6 \) and \( P(-3)=-6 \).
2. Let \( g(x)=mx+n \). Then \( 3m+n=6 \) and \( -3m+n=-6 \), so \( n=0,\ m=2 \).
3. \( g(2)=4 \).

Teacher's Note:
1. The remainder on dividing by a quadratic is linear.
2. Use both roots x = 3 and x = −3 of the divisor.