JEE Mathematics Quadratic Equations MCQs Set 12

Answer: (a) \( \left(-\infty,-\frac14\right)\cup\{2\}\cup(5,6] \)

Explanation:
1. a = 2: the only zero is x = −5/4, which lies in (−2, 1), so a = 2 works.
2. For a ≠ 2 we need real zeros: \( D=4(6-a)\ge0\Rightarrow a\le6 \), together with \( (a-2)f(-2)=(a-2)(a-5)\gt0 \), \( (a-2)f(1)=(a-2)(4a+1)\gt0 \) and axis \( -2\lt-\frac a{a-2}\lt1 \).
3. These give a < −1/4 or 5 < a ≤ 6. Including a = 2, the set is the first option.

Teacher's Note:
1. Treat a = 2 separately because the equation becomes linear.
2. For a ≠ 2 use D ≥ 0, sign conditions at −2 and 1, and the axis.

Answer: (a) \( K\le-1 \)

Explanation:
1. Case 1: product \( K+5\lt0 \), i.e. K < −5: the roots have opposite signs, so one is positive.
2. Case 2: both roots non-negative with at least one positive: \( D=4(K-1)^2-4(K+5)=4(K-4)(K+1)\ge0 \), sum \( -2(K-1)\gt0 \) and product \( K+5\ge0 \) gives −5 ≤ K ≤ −1.
3. Union of the cases: K ≤ −1.

Teacher's Note:
1. Split into the cases of opposite-sign roots and of both roots non-negative.
2. Take the union of the two solution sets.

Answer: (a) 4

Explanation:
1. P divides every combination of the two polynomials, so it divides their GCD.
2. Their GCD is \( x^2-2x+5 \) (check: \( x^4+6x^2+25=(x^2-2x+5)(x^2+2x+5) \)).
3. Hence \( P(x)=x^2-2x+5 \) and \( P(1)=4 \).

Teacher's Note:
1. A common factor must divide the GCD of the polynomials.
2. Factorise the quartic x⁴ + 6x² + 25 as a difference of squares.

Answer: (a) \( [1,\infty) \)

Explanation:
1. Put \( t=3^x\gt0 \): \( at^2+4(a-1)t+(a-1)\gt0 \) for all t > 0.
2. If a < 1 the terms with (a − 1) are negative, and for small t the expression is close to a − 1 < 0, so it fails (a ≤ 0 fails for large t as well).
3. If a ≥ 1 every term is non-negative, and at least a·t² > 0, so it holds. Answer: [1, ∞).

Teacher's Note:
1. Substitute t = 3ˣ, which ranges over all positive reals.
2. Test a below 1 near t → 0 and a at least 1 directly.

Answer: (a) 3 and 5

Explanation:
1. Write it as \( x^2-2mx+(8m-15)=0 \). Equal roots need \( D=4m^2-4(8m-15)=0 \).
2. \( m^2-8m+15=0\Rightarrow(m-3)(m-5)=0 \).

Teacher's Note:
1. Rearrange into standard form before taking the discriminant.
2. Equal roots mean D = 0.

Answer: (a) \( 2,\ -\frac{10}9 \)

Explanation:
1. \( \frac D4=(1+3m)^2-7(3+2m)=9m^2-8m-20=0 \).
2. \( (m-2)(9m+10)=0 \), so m = 2 or m = −10/9.

Teacher's Note:
1. Use D/4 = (b/2)² − ac when the middle coefficient is even.
2. Factorise the resulting quadratic in m.

Answer: (a) \( \frac{b^2(b^2-4ac)}{a^2c^2} \)

Explanation:
1. \( \left(\frac\alpha\beta-\frac\beta\alpha\right)^2=\frac{(\alpha^2-\beta^2)^2}{\alpha^2\beta^2}=\frac{(\alpha+\beta)^2(\alpha-\beta)^2}{(\alpha\beta)^2} \).
2. Using \( (\alpha+\beta)^2=\frac{b^2}{a^2} \), \( (\alpha-\beta)^2=\frac{b^2-4ac}{a^2} \) and \( (\alpha\beta)^2=\frac{c^2}{a^2} \): result \( \frac{b^2(b^2-4ac)}{a^2c^2} \).

Teacher's Note:
1. Factorise α² − β² as (α + β)(α − β).
2. Use (α − β)² = (α + β)² − 4αβ.

Answer: (a) 7

Explanation:
1. From \( x=1+2i \): \( (x-1)^2=-4 \), so \( x^2-2x+5=0 \).
2. Divide: \( x^3+x^2-x+22=(x^2-2x+5)(x+3)+7 \).
3. Hence the value is 7.

Teacher's Note:
1. Form the quadratic with real coefficients satisfied by x.
2. Divide the cubic by it and read off the remainder.

Answer: (a) −15

Explanation:
1. From \( x=3+i \): \( (x-3)^2=-1 \), so \( x^2-6x+10=0 \).
2. Divide: \( x^3-3x^2-8x+15=(x^2-6x+10)(x+3)-15 \).
3. So the value is −15.

Teacher's Note:
1. Use the real-coefficient quadratic satisfied by x.
2. The remainder after division is the required value.

Answer: (a) 0

Explanation:
1. Let \( x=at \) with \( t=1-i\sqrt3 \): then \( (t-1)^2=-3 \), i.e. \( t^2-2t+4=0 \).
2. The expression is \( a^3(t^3-t^2+2t+4)=a^3(t+1)(t^2-2t+4)=0 \) (check: (t+1)(t² − 2t + 4) = t³ − t² + 2t + 4).

Teacher's Note:
1. Substitute x = at and factor the cubic in t.
2. The factor t² − 2t + 4 vanishes for this t.

Answer: (a) \( \frac13 \)

Explanation:
1. Put y for the expression: \( 2yx^2+(3y-1)x+(6y-2)=0 \). Real x needs \( (3y-1)^2-8y(6y-2)\ge0 \).
2. \( 39y^2-10y-1\le0\Rightarrow(3y-1)(13y+1)\le0 \), so \( -\frac1{13}\le y\le\frac13 \). The greatest value is 1/3.

Teacher's Note:
1. Treat the value as a parameter y and require the discriminant to be non-negative.
2. Solve the resulting quadratic inequality in y.

Answer: (a) \( 20\pm6\sqrt5 \)

Explanation:
1. Rearrange: \( 9x^2+(a-2)x+(a-6)=0 \).
2. Equal roots: \( (a-2)^2-36(a-6)=a^2-40a+220=0 \), so \( a=20\pm6\sqrt5 \).

Teacher's Note:
1. Bring all terms to one side first.
2. Equal roots means D = 0.

Answer: (a) \( 2,\ -\frac{22}3 \)

Explanation:
1. \( D=(k+4)^2-4(k-1)(k+7)=-3k^2-16k+44=-(k-2)(3k+22) \).
2. Equal roots: D = 0, so k = 2 or k = −22/3 (both give k ≠ 1).

Teacher's Note:
1. Expand the discriminant and factorise it.
2. Make sure the leading coefficient k − 1 is not zero.

Answer: (a) 4

Explanation:
1. \( D=4(a-1)^2-12(2a-5)=4(a^2-8a+16)=4(a-4)^2 \).
2. D = 0 gives a = 4 (and 2a − 5 = 3 ≠ 0).

Teacher's Note:
1. Expand D carefully; it is a perfect square.
2. A single value of a results.

Answer: (a) \( m\lt-\frac12\ \text{or}\ m\gt\frac12 \)

Explanation:
1. \( D=1-4m^2\lt0\Rightarrow m^2\gt\frac14 \).
2. So \( |m|\gt\frac12 \).

Teacher's Note:
1. No real roots means D < 0.
2. m² > 1/4 gives both signs of m.

Answer: (a) \( \left(-\infty,-\frac17\right)\cup(1,\infty) \)

Explanation:
1. \( D=(m+1)^2-4m(2m-1)=-7m^2+6m+1\lt0\Rightarrow7m^2-6m-1\gt0 \).
2. \( (7m+1)(m-1)\gt0\Rightarrow m\lt-\frac17\ \text{or}\ m\gt1 \).

Teacher's Note:
1. Expand and collect the discriminant.
2. Solve the quadratic inequality using the roots −1/7 and 1.

Answer: (a) \( 2\lt c\lt4 \)

Explanation:
1. For a quadratic we need c ≠ 2, and no real roots means \( D=4(c-2)^2-8(c-2)=4(c-2)(c-4)\lt0 \).
2. Hence \( 2\lt c\lt4 \).

Teacher's Note:
1. Keep the leading coefficient non-zero so that the equation is quadratic.
2. No real roots means D < 0.

Answer: (a) 13

Explanation:
1. \( D=4(k-12)^2-8(k-12)=4(k-12)(k-14)\lt0\Rightarrow12\lt k\lt14 \).
2. The only integer strictly between 12 and 14 is 13.

Teacher's Note:
1. The equation is quadratic only if k ≠ 12.
2. Solve D < 0 and list the integers.

Answer: (a) \( \pm10 \)

Explanation:
1. Touching the x-axis means a double root: \( D=a^2-100=0 \).
2. So a = ±10.

Teacher's Note:
1. A curve touches the x-axis when the quadratic has equal roots.
2. Set the discriminant to zero.

Answer: (a) \( 28x^2-20x+1=0 \)

Explanation:
1. \( \sqrt{72}=6\sqrt2 \). Sum \( =\frac{(10+6\sqrt2)+(10-6\sqrt2)}{100-72}=\frac{20}{28}=\frac57 \).
2. Product \( =\frac1{100-72}=\frac1{28} \).
3. Equation: \( x^2-\frac57x+\frac1{28}=0\Rightarrow28x^2-20x+1=0 \).

Teacher's Note:
1. Use √72 = 6√2 to see that the roots are reciprocals of conjugates.
2. Form the equation from the sum and product.

Answer: (a) \( -3\lt k\lt5 \)

Explanation:
1. The leading coefficient is positive, so we need \( D\lt0 \).
2. \( D=(k-3)^2+4(k-6)=k^2-2k-15=(k-5)(k+3)\lt0 \), giving −3 < k < 5.

Teacher's Note:
1. A positive-leading quadratic is positive for all x exactly when D < 0.
2. Factorise D to solve the inequality.

Answer: (a) \( 0\le a\lt\frac12 \)

Explanation:
1. If a = 0 the expression is 0.5 > 0, true for all x.
2. If a ≠ 0 we need a > 0 and \( D=4a^2-2a\lt0 \), i.e. \( 0\lt a\lt\frac12 \).
3. (At a = 1/2 the expression is \( \frac12(x+1)^2 \), which is 0 at x = −1.) So 0 ≤ a < 1/2.

Teacher's Note:
1. Treat a = 0 separately.
2. The endpoint a = 1/2 must be tested directly because D = 0 there.