JEE Mathematics Quadratic Equations MCQs Set 19

Answer: (b) \( a=1 \) and \( a=2 \)

Explanation:
1. Put \( t=x^2 \): \( t^2-(a-2)(a-3)t-(a-1)(a-2)=0 \). All roots of the quartic are real only if both values of t are non-negative.
2. Product of the t-roots \( =-(a-1)(a-2)\ge0\Rightarrow1\le a\le2 \), so the only integers are 1 and 2.
3. For \( a=1 \): \( t^2-2t=0 \Rightarrow t=0,2 \). For \( a=2 \): \( t^2=0 \). Both give real x.

Teacher's Note:
1. Substitute \( t=x^2 \) and demand non-negative t-roots.
2. Check the integer candidates directly.

Answer: (a) (B), (C) and (D) only

Explanation:
1. (A): \( (7-\alpha)(7-\beta)=49-7(\alpha+\beta)+\alpha\beta=49+21+1=71\ne0 \). False.
2. (B): \( (2-\alpha)(2-\beta)=4-2(\alpha+\beta)+\alpha\beta=4+6+1=11 \). True.
3. (C): \( 3\alpha+1=-\alpha^2 \), so each term equals \( -1 \) and the sum is \( -2 \). True.
4. (D): \( \alpha\beta=1 \), so \( \dfrac{\alpha}{1+\beta}=\dfrac{1}{\beta(1+\beta)}=\dfrac{1}{\beta^2+\beta}=-\dfrac{1}{2\beta+1} \), and similarly \( \dfrac{\beta}{1+\alpha}=-\dfrac{1}{2\alpha+1} \). Now \( (2\alpha+1)(2\beta+1)=4-6+1=-1 \) and \( (2\alpha+1)^2+(2\beta+1)^2=4\cdot7-12+2=18 \), so the sum of squares is \( \dfrac{18}{1}=18 \). True.

Teacher's Note:
1. Use the equation satisfied by each root to simplify the denominators.
2. Evaluate statement (D) numerically if the algebra looks long.

Answer: (b) 1

Explanation:
1. Put \( t=e^x>0 \): \( t^4+t^3-4t^2+t+1=(t-1)^2(t^2+3t+1)=0 \).
2. \( t^2+3t+1=0 \) has only negative roots, which are rejected, so \( t=1 \) and \( x=0 \).
3. There is exactly one real root.

Teacher's Note:
1. Substitute \( t=e^x \) and remember that t must be positive.
2. Factor the quartic by noting that t = 1 is a repeated root.

Answer: (b) \( -29 \)

Explanation:
1. Each root satisfies \( t^2=-t+1 \), so \( V_n=-V_{n-1}+V_{n-2} \) with \( V_0=2,\ V_1=-1 \).
2. \( V_2=3,\ V_3=-4,\ V_4=7,\ V_5=-11,\ V_6=18,\ V_7=-29 \).

Teacher's Note:
1. Build the recurrence from the equation and iterate.
2. Alternate signs appear because the sum of roots is negative.

Answer: (c) 3

Explanation:
1. Each root satisfies \( t^2=6t+2 \). Multiply by \( t^{8} \): \( t^{10}=6t^9+2t^8 \).
2. Subtracting for \( \alpha \) and \( \beta \): \( a_{10}=6a_9+2a_8 \), so \( a_{10}-2a_8=6a_9 \).
3. The ratio is \( \dfrac{6a_9}{2a_9}=3 \).

Teacher's Note:
1. Use the quadratic to build a recurrence for \( a_n \).
2. The ratio then no longer depends on the large powers.

Answer: (b) (A) and (D) only

Explanation:
1. \( r=\alpha^4+\beta^4>0 \) and \( q=\alpha\beta \).
2. Product of the roots \( =2q^2-r=2\alpha^2\beta^2-\alpha^4-\beta^4=-(\alpha^2-\beta^2)^2<0 \) because \( \alpha\neq\pm\beta \).
3. A negative product means one positive and one negative root, so the roots are real; (A) and (D) are true.

Teacher's Note:
1. A negative product of roots already guarantees real roots of opposite sign.
2. Write the product as a negative perfect square.

Answer: (b) (A), (B) and (D) only

Explanation:
1. \( \alpha^2=a\alpha+a+b \), so \( \alpha^2-a\alpha=a+b \); likewise for \( \beta \). The expression is \( \dfrac{1}{a+b}+\dfrac{1}{a+b}-\dfrac{2}{a+b}=0 \), so (A) is true.
2. \( f(a)+a+b=(a^2-a^2-a-b)+a+b=0 \), so (B) is true.
3. \( f(b)+a+b=b^2-ab=b(b-a)\neq0 \) in general, so (C) is false.
4. \( f(a/2)+a^2/4+a+b=\dfrac{a^2}{4}-\dfrac{a^2}{2}-a-b+\dfrac{a^2}{4}+a+b=0 \), so (D) is true.

Teacher's Note:
1. Replace \( \alpha^2-a\alpha \) using the equation satisfied by the root.
2. Then test every option by direct substitution.

Answer: (c) (A), (B), (C) and (D)

Explanation:
1. \( f(x)-x^2 \) vanishes at 1, 2, 3 and is monic cubic, so \( f(x)=x^2+(x-1)(x-2)(x-3) \).
2. (A) \( f(4)=16+6=22 \). True.
3. (B) and (C): \( f(x)-x^3=-5x^2+11x-6=-(x-1)(5x-6) \), so \( f(x)=x^3 \) at exactly \( x=1 \) and \( x=\dfrac65 \). Both true.
4. (D) \( f(0)=-6<0 \) and \( f(1)=1>0 \), so a root lies in \( (0,1) \). True.

Teacher's Note:
1. Subtract the obvious polynomial \( x^2 \) to expose the factors.
2. Use the sign change test for the last statement.

Answer: (d) (B) and (C) only

Explanation:
1. The roots are primes with sum 32, and the only such pairs are \( (3,29) \) and \( (13,19) \).
2. Hence \( c=3\times29=87 \) or \( c=13\times19=247 \).

Teacher's Note:
1. List all ways to write 32 as a sum of two primes.
2. The product of each pair gives a possible c.

Answer: (d) (B), (C) and (D) only

Explanation:
1. (A): \( \alpha^2+\beta^2=a^2-2b \) and \( \alpha^2\beta^2=b^2 \), so the equation is \( x^2-(a^2-2b)x+b^2=0 \). The given one is different, so (A) is false.
2. (B): sum \( =\dfrac{\alpha+\beta}{\alpha\beta}=-\dfrac ab \), product \( =\dfrac1b \), giving \( bx^2+ax+1=0 \). True.
3. (C): sum \( =\dfrac{a^2-2b}{b} \), product 1, giving \( bx^2+(2b-a^2)x+b=0 \). True.
4. (D): replace x by \( x+1 \): \( (x+1)^2+a(x+1)+b=x^2+(a+2)x+1+a+b \). True.

Teacher's Note:
1. Test each statement by finding the new sum and product.
2. Statement (A) mixes up the sign of the middle coefficient.

Answer: (a) (B) and (C) only

Explanation:
1. Sum of new roots: \( -\dfrac BA=-\dfrac ba+2\delta \), so \( \delta=\dfrac12\left(\dfrac ba-\dfrac BA\right) \). This is (B), and (A) has the opposite sign.
2. Shifting both roots by the same amount keeps their difference, so \( \dfrac{\sqrt D}{|a|} \) is unchanged: \( \dfrac{b^2-4ac}{a^2}=\dfrac{B^2-4AC}{A^2} \). This is (C).
3. Option (D) has plus signs and is false.

Teacher's Note:
1. The square of the difference of the roots is \( D/a^2 \).
2. A shift of roots changes the sum but not the difference.

Answer: (a) 11

Explanation:
1. \( p_1=1,\ p_2=3 \) and \( p_k=p_{k-1}+p_{k-2} \).
2. \( p_3=4,\ p_4=7,\ p_5=11 \).

Teacher's Note:
1. The power sums follow the same recurrence as the equation.
2. Compute step by step from p1 and p2.

Answer: (d) (A) and (D) only

Explanation:
1. Write \( ax^3+bx^2+cx+d=(x^2+x+1)(ax+e) \). Comparing coefficients: \( b=a+e \) and \( d=e \).
2. The real root is \( x=-\dfrac ea \). Since \( e=d \), \( x=-\dfrac da \), and since \( e=b-a \), \( x=\dfrac{a-b}{a} \).

Teacher's Note:
1. Factor the cubic with the unknown linear factor and compare coefficients.
2. Two different expressions for e give two equivalent forms of the root.

Answer: (c) (A) and (C) only

Explanation:
1. The roots are \( \dfrac{-1\pm\sqrt5}{4} \), so (A) is true and (B) is false.
2. Since \( 4\alpha^2=1-2\alpha \), \( 4\alpha^3=\alpha-2\alpha^2=\alpha-\dfrac{1-2\alpha}{2}=2\alpha-\dfrac12 \). So \( 4\alpha^3-3\alpha=-\alpha-\dfrac12 \), which equals the other root because the sum of roots is \( -\dfrac12 \). Hence (C) is true and (D) is false.

Teacher's Note:
1. Reduce powers of the root using the equation it satisfies.
2. Compare the result with the other root found from the sum of roots.

Answer: (d) (A), (B) and (C) only

Explanation:
1. \( a_n \) are the Fibonacci numbers \( 1,1,2,3,5,\dots \) and \( b_n=a_{n-1}+a_{n+1} \) are the Lucas numbers \( 1,3,4,7,\dots=\alpha^n+\beta^n \). So (B) is true.
2. (A) \( \sum a_nx^n=\dfrac{x}{1-x-x^2} \); at \( x=\dfrac1{10} \) this is \( \dfrac{0.1}{0.89}=\dfrac{10}{89} \). True.
3. (C) \( a_1+\dots+a_n=a_{n+2}-1 \) is the standard Fibonacci sum identity. True.
4. (D) \( \sum b_nx^n=\dfrac{x+2x^2}{1-x-x^2} \); at \( x=\dfrac1{10} \) this is \( \dfrac{0.12}{0.89}=\dfrac{12}{89}\neq\dfrac{8}{89} \). False.

Teacher's Note:
1. Recognise the Fibonacci and Lucas sequences and use their generating functions.
2. Check the last statement numerically.

Answer: (a) 1 : 2 : 3

Explanation:
1. \( x^2+2x+3=0 \) has imaginary roots (\( D=-8 \)).
2. A real quadratic with a common imaginary root has both roots common, so the equations are proportional: \( a:b:c=1:2:3 \).

Teacher's Note:
1. Imaginary roots of real equations occur in conjugate pairs.
2. One common non-real root forces both roots to be common.

Answer: (b) (A), (C) and (D) only

Explanation:
1. A quadratic equation has more than two roots only when it is an identity, i.e. all three coefficients vanish.
2. \( p^2-3p+2=0\Rightarrow p=1,2 \); \( p^2-5p+4=0\Rightarrow p=1,4 \); \( p-p^2=0\Rightarrow p=0,1 \). All vanish only for \( p=1 \).
3. So every value except \( p=1 \) is allowed, and from the given list these are 0, 2 and 4.

Teacher's Note:
1. A polynomial equation of degree at most 2 with more than two roots must be the zero polynomial.
2. Find the common solution of the three coefficient equations.

Answer: (d) 18

Explanation:
1. \( p \) is a quadratic with a single (double) root at \( -1 \), so \( p(x)=k(x+1)^2 \).
2. \( p(-2)=k=2 \), hence \( p(2)=2\times9=18 \).

Teacher's Note:
1. Only one zero for a quadratic means it is a repeated root.
2. Use the given value to find the constant multiplier.

Answer: (c) (A) and (D) only

Explanation:
1. Distinct real roots: \( 1-4\alpha^2>0\Rightarrow|\alpha|<\dfrac12 \).
2. \( |x_1-x_2|=\dfrac{\sqrt{1-4\alpha^2}}{|\alpha|}<1\Rightarrow1-4\alpha^2<\alpha^2\Rightarrow\alpha^2>\dfrac15 \).
3. So \( S=\left(-\dfrac12,-\dfrac1{\sqrt5}\right)\cup\left(\dfrac1{\sqrt5},\dfrac12\right) \), which contains (A) and (D).

Teacher's Note:
1. Convert the condition on the difference of roots into a condition on the discriminant.
2. Solve the two inequalities together.

Answer: (a) (A), (B), (C) and (D)

Explanation:
1. For \( x\le0 \) every term of P is non-negative and the constant term is 1, so \( P(x)\ge1 \). For \( x\ge1 \) group as \( (x^{32}-x^{25})+(x^{18}-x^{11})+(x^4-x^3)+1\ge1 \). For \( 0<x<1 \) group as \( x^{32}+(x^{18}-x^{25})+(x^4-x^{11})+(1-x^3)>0 \). So P has no real roots.
2. A polynomial of degree 32 with no real roots has 32 imaginary roots, so (A), (B) and (C) are true.
3. \( P(x)+P(-x)=2(x^{32}+x^{18}+x^4+1)>0 \) for all real x, so all 32 roots are imaginary. (D) is true.

Teacher's Note:
1. Show that the polynomial is positive for every real x by grouping terms.
2. A polynomial of even degree with no real root has all roots imaginary.

Answer: (c) \( (-\infty,10) \)

Explanation:
1. Write \( f(x)=x^2+3x+m(1-x) \). At \( x=1 \), \( f=4>0 \) for every m.
2. For \( 1<x\le2 \) we need \( m<\dfrac{x^2+3x}{x-1} \). The right side \( g(x)=\dfrac{x^2+3x}{x-1}=x+4+\dfrac{4}{x-1} \) is decreasing on \( (1,2] \), with least value \( g(2)=10 \).
3. Hence \( m<10 \).

Teacher's Note:
1. Separate m from x and bound it by the least value of the resulting function.
2. Treat the end point x = 1 separately because it makes the coefficient of m vanish.