JEE Mathematics Quadratic Equations MCQs Set 21

Answer: (b) \(acx^2-bx+1=0\)

Explanation:
1. From \(a\alpha^2+b\alpha+c=0\): \(a\alpha+b=-\dfrac c\alpha\), so \(\dfrac1{a\alpha+b}=-\dfrac\alpha c\); similarly for \(\beta\).
2. Sum \(=-\dfrac{\alpha+\beta}c=\dfrac b{ac}\); product \(=\dfrac{\alpha\beta}{c^2}=\dfrac1{ac}\).
3. The equation is \(x^2-\dfrac b{ac}x+\dfrac1{ac}=0\), i.e. \(acx^2-bx+1=0\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (c) 1, -4

Explanation:
1. First equation: sum \(0=-p\), product \(c=-4\).
2. Second equation: sum \(-3=-b\), so \(b=3\).
3. \(x^2+3x-4=(x+4)(x-1)=0\) gives 1 and -4.

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (c) \(a^2+b^2\)

Explanation:
1. Let \(t=x+\sqrt{x^2+b^2}>0\); then \(x=\dfrac{t^2-b^2}{2t}\).
2. \(y=2at-(t^2-b^2)=-(t-a)^2+a^2+b^2\).
3. This is largest at \(t=a>0\) (attainable), giving \(a^2+b^2\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (a) an even integer

Explanation:
1. \(\alpha^n,\beta^n\) are the roots of \(y^2+p^ny+q^n=0\), so \(\alpha^n+\beta^n=-p^n\).
2. \(\alpha/\beta\) being a root of \(x^n+1+(x+1)^n=0\) means \(\alpha^n+\beta^n+(\alpha+\beta)^n=0\), i.e. \(-p^n+(-p)^n=0\).
3. This holds only when n is even.

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (c) real and unequal roots

Explanation:
1. Let \(a=\sin\alpha<b=\sin\beta<c=\sin\gamma\), and call the left side f(x). Then \(f(a)=(a-b)(a-c)>0\), \(f(b)=(b-a)(b-c)<0\), \(f(c)=(c-a)(c-b)>0\).
2. The sign change between a and b and between b and c gives two real, distinct roots (both below 1).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (b) \(\left[\dfrac73,\infty\right)\)

Explanation:
1. The inequality rearranges to \(a(x^2-x+1)\ge x^2+x+1\), so \(a\ge g(x)=\dfrac{x^2+x+1}{x^2-x+1}=1+\dfrac{2x}{x^2-x+1}\) for all \(x\ge2\).
2. the derivative of g has the sign of \(1-x^2<0\), so g decreases for \(x\ge2\) with maximum \(g(2)=\dfrac73\).
3. So \(a\ge\dfrac73\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (d) 45

Explanation:
1. The other root is the conjugate \(2+i\). Sum of roots \(4=-\dfrac{12}a\) gives \(a=-3\).
2. Product \(5=\dfrac ba\) gives \(b=-15\), so \(ab=45\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (a) 0

Explanation:
1. Subtracting the equations: \((a-c)x^2+(c-a)=0\), so \(x^2=1\).
2. The negative common root is \(x=-1\); substituting gives \(a-b+c=0\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (b) 4

Explanation:
1. \(q=\dfrac{p+r}2\); real roots need \(q^2-4pr\ge0\), i.e. \((p+r)^2\ge16pr\).
2. With \(t=r/p\): \(t^2-14t+1\ge0\), i.e. \((t-7)^2\ge48\) and \(|t-7|\ge4\sqrt3\). So k = 4.

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (c) \(\left(-1,\dfrac{5-3\sqrt5}2\right]\cup\left[\dfrac{5+3\sqrt5}2,6\right)\)

Explanation:
1. Roots -5 and 1 give sum \(-4=-[a^2-5a-1]\), so \([a^2-5a-1]=4\), i.e. \(4\le a^2-5a-1<5\).
2. \(a^2-5a-5\ge0\) gives \(a\le\tfrac{5-3\sqrt5}2\) or \(a\ge\tfrac{5+3\sqrt5}2\); \(a^2-5a-6<0\) gives \(-1<a<6\).
3. Intersecting gives the stated set.

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (d) \(p\in(0,2)\)

Explanation:
1. Roots are \(p\pm1\).
2. \(p-1<1<p+1\) gives \(0<p<2\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (a) \(p<3\)

Explanation:
1. The larger root is \(p+1\).
2. Both roots are below 4 exactly when p + 1 < 4, i.e. p < 3.

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (d) \(one real root in (e, pi) and another in (pi, pi + e)\)

Explanation:
1. All three numerators are positive. The terms have simple poles at \(e<\pi<\pi+e\).
2. Just right of \(e\) the function tends to \(+\infty\); just left of \(\pi\) it tends to \(-\infty\): a root in \((e,\pi)\). Just right of \(\pi\) it is \(+\infty\) and just left of \(\pi+e\) it is \(-\infty\): a root in \((\pi,\pi+e)\).
3. It is a quadratic after clearing denominators, so these are the only two roots.

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (c) 2

Explanation:
1. \(a+b=10c\) and \(c+d=10a\). Since a and c are roots of their own quadratics: \(a^2-10ca-11d=0\) and \(c^2-10ac-11b=0\).
2. Subtract: \(a^2-c^2=11(d-b)\). With \(b=10c-a\) and \(d=10a-c\), \(d-b=11(a-c)\), so \(a+c=121\) (as a is not c).
3. \(a+b+c+d=10c+10a=1210\) and \(1210/605=2\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (a) 2

Explanation:
1. The product of roots of the second equation is \(\dfrac{c^5}{a^5}=4\cdot8=32\).
2. So \(\dfrac ca=2\), and \(\alpha\beta=\dfrac ca=2\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.

Answer: (b) 2

Explanation:
1. Since \(2\omega^2=\dfrac2\omega\) and \(2\omega=\dfrac2{\omega^2}\), both \(x=\omega\) and \(x=\omega^2\) satisfy \(\dfrac1{x+a}+\dfrac1{x+b}+\dfrac1{x+c}=\dfrac2x\).
2. With \(P(x)=(x+a)(x+b)(x+c)=x^3+e_1x^2+e_2x+e_3\), the equation is \(xP^{\prime}(x)=2P(x)\), i.e. \(x^3-e_2x-2e_3=0\), a cubic with no \(x^2\) term.
3. Its roots sum to 0, so with roots \(\omega,\omega^2\) the third root is \(-(\omega+\omega^2)=1\). Thus \(x=1\) satisfies the equation: \(\sum\dfrac1{a+1}=\dfrac21=2\).

Teacher's Note:
1. Use the sum and product of the roots and the sign of the quadratic at key points.
2. Check the discriminant before concluding that roots exist.