JEE Mathematics Quadratic Equations MCQs Set 17

Answer: (d) 4

Explanation:
1. \( 2x-3=5i \Rightarrow 4x^2-12x+34=0 \), i.e. \( 2x^2-6x+17=0 \).
2. Divide: \( 2x^3+2x^2-7x+72=(2x^2-6x+17)(x+4)+4 \).
3. So the value is 4.

Teacher's Note:
1. Reduce the polynomial by dividing it by the quadratic satisfied by x.
2. The remainder is the required value.

Answer: (d) \( a=0 \) or \( a=24 \)

Explanation:
1. Let the common root be t. Subtracting the two equations gives \( 3t-a=0 \), so \( a=3t \).
2. Then \( t^2-11t+3t=0 \Rightarrow t(t-8)=0 \), so \( t=0 \) or \( 8 \); hence \( a=0 \) or \( 24 \).

Teacher's Note:
1. Subtracting equations is the fastest way to eliminate the squared term.
2. Check both values by substitution.

Answer: (a) real roots

Explanation:
1. The condition \( 4a+2b+c=0 \) says \( f(2)=0 \), so 2 is a real root of the equation.
2. A real quadratic with one real root has both roots real, so the roots are real.

Teacher's Note:
1. Spot the value of x that makes the condition equal to f(x) = 0.
2. A real root of a real quadratic forces the other root to be real too.

Answer: (c) \( K\in(-2,3) \)

Explanation:
1. 2 must lie between the roots, so \( f(2)<0 \).
2. \( f(2)=4-2(K+1)+K^2+K-8=K^2-K-6=(K-3)(K+2)<0 \), hence \( -2<K<3 \).

Teacher's Note:
1. If a number lies between the roots of an upward parabola, the function value there is negative.
2. No discriminant condition is needed once f(2) is negative.

Answer: (c) \( x^3-15x^2+67x-77=0 \)

Explanation:
1. Put \( y=2x+1 \), so \( x=\dfrac{y-1}{2} \), in the given equation.
2. \( \dfrac{(y-1)^3}{8}-\dfrac{6(y-1)^2}{4}+\dfrac{10(y-1)}{2}-3=0 \). Multiply by 8 and expand: \( y^3-15y^2+67y-77=0 \).
3. Check: new sum \( =2(6)+3=15 \), new product \( =(2\alpha+1)(2\beta+1)(2\gamma+1)=8(3)+4(10)+2(6)+1=77 \).

Teacher's Note:
1. For a linear transformation of the roots, substituting the inverse relation is quicker than finding all symmetric sums.
2. Verify the answer using the sum of the new roots.

Answer: (b) \( \left(-\infty,-\dfrac12\right) \)

Explanation:
1. The denominator \( x^2-8x+17=(x-4)^2+1>0 \), so the numerator must be negative for all x.
2. This needs \( k<0 \) and \( D<0 \): \( 4(k+1)^2-4k(9k+4)<0 \Rightarrow 8k^2+2k-1>0 \Rightarrow (4k-1)(2k+1)>0 \).
3. With \( k<0 \) this gives \( k<-\dfrac12 \).

Teacher's Note:
1. Check the sign of the denominator first and then control the numerator.
2. A quadratic is negative for all x only if its leading coefficient is negative and D is negative.

Answer: (c) \( -\dfrac94 \)

Explanation:
1. Sum: \( a+b=-a \) and product: \( ab=b \). Since \( b\neq0 \), \( a=1 \) and then \( b=-2 \).
2. The least value is \( b-\dfrac{a^2}{4}=-2-\dfrac14=-\dfrac94 \).

Teacher's Note:
1. Use both the sum and the product of the roots to find a and b.
2. The minimum of an upward quadratic is -D/4a.

Answer: (c) \( pq=r \)

Explanation:
1. Let the roots be \( s,-s,t \). Then \( t=p \) (sum of roots) and \( \sum\alpha\beta=-s^2=q \).
2. Product \( =-s^2t=r \Rightarrow qp=r \).

Teacher's Note:
1. Choose a convenient form for the roots and use all three relations.
2. Eliminate s to reach a relation among p, q, r.

Answer: (b) 2

Explanation:
1. Sum \( =-\dfrac{2a+3}{a+1}=-1 \Rightarrow 2a+3=a+1 \Rightarrow a=-2 \).
2. Product \( =\dfrac{3a+4}{a+1}=\dfrac{-2}{-1}=2 \).

Teacher's Note:
1. First find the unknown parameter from the given condition.
2. Then substitute it in the formula for the product.

Answer: (a) \( 4x^2-12x-7=0 \)

Explanation:
1. \( \alpha=\dfrac72 \) and \( \beta=-\dfrac12 \).
2. Sum \( =3 \), product \( =-\dfrac74 \), so \( x^2-3x-\dfrac74=0 \), i.e. \( 4x^2-12x-7=0 \).

Teacher's Note:
1. Solve the two linear equations first to get the roots.
2. Then form the quadratic from the sum and product.

Answer: (a) \( \left(-\infty,-\dfrac45\right]\cup(1,\infty) \)

Explanation:
1. Let \( u=x^2-2x=(x-1)^2-1\ge-1 \). Then \( y=\dfrac{u+9}{u-9}=1+\dfrac{18}{u-9} \).
2. For \( -1\le u<9 \): \( u-9\in[-10,0) \), so \( \dfrac{18}{u-9}\in\left(-\infty,-\dfrac95\right] \) and \( y\in\left(-\infty,-\dfrac45\right] \).
3. For \( u>9 \): \( \dfrac{18}{u-9}>0 \), so \( y>1 \).

Teacher's Note:
1. Substituting u for the repeated expression turns this into a simple fraction.
2. Remember that the value u = 9 makes the denominator zero.

Answer: (b) \( a\in(-2,2) \)

Explanation:
1. Opposite signs need a negative product: \( \dfrac{9(a^4-16)}{a^2-a+2}<0 \).
2. \( a^2-a+2>0 \) for every real a, so \( a^4<16 \), i.e. \( |a|<2 \).

Teacher's Note:
1. The sign of the product of roots decides opposite signs; no other condition is needed.
2. Check that the leading coefficient never vanishes.

Answer: (a) \( -3 \)

Explanation:
1. The sum over all pairs equals \( \displaystyle\sum_{i\ne j}\dfrac{\alpha_i}{\alpha_j}=(\alpha+\beta+\gamma)\left(\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma\right)-3 \).
2. Here \( \sum\alpha=-\dfrac12 \) and \( \sum\alpha\beta=0 \), so \( \sum\dfrac1\alpha=\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}=0 \).
3. The required value is \( 0-3=-3 \).

Teacher's Note:
1. Writing the sum of ratios as a product of two symmetric sums avoids solving the cubic.
2. Remember to subtract the three terms \( \alpha/\alpha \) that appear in the product.

Answer: (b) \( -\dfrac{7}{4} \)

Explanation:
1. \( \alpha+\beta=-\dfrac32 \) and \( \alpha\beta=2 \).
2. \( \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\dfrac94-4=-\dfrac74 \).

Teacher's Note:
1. Always rewrite symmetric expressions through the sum and product of the roots.
2. Roots may be complex here, but the sum of squares still comes out real.

Answer: (b) \( \dfrac12,\ \dfrac12,\ -6 \)

Explanation:
1. A repeated root is also a root of the derivative \( 12x^2+40x-23=0 \), i.e. \( (2x-1)(6x+23)=0 \).
2. Testing \( x=\dfrac12 \): \( 4\cdot\dfrac18+20\cdot\dfrac14-\dfrac{23}{2}+6=0 \), so it is a repeated root.
3. The sum of roots is \( -5 \), so the third root is \( -5-1=-6 \).

Teacher's Note:
1. A repeated root of f(x) is a common root of f(x) and f'(x).
2. Use the sum of roots to find the remaining root quickly.

Answer: (c) \( 3\pm2\sqrt2 \)

Explanation:
1. A perfect square needs \( D=0 \): \( (k+1)^2-8k=0 \).
2. \( k^2-6k+1=0 \Rightarrow k=3\pm2\sqrt2 \).

Teacher's Note:
1. Perfect square means equal roots, hence zero discriminant.
2. Also check that the leading coefficient k is not zero.

Answer: (c) 3 and 8

Explanation:
1. The constant term was copied correctly: \( q=4\times6=24 \). The correct coefficient is \( p=-11 \).
2. Correct equation: \( x^2-11x+24=0 \), i.e. \( (x-3)(x-8)=0 \), so \( x=3,8 \).

Teacher's Note:
1. The wrongly written coefficient does not affect the product of the roots, so the constant term stays reliable.
2. Check by confirming sum 11 and product 24.

Answer: (a) \( 3x^2-19x+3=0 \)

Explanation:
1. Both \( \alpha,\beta \) satisfy \( t^2-5t+3=0 \) and are different, so \( \alpha+\beta=5,\ \alpha\beta=3 \).
2. Sum of new roots \( =\dfrac{\alpha^2+\beta^2}{\alpha\beta}=\dfrac{25-6}{3}=\dfrac{19}{3} \); product \( =1 \).
3. Equation: \( x^2-\dfrac{19}{3}x+1=0 \), i.e. \( 3x^2-19x+3=0 \).

Teacher's Note:
1. When two numbers satisfy the same quadratic and are unequal, they are its two roots.
2. The product of a number and its reciprocal pair is always 1.

Answer: (d) \( -\dfrac{7}{8} \)

Explanation:
1. \( \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}=\dfrac{\alpha^2+\beta^2}{\alpha\beta} \).
2. Using \( \alpha^2+\beta^2=-\dfrac74 \) and \( \alpha\beta=2 \), the value is \( -\dfrac{7}{8} \).

Teacher's Note:
1. Combine the fractions first, then substitute the known symmetric values.
2. Do not forget to divide by the product of the roots.

Answer: (d) \( [3,\ 6) \)

Explanation:
1. The vertex is at \( x=2 \), outside \( (0,1] \), so f is decreasing on the interval.
2. As \( x\to0^+ \), \( f\to6 \) (not attained); at \( x=1 \), \( f=3 \).
3. Range \( =[3,6) \).

Teacher's Note:
1. Check where the vertex lies before deciding the range on a restricted domain.
2. An open end of the interval gives an open end of the range.

Answer: (a) \( x=2 \)

Explanation:
1. Put \( t=2^x>0 \): \( t^2+4t-32=0 \Rightarrow (t+8)(t-4)=0 \).
2. \( t=-8 \) is impossible, so \( 2^x=4 \) and \( x=2 \).

Teacher's Note:
1. Exponential equations of this type become quadratics after a substitution.
2. Reject negative values of t because an exponential is always positive.

Answer: (d) \( (-4,\ 7) \)

Explanation:
1. Complex roots of a real quadratic are conjugates, so the roots are \( 2\pm i\sqrt3 \).
2. Sum \( =4=-p \Rightarrow p=-4 \); product \( =4+3=7=q \).

Teacher's Note:
1. Use the conjugate root theorem before forming the sum and product.
2. The product of conjugates is the square of the modulus.