Question

Find the value of a for which the inequality \[co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)cotx - \left( {a - 3} \right) < 0,\] is true for at least one $ x \in (0,\dfrac{\pi }{2}) $
(A) $ a \in ( - \infty , - 1) $
(B) $ a \in \{ ( - 3, - 2\sqrt 5 ),3\} $
(C) $ \alpha \in ( - \infty , - 3 - 2\sqrt 5 ) \cup (3,\infty ) $
(D) None of these

Answer 1

Hint: To solve this question, we will start with assuming \[\cot x{\text{ }} = {\text{ }}t\], then we get our given equation in the form $ {t^2} + (a + 1)t - (a - 3) < 0 $ for at least one $ t \in (0,\infty ) $ . From here we will consider two cases, and then we will get our required answer.

Complete step-by-step answer:
We have been given an inequality \[co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)cotx - \left( {a - 3} \right) < 0,\] and we need to find the value of a for which the inequality is true for at least one $ x \in (0,\dfrac{\pi }{2}) $ .
\[co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)cotx - \left( {a - 3} \right) < 0,\] for at least one $ x \in (0,\dfrac{\pi }{2}) $ .
Now, let \[\cot x{\text{ }} = {\text{ }}t\]
We know that, $ \cot 0 = \infty $ and $ \cot \dfrac{\pi }{2} = 0 $
Since, $ x \in (0,\dfrac{\pi }{2}) $ , therefore, on putting $ \cot 0 = \infty $ and $ \cot \dfrac{\pi }{2} = 0 $ , we get $ t \in (0,\infty ) $
Now on putting the value \[\cot x{\text{ }} = {\text{ }}t\], we get our above equation \[co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)cotx - \left( {a - 3} \right) < 0,\] in the form $ {t^2} + (a + 1)t - (a - 3) < 0. $
Now, let $ f(t) = {t^2} + (a + 1)t - (a - 3) < 0 $ , for at least one $ t \in (0,\infty ) $
So, we have two following possibilities for f(t). Let us see the cases mentioned below.
Case 1:
 $
f(0) < 0 \\
f(0) = - (a - 3) < 0 \\
= - a + 3 < 0 \\
= - a < - 3 \\
= a > 3 \\
 $
Case 2:
We have been given a quadratic equation, so, we will consider, \[D > {\text{ }}0\], i.e., the equation will have distinct roots.

 $ i.e.,{b^2} - 4ac > 0 $
 $
\Rightarrow {(a + 1)^2} + 4(a - 3) > 0 \\
\Rightarrow {a^2} + 6a - 11 > 0 \\
\Rightarrow {(a + 3)^2} > 20 \\
\Rightarrow a + 3 < - 2\sqrt 5 {\text{ or }}a + 3 > 2\sqrt 5 \\
\Rightarrow a < - 3 - 2 - \sqrt 2 {\text{ or a}} > - 3 + 2\sqrt 5 ...eq.(1) \\
\\
 $
We had considered earlier, \[D > 0,\]then
 $
\dfrac{{ - b}}{{2a}} > 0 \\
\Rightarrow - (a + 1) > 0 \\
\Rightarrow a < - 1...eq.(2) \\
\Rightarrow a(0) \geqslant 0 \\
\Rightarrow - a + 3 \geqslant 0 \\
\Rightarrow a \leqslant 3...eq.(3) \\
 $
From \[eq.\left( 1 \right),{\text{ }}\left( 2 \right)\] and \[\left( 3 \right),\] we get
 $ a \in ( - \infty , - 3 - 2\sqrt 5 ) $
So, from both the cases, we get
 $ \alpha \in ( - \infty , - 3 - 2\sqrt 5 ) \cup (3,\infty ) $ $ $
Hence, option (C), $ \alpha \in ( - \infty , - 3 - 2\sqrt 5 ) \cup (3,\infty ) $ is correct.
So, the correct answer is “Option C”.

Note: In the question, we have been given a quadratic equation. The standard form of quadratic equation is \[a{x^2} + bx + c = 0.\]
Discriminant, \[\;D{\text{ }} = {\text{ }}{b^2} - 4ac\]
The value of D tells us the kind of root equation has. If \[D > 0,\]then the equation has two distinct roots.
If \[D = 0,\] then the equation has one real root.
If \[D < 0,\] then the equation has no real roots, i.e., two imaginary roots will form.