JEE Mathematics Quadratic Equations MCQs Set 06

Answer: (a) 7

Explanation:
1. For each root r: \( r-2=\frac1r \), 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 two distinct numbers satisfy the same quadratic, that quadratic is the required equation.
2. Convert the verbal condition on each root into one equation first.

Answer: (b) \( p\ge -\tfrac14 \)

Explanation:
1. Put \( u=\sqrt[3]{x} \). Then \( \sqrt[3]{x+3p+1}=1+u \).
2. Cube: \( x+3p+1=1+3u+3u^2+u^3 \); since \( x=u^3 \), \( p=u^2+u \).
3. \( u^2+u=\left(u+\tfrac12\right)^2-\tfrac14\ge-\tfrac14 \), and every such value is attained for real u, so \( p\ge-\tfrac14 \).

Teacher's Note:
1. Cube-root equations are best handled by substituting the cube root of the simplest term.
2. The range of a quadratic in u is found by completing the square.

Answer: (c) 12

Explanation:
1. The product of the five roots is \( (-1)^5(-420)=420=2^2\cdot3\cdot5\cdot7 \).
2. Five integers each greater than 1 with this product must be 2, 2, 3, 5, 7.
3. \( f(4)=(4-2)^2(4-3)(4-5)(4-7)=4\cdot1\cdot(-1)(-3)=12 \).

Teacher's Note:
1. Use the relation between the product of roots and the constant term.
2. A prime factorisation with exactly five prime factors (counting repeats) leaves no other choice for the roots.

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

Explanation:
1. Vertical distance \( d(x)=(-2x^2+4x+3)-(x-2)=-2x^2+3x+5 \).
2. The parabola is above the line between the roots \( x=-1 \) and \( x=\tfrac52 \).
3. Maximum at the vertex \( x=\tfrac34 \): \( d=-2\cdot\tfrac9{16}+\tfrac94+5=\tfrac{49}{8} \).

Teacher's Note:
1. The vertical gap between two graphs is the difference of their equations.
2. A downward quadratic is greatest at its vertex.

Answer: (c) 16

Explanation:
1. Numerator: \( x^4-7x^2+9=(x^2-3)^2-x^2=(x^2-x-3)(x^2+x-3) \).
2. Denominator: \( x-\tfrac3x+1=\dfrac{x^2+x-3}{x} \), so \( f(x)=x(x^2-x-3) \) for \( x^2+x-3\ne0,\ x\ne0 \).
3. Zeroes: \( x=\dfrac{1\pm\sqrt{13}}{2} \) (neither makes the denominator zero). Hence \( a+b+c=1+13+2=16 \).

Teacher's Note:
1. Factorise before cancelling and keep track of the values excluded from the domain.
2. Check that the zeroes found do not coincide with roots of the denominator.

Answer: (a) 0

Explanation:
1. \( \sqrt{x^2-6x+9}=|x-3| \) and \( x^2-6x+6=(x-3)^2-3 \).
2. Let \( t=|x-3| \). The second root needs \( t^2\ge3 \), so \( t\ge\sqrt3>1 \).
3. Then \( t+\sqrt{t^2-3}\ge\sqrt3>1 \), so the sum can never equal 1. No real solution.

Teacher's Note:
1. Rewrite quadratics under radicals as perfect squares plus a constant.
2. Domain restrictions can make an equation impossible before any algebra.

Answer: (c) 3

Explanation:
1. If the roots are \( \alpha,\beta \), squaring must reproduce the same pair of roots: \( \{\alpha^2,\beta^2\}=\{\alpha,\beta\} \).
2. Either each root satisfies \( r^2=r \), so each root is 0 or 1; or the roots are swapped, \( \alpha^2=\beta,\ \beta^2=\alpha \), which gives non-real values (or \( \alpha=\beta=1,0 \)).
3. Real possibilities: roots (0,0), (0,1), (1,1), i.e. \( x^2=0 \), \( x^2-x=0 \), \( x^2-2x+1=0 \). Three equations.

Teacher's Note:
1. Think of the root set, not each root separately.
2. Real numbers with r^2=r are only 0 and 1.

Answer: (b) \( \pi \)

Explanation:
1. No intersection means \( x^2+2bx+1-2ax-2ab=0 \) has no real root.
2. This is \( x^2+2(b-a)x+(1-2ab)=0 \) with discriminant \( 4(b-a)^2-4(1-2ab)<0 \).
3. Simplify: \( a^2+b^2<1 \). S is the interior of the unit circle, of area \( \pi \).

Teacher's Note:
1. Equate the two functions and demand a negative discriminant.
2. An inequality of the form a^2+b^2 less than r^2 describes the inside of a circle.

Answer: (a) -11

Explanation:
1. The y-intercept is \( c=2 \).
2. Product of zeroes \( =-c=-2 \). Mean of zeroes \( =-\tfrac a3=-2 \), so \( a=6 \).
3. Sum of coefficients \( 1+a+b+c=-2 \): \( 1+6+b+2=-2 \), so \( b=-11 \).

Teacher's Note:
1. P(0) is the y-intercept and equals the constant term.
2. Vieta's relations for a cubic: sum = -a, product = -c.

Answer: (b) 1

Explanation:
1. Clearing the denominator: \( kx^2-2x-2=0 \) with the restriction \( kx\ne1 \).
2. If \( k=0 \): \( x=-1 \), which is valid (denominator is -1). One solution.
3. If \( k\ne0 \): exactly one solution needs a zero discriminant \( 4+8k=0 \), so \( k=-\tfrac12 \), giving \( x=-2 \) which makes \( kx-1=0 \) (rejected), or one of two roots must equal \( 1/k \), which also forces \( k=-\tfrac12 \). Neither gives a valid solution.
4. Only \( k=0 \) works: 1 value.

Teacher's Note:
1. Always check solutions of a rational equation against the denominator.
2. Treat k = 0 separately because the equation then stops being quadratic.

Answer: (c) real and unequal roots

Explanation:
1. Since 1 is a root, \( a+b+c=0 \), so \( b=-(a+c) \).
2. Discriminant of the new equation: \( 9b^2-32ac=9(a+c)^2-32ac=9a^2-14ac+9c^2 \).
3. As a quadratic in \( a/c \) its own discriminant is \( 196-324<0 \) and the leading coefficient is positive, so it is always positive. Roots are real and unequal.

Teacher's Note:
1. Use the sum of coefficients to eliminate b.
2. If a quadratic form in two variables has negative discriminant, it keeps one sign for all non-zero values.

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

Explanation:
1. Let \( f(x)=x^2+(4-2k)x+k^2-3k-1 \). Both roots below 3 needs: \( D\ge0 \), axis \( <3 \), and \( f(3)>0 \).
2. D: \( (4-2k)^2-4(k^2-3k-1)=20-4k\ge0\Rightarrow k\le5 \).
3. Axis \( x=k-2<3\Rightarrow k<5 \).
4. f(3): \( k^2-9k+20=(k-4)(k-5)>0\Rightarrow k<4\ \text{or}\ k>5 \).
5. Combining: \( k<4 \).

Teacher's Note:
1. For roots on one side of a number, check D, the axis position and the sign of f at that number.
2. Intersect all three conditions.

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. \( a^2-ac+c^2=\left(a-\tfrac c2\right)^2+\tfrac34c^2>0 \) (a is non-zero), so the roots are real and unequal.

Teacher's Note:
1. Replace b using the A.P. condition.
2. a^2-ac+c^2 is positive for all real a, c not both zero.

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. Subtract: \( 4d=16 \), so \( d=4 \), \( a=-1 \). The terms are -1, 3, 7, 11.
3. Hence \( A=pq=-3 \) and \( B=rs=77 \), and \( A+B=74 \).

Teacher's Note:
1. Write the four terms of the A.P. with a and d, then use the two sums of roots.
2. The products of roots then give A and B.

Answer: (a) 3

Explanation:
1. Exactly one real point means the quadratic expression, being a sum of non-negative squares plus a constant, has a single minimum point where it is zero.
2. Set the partial derivatives to zero: \( 4x+4y-12=0 \) and \( 4x+14y-2=0 \). Solving: \( y=-1,\ x=4 \).
3. So \( x+y=3 \) (and \( t=23 \)).

Teacher's Note:
1. A single real solution corresponds to the minimum point of a positive definite quadratic form.
2. Setting both partial derivatives to zero locates that point.

Answer: (c) \( -\tfrac12\le k\le4 \)

Explanation:
1. Put \( t=\cos x\in[-1,1] \): need \( g(t)=kt^2-kt+1\ge0 \) on \( [-1,1] \).
2. If \( k>0 \): minimum at \( t=\tfrac12 \) is \( 1-\tfrac k4\ge0 \), so \( k\le4 \).
3. If \( k<0 \): the parabola opens downward, so the minimum is at an endpoint; \( g(-1)=2k+1\ge0 \) gives \( k\ge-\tfrac12 \). For \( k=0 \) the inequality holds.
4. Together \( -\tfrac12\le k\le4 \).

Teacher's Note:
1. Substitute t = cos x and restrict t to [-1, 1].
2. Check the sign of k: it decides whether the minimum is at the vertex or at an endpoint.

Answer: (c) -3

Explanation:
1. \( x^2-4x=(x-2)^2-4 \) is decreasing on \( [0,1] \), so its least value is at \( x=1 \): \( -3 \).
2. \( y^2+6y=(y+3)^2-9 \) is increasing on \( [0,1] \), so its least value is at \( y=0 \): 0.
3. Minimum \( -3+0=-3 \).

Teacher's Note:
1. The function separates into a part in x and a part in y, so minimise each separately.
2. Compare the vertex position with the interval to decide which endpoint wins.

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

Explanation:
1. Since 1 lies strictly between the roots and the leading coefficient is positive, \( f(1)<0 \).
2. \( f(1)=1-3+\lambda=\lambda-2<0 \), so \( \lambda<2 \).
3. This also makes the discriminant \( 9-4\lambda>0 \) automatically.

Teacher's Note:
1. A number lies between the roots exactly when f at that number has the opposite sign to the leading coefficient.
2. No separate discriminant check is needed in this case.

Answer: (c) 5

Explanation:
1. Symmetry about \( x=\tfrac74 \) means \( -\tfrac b{2a}=\tfrac74 \), so \( b=-\tfrac{7a}2 \).
2. \( f(x)=7x+a \) gives \( ax^2+(b-7)x=0 \), whose roots are 0 and \( \tfrac{7-b}{a} \). A single solution needs \( b=7 \).
3. Then \( a=-2 \) and \( a+b=5 \).

Teacher's Note:
1. f(c+t)=f(c-t) tells you the axis of the parabola.
2. A quadratic with no constant term always has root 0; one solution means the other root is 0 too.

Answer: (c) 128

Explanation:
1. \( x^3=8-4x \), so \( x^4=8x-4x^2 \).
2. \( x^7=x^3\cdot x^4=(8-4x)(8x-4x^2)=64x-64x^2+16x^3 \).
3. Replace \( 16x^3=128-64x \): \( x^7=128-64x^2 \).
4. Hence \( x^7+64x^2=128 \).

Teacher's Note:
1. Reduce high powers using the given cubic instead of solving it.
2. Multiply the reduced forms of lower powers to reach x^7.