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Question: Find the value of a for which the inequality \[co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)...

Find the value of a for which the inequality cot2x + (a+1)cotx−(a−3)<0,co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)cotx - \left( {a - 3} \right) < 0, is true for at least one x∈(0,π2)x \in (0,\dfrac{\pi }{2})
(A) a∈(−∞,−1)a \in ( - \infty , - 1)
(B) a∈(−3,−25),3a \in \\{ ( - 3, - 2\sqrt 5 ),3\\}
(C) α∈(−∞,−3−25)∪(3,∞)\alpha \in ( - \infty , - 3 - 2\sqrt 5 ) \cup (3,\infty )
(D) None of these

Explanation

Solution

Hint : To solve this question, we will start with assuming cot⁡x = t\cot x{\text{ }} = {\text{ }}t, then we get our given equation in the form t2+(a+1)t−(a−3)<0{t^2} + (a + 1)t - (a - 3) < 0 for at least one t∈(0,∞)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 cot2x + (a+1)cotx−(a−3)<0,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∈(0,π2)x \in (0,\dfrac{\pi }{2}) .
cot2x + (a+1)cotx−(a−3)<0,co{t^2}x{\text{ }} + {\text{ }}\left( {a + 1} \right)cotx - \left( {a - 3} \right) < 0, for at least one x∈(0,π2)x \in (0,\dfrac{\pi }{2}) .
Now, let cot⁡x = t\cot x{\text{ }} = {\text{ }}t
We know that, cot⁡0=∞\cot 0 = \infty and cot⁡π2=0\cot \dfrac{\pi }{2} = 0
Since, x∈(0,π2)x \in (0,\dfrac{\pi }{2}) , therefore, on putting cot⁡0=∞\cot 0 = \infty and cot⁡π2=0\cot \dfrac{\pi }{2} = 0 , we get t∈(0,∞)t \in (0,\infty )
Now on putting the value cot⁡x = t\cot x{\text{ }} = {\text{ }}t, we get our above equation cot2x + (a+1)cotx−(a−3)<0,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)=t2+(a+1)t−(a−3)<0f(t) = {t^2} + (a + 1)t - (a - 3) < 0 , for at least one t∈(0,∞)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  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> 0D > {\text{ }}0, i.e., the equation will have distinct roots.

i.e.,b2−4ac>0i.e.,{b^2} - 4ac > 0
⇒(a+1)2+4(a−3)>0 ⇒a2+6a−11>0 ⇒(a+3)2>20 ⇒a+3<\-25 or a+3>25 ⇒a<\-3−2−2 or a>−3+25...eq.(1)   \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,D > 0,then
−b2a>0 ⇒−(a+1)>0 ⇒a<\-1...eq.(2) ⇒a(0)⩾0 ⇒−a+3⩾0 ⇒a⩽3...eq.(3)  \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.(1), (2)eq.\left( 1 \right),{\text{ }}\left( 2 \right) and (3),\left( 3 \right), we get
a∈(−∞,−3−25)a \in ( - \infty , - 3 - 2\sqrt 5 )
So, from both the cases, we get
α∈(−∞,−3−25)∪(3,∞)\alpha \in ( - \infty , - 3 - 2\sqrt 5 ) \cup (3,\infty )
Hence, option (C), α∈(−∞,−3−25)∪(3,∞)\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 ax2+bx+c=0.a{x^2} + bx + c = 0.
Discriminant,   D = b2−4ac\;D{\text{ }} = {\text{ }}{b^2} - 4ac
The value of D tells us the kind of root equation has. If D>0,D > 0,then the equation has two distinct roots.
If D=0,D = 0, then the equation has one real root.
If D<0,D < 0, then the equation has no real roots, i.e., two imaginary roots will form.