JEE Mathematics Quadratic Equations MCQs Set 13

Answer: (a) 3

Explanation:
1. \( \frac D4=(4k-1)^2-(15k^2-2k-7)=k^2-6k+8 \).
2. Need D < 0: \( (k-2)(k-4)\lt0\Rightarrow2\lt k\lt4 \), so the integer is k = 3.

Teacher's Note:
1. Use D/4 and require it to be negative.
2. List the integers in the open interval (2, 4).

Answer: (a) 3

Explanation:
1. \( \frac D4=(k+2)^2-(12+k^2)=4k-8\gt0\Rightarrow k\gt2 \).
2. The least integer greater than 2 is 3.

Teacher's Note:
1. Distinct real roots need D > 0.
2. Pick the first integer strictly greater than 2.

Answer: (a) −2 and 1

Explanation:
1. Sum of roots \( =-(2-a-a^2)=a^2+a-2=0 \).
2. \( (a+2)(a-1)=0 \), so a = −2 or a = 1. The equation then is \( x^2-4=0 \) or \( x^2-1=0 \), both with real roots.

Teacher's Note:
1. The sum of roots is minus the coefficient of x.
2. Solve the resulting quadratic in a.

Answer: (a) \( -6\lt a\lt3 \)

Explanation:
1. Intersections satisfy \( (a-6)x^2-2ax-3=0 \). The graphs do not meet when this has no real root.
2. For a ≠ 6: \( D=4a^2+12(a-6)=4(a^2+3a-18)=4(a+6)(a-3)\lt0 \), so −6 < a < 3. (For a = 6 the line meets the horizontal line y = −2, so a = 6 is excluded, and it is outside the interval anyway.)

Teacher's Note:
1. Equate the two expressions to get a quadratic in x.
2. No intersection means no real root.

Answer: (a) −4

Explanation:
1. Let the roots be r and 2r: \( 3r=1-2a \), \( 2r^2=a^2+2 \).
2. Substituting \( r=\frac{1-2a}3 \): \( 2(1-2a)^2=9a^2+18\Rightarrow a^2+8a+16=0 \), so a = −4.
3. Check: a = −4 gives \( x^2-9x+18=0 \) with roots 3 and 6.

Teacher's Note:
1. Use the sum and product with roots r and 2r.
2. Eliminate r to get an equation in a.

Answer: (a) \( -\frac32,\ 6 \)

Explanation:
1. Roots r and 2r: \( 3r=-a \), \( 2r^2=a+2 \).
2. Substituting \( r=-\frac a3 \): \( \frac{2a^2}9=a+2\Rightarrow2a^2-9a-18=0\Rightarrow(2a+3)(a-6)=0 \).
3. So a = −3/2 or a = 6.

Teacher's Note:
1. Take roots r and 2r and use both sum and product.
2. Solve the quadratic in a.

Answer: (a) \( 2,\ \frac92 \)

Explanation:
1. Roots r and 1.5r: \( 2.5r=\frac{a+3}a \) and \( 1.5r^2=\frac3a \).
2. Eliminating r: \( 1.5\cdot\frac{4(a+3)^2}{25a^2}=\frac3a\Rightarrow2(a+3)^2=25a\Rightarrow2a^2-13a+18=0 \).
3. \( (a-2)(2a-9)=0 \), so a = 2 or a = 9/2.

Teacher's Note:
1. Take roots r and 1.5r and use both sum and product.
2. Eliminate r to reach a quadratic in a.

Answer: (a) \( 6,\ -\frac6{19} \)

Explanation:
1. Let the roots be 9r and r: \( 10r=3a+2 \), \( 9r^2=a^2\Rightarrow r=\pm\frac a3 \).
2. If r = a/3: \( \frac{10a}3=3a+2\Rightarrow a=6 \) (roots 2, 18). If r = −a/3: \( -\frac{10a}3=3a+2\Rightarrow a=-\frac6{19} \).
3. So a = 6 or a = −6/19.

Teacher's Note:
1. Use the sum for 10r and the product for r².
2. Keep both signs of r.

Answer: (a) \( \frac{27}8\ \text{or}\ -\frac{125}8 \)

Explanation:
1. Let the roots be r and r²: \( r+r^2=\frac{15}4 \), \( r^3=a \).
2. \( 4r^2+4r-15=0\Rightarrow(2r-3)(2r+5)=0 \), so r = 3/2 or r = −5/2.
3. a = r³ = 27/8 or −125/8.

Teacher's Note:
1. Use the sum of roots to get r first.
2. Then the product gives a = r³.

Answer: (a) \( \pm7 \)

Explanation:
1. \( (x_2-x_1)^2=p^2-48=1 \), so \( p^2=49 \) and p = ±7.
2. (Check: p = 7 gives roots −3, −4.)

Teacher's Note:
1. Use (x₂ − x₁)² = (x₁ + x₂)² − 4x₁x₂.
2. Solve p² = 49.

Answer: (a) \( \pm3\sqrt5 \)

Explanation:
1. \( (x_1-x_2)^2=\frac{k^2}{25}-\frac45=1 \).
2. \( k^2=45\Rightarrow k=\pm3\sqrt5 \).

Teacher's Note:
1. Use the identity for the square of the difference of roots.
2. Solve k² = 45.

Answer: (a) \( \frac32,\ 3 \)

Explanation:
1. \( (x_1-x_2)^2=\frac{(a-4)^2}{(a-2)^2}+\frac8{a-2}=9 \).
2. Multiply by \( (a-2)^2 \): \( (a-4)^2+8(a-2)=9(a-2)^2\Rightarrow8a^2-36a+36=0 \), i.e. \( 2a^2-9a+9=0 \).
3. \( (2a-3)(a-3)=0 \), so a = 3/2 or 3.

Teacher's Note:
1. Write (x₁ − x₂)² using sum and product and clear denominators.
2. Factorise the resulting quadratic in a.

Answer: (a) −13

Explanation:
1. From the equation: \( x_1x_2=-\frac{28}5 \). Solve with \( 5x_1+2x_2=1 \): \( x_2=\frac{1-5x_1}2 \), so \( x_1(1-5x_1)=-\frac{56}5 \), i.e. \( 25x_1^2-5x_1-56=0 \).
2. This gives \( x_1=\frac85 \) or \( -\frac75 \).
3. Then \( x_2=-\frac72 \) or 4, and \( b=-5(x_1+x_2)=\frac{19}2 \) or −13. Only b = −13 is an integer.

Teacher's Note:
1. Eliminate x₂ using the given linear relation.
2. Compute b from the sum of the roots and keep only the integer value.

Answer: (a) \( \frac12,\ 1 \)

Explanation:
1. The equation is \( x^2-2ax+2a-1=0 \): sum = 2a, product = 2a − 1.
2. Sum of squares \( =4a^2-2(2a-1)=4a^2-4a+2 \).
3. \( 2a=4a^2-4a+2\Rightarrow2a^2-3a+1=0\Rightarrow(2a-1)(a-1)=0 \).

Teacher's Note:
1. Expand the equation to standard form first.
2. Compare sum with sum of squares.

Answer: (a) −2

Explanation:
1. Subtract: \( (a-1)x+(1-a)=0\Rightarrow(a-1)(x-1)=0 \). For a ≠ 1 the common root is x = 1.
2. Substitute x = 1: \( 1+a+1=0\Rightarrow a=-2 \).
3. (For a = 1 both equations coincide and have no real root.)

Teacher's Note:
1. Subtract the equations to find the common root.
2. Substitute it back to find a.

Answer: (a) −2

Explanation:
1. Let r be a root of the first: \( r^2-r+m=0 \). Then 2r satisfies the second: \( 4r^2-2r+3m=0 \).
2. Eliminate m (from the first, \( m=r-r^2 \)): \( 4r^2-2r+3r-3r^2=r^2+r=0 \), so r = −1 (r = 0 gives m = 0, excluded).
3. Then m = −1 − 1 = −2.

Teacher's Note:
1. Substitute the double root in the second equation.
2. Eliminate m using the first equation.

Answer: (a) negative

Explanation:
1. No real roots means the sign of the trinomial never changes, so it has the sign of a for all x.
2. At x = 1 the value is a + b + c < 0, so the trinomial is negative everywhere.
3. At x = 0 the value is c, so c < 0.

Teacher's Note:
1. A quadratic without real roots keeps one sign for all x.
2. Use the value at x = 1 to decide that sign.

Answer: (a) \( \frac{a(a^2-18a+9)}{27} \)

Explanation:
1. Sum \( =\frac a3 \), product \( =\frac{2a-1}3 \).
2. \( x_1^3+x_2^3=(x_1+x_2)^3-3x_1x_2(x_1+x_2)=\frac{a^3}{27}-\frac{a(2a-1)}{3} \).
3. \( =\frac{a^3-18a^2+9a}{27}=\frac{a(a^2-18a+9)}{27} \).

Teacher's Note:
1. Use the cube identity in terms of sum and product.
2. Put everything over 27.

Answer: (a) \( 2\lt a\le\frac94 \)

Explanation:
1. Rewrite as \( x^2-x+(a-2)=0 \).
2. Real roots: \( D=1-4(a-2)=9-4a\ge0\Rightarrow a\le\frac94 \). Positive roots: sum 1 > 0 and product \( a-2\gt0 \).
3. Hence \( 2\lt a\le\frac94 \).

Teacher's Note:
1. Bring the equation to standard form first.
2. For both roots positive we need D ≥ 0, sum > 0 and product > 0.

Answer: (a) \( 3\lt p\le\frac{15}4 \)

Explanation:
1. \( D=4p^2-20p(p-3)=4p(15-4p)\ge0\Rightarrow0\le p\le\frac{15}4 \).
2. Sum \( \frac{2p}{p-3}\gt0 \) and product \( \frac{5p}{p-3}\gt0 \) need \( p\lt0 \) or \( p\gt3 \).
3. Together: 3 < p ≤ 15/4 (at p = 15/4 the two equal roots are positive, so the end value is included).

Teacher's Note:
1. Apply D ≥ 0, sum > 0 and product > 0.
2. Check the end value where the roots coincide.

Answer: (a) \( a\lt-6 \)

Explanation:
1. We need a negative leading coefficient and D < 0: \( a+4\lt0 \) and \( D=4a^2-4(a+4)(2a-6)=-4(a-4)(a+6)\lt0 \).
2. \( (a-4)(a+6)\gt0\Rightarrow a\lt-6 \) or \( a\gt4 \); with a + 4 < 0 we get a < −6.

Teacher's Note:
1. Always negative needs the leading coefficient negative and D < 0.
2. Intersect the two conditions.

Answer: (a) \( a\gt6 \)

Explanation:
1. We need \( a-3\gt0 \) and \( D=4a^2-4(a-3)(3a-6)=-4(a-6)(2a-3)\lt0 \).
2. \( (a-6)(2a-3)\gt0\Rightarrow a\lt\frac32 \) or \( a\gt6 \); with a > 3 this leaves a > 6.

Teacher's Note:
1. Positive for all x needs a positive leading coefficient and D < 0.
2. Intersect the two conditions.

Answer: (a) \( a\gt\frac53 \)

Explanation:
1. We need \( a-1\gt0 \) and \( D=(a+1)^2-4(a-1)(a+1)=-(a+1)(3a-5)\lt0 \).
2. \( (a+1)(3a-5)\gt0\Rightarrow a\lt-1 \) or \( a\gt\frac53 \); with a > 1 this gives a > 5/3.

Teacher's Note:
1. Check the sign of the leading coefficient and the discriminant together.
2. Intersect the intervals.

Answer: (a) 5

Explanation:
1. Need \( k-2\gt0 \) and \( D=64-4(k-2)(k+4)=-4(k-4)(k+6)\lt0 \).
2. \( (k-4)(k+6)\gt0\Rightarrow k\gt4 \) (as k > 2). The least integer is 5.

Teacher's Note:
1. Positive for all x needs a positive leading coefficient and D < 0.
2. k = 4 gives D = 0, which is not allowed.