Question

If the quadratic equation $ a{{x}^{2}}+bx+c=0\left( a > 0 \right) $ has $ {{\sec }^{2}}\theta $ and $ {{\operatorname{cosec}}^{2}}\theta $ as its roots, then which of the following must hold good? \[\]
A. $ b+c=0 $ \[\]
B. $ {{b}^{2}}-4ac\ge 0 $ \[\]
C. $ c\ge 4a $ \[\]
D. $ 4a+b=0 $ \[\]

Answer 1

Hint: We see that since the roots $ {{\sec }^{2}}\theta ,{{\operatorname{cosec}}^{2}}\theta $ are well-defined we shall have discriminant $ D\ge 0 $ . We use the sum of roots of quadratic equation $ a{{x}^{2}}+bx+c=0 $ as $ \dfrac{-b}{a} $ and product of the roots as $ \dfrac{c}{a} $ . We convert $ {{\sec }^{2}}\theta $ and $ {{\operatorname{cosec}}^{2}}\theta $ into sine and cosine and find the relations between $ a,c $ and then between $ b,c $ . \[\]

Complete step by step answer:
We know that the general form of a quadratic equation is given by $ a{{x}^{2}}+bx+c=0 $ where $ a,b,c $ are real numbers and $ a $ cannot be zero. A quadratic equation always has two roots and their sum is given as the negative ratio of coefficient of $ x $ to coefficient of $ {{x}^{2}} $ . If $ {{x}_{1}},{{x}_{2}} $ x are two roots then their sum is given by
\[{{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}\]
The product of their roots is given as the ratio of constant term to coefficient of $ {{x}^{2}} $ which means
\[{{x}_{1}}{{x}_{2}}=\dfrac{c}{a}\]
We are give in the question that $ {{\sec }^{2}}\theta $ and $ {{\operatorname{cosec}}^{2}}\theta $ are the roots of the quadratic equation. We see that because of square $ {{\sec }^{2}}\theta $ and $ {{\operatorname{cosec}}^{2}}\theta $ are real positive numbers can be equal but never zero. Hence they can be either distinct or equal. We know that for real roots the discriminant of quadratic equation has the condition
\[D={{b}^{2}}-4ac\ge 0.......\left( 1 \right)\]
We use the formula for sum of roots and have
\[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =\dfrac{-b}{a}\]
We convert the secant and cosecant to sine and cosine using the reciprocal relations $ \sec \theta =\dfrac{1}{\cos \theta },cosec\theta =\dfrac{1}{\sin \theta } $ in the above step to have;

\[\begin{align}
  & \Rightarrow \dfrac{1}{{{\cos }^{2}}\theta }+\dfrac{1}{{{\sin }^{2}}\theta }=\dfrac{-b}{a} \\
 & \Rightarrow \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=\dfrac{-b}{a} \\
\end{align}\]
We use the Pythagorean trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=\dfrac{-b}{a} \\
 & \Rightarrow \dfrac{-a}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=b..........\left( 2 \right) \\
\end{align}\]
We use the formula for product of roots and have
\[{{\sec }^{2}}\theta \cdot \cos e{{c}^{2}}\theta =\dfrac{c}{a}\]
We convert the secant and cosecant to sine and cosine using the reciprocal relations $ \sec \theta =\dfrac{1}{\cos \theta },cosec\theta =\dfrac{1}{\sin \theta } $ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{\cos }^{2}}\theta }\cdot \dfrac{1}{{{\sin }^{2}}\theta }=\dfrac{c}{a} \\
 & \Rightarrow \dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=\dfrac{c}{a}.....\left( 3 \right) \\
 & \Rightarrow \dfrac{a}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=c.....\left( 4 \right) \\
\end{align}\]
We add respective sides of equation (3) and (4) to have
\[\begin{align}
  & \dfrac{-a}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }+\dfrac{a}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=b+c \\
 & \Rightarrow b+c=0......\left( 5 \right) \\
\end{align}\]

We multiply and divide 4 in the numerator and denominator of the left hand side of equation (3) to have;
\[\begin{align}
  & \Rightarrow \dfrac{4}{4{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=\dfrac{c}{a} \\
 & \Rightarrow \dfrac{4}{{{\left( 2\sin \theta \cos \theta \right)}^{2}}}=\dfrac{c}{a} \\
\end{align}\]
We use sine double angle formula $ \sin 2\theta =2\sin \theta \cos \theta $ in the denominator of left hand side to have;
\[\begin{align}
  & \Rightarrow \dfrac{4}{{{\sin }^{2}}2\theta }=\dfrac{c}{a} \\
 & \Rightarrow {{\sin }^{2}}2\theta =\dfrac{c}{4a} \\
\end{align}\]
We know that the maximum sine function is 1. So we have;
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}2\theta \le 1 \\
 & \Rightarrow \dfrac{c}{4a}\le 1 \\
 & \Rightarrow c\le 4a...\left( 6 \right) \\
\end{align}\]
We have results (1), (4) and (6) that the correct options are A,B and C.

Note:
We note that $ y=a{{x}^{2}}+bx+c $ with the given condition $ a > 0 $ is a parabola opens upward which may cut twice the positive $ x- $ axis to give two distinct roots or touch once to give unique rots. It cannot cut the negative $ x- $ axis since $ {{\sec }^{2}}\theta ,\text{cose}{{\text{c}}^{2}}\theta $ are always positive. We should remember that the domain and range function $ \sec x,\operatorname{cosec}x $ as $ \sec x:R-\left( 2n+1 \right)\dfrac{\pi }{2}\to R-\left( -1,1 \right) $ and $ \operatorname{cosec}x:R-\dfrac{n\pi }{2}\to R-\left( -1,1 \right) $ . So $ \theta $ in this problem cannot be odd integral multiple of $ \dfrac{\pi }{2} $ or integral multiple of $ \pi $ .