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I wanted to know, how can I find the value of $a$ for which the inequality $\tan^2x + (a+1)\tan x-(a-3)<0$ is true for at least one $x\in(0,\pi/2)$.

I don't know how to proceed, any help is appreciated.

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up vote 4 down vote accepted

Set $\tan x = y$. Note that $y$ can take only positive values since $x\in (0,\dfrac{\pi}{2})$.

Then $$y^2 + (a+1)y - (a-3) <0$$ $$\iff\left(y+\dfrac{a+1}2 \right)^2 - \dfrac{(a+1)^2}4- (a-3) <0\quad \text{(completing the square)}$$ $$\iff \left(y+\dfrac{a+1}2 \right)^2 < \dfrac{a^2+6a-11}4 = \dfrac{(a+3)^2-20}4$$


$$ -\sqrt{\dfrac{(a+3)^2-20}4}<y+\dfrac{a+1}2 < \sqrt{\dfrac{(a+3)^2-20}4}$$

$$\iff -\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2<y < \sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2$$

We just need to make sure that there exists a positive $y$ that satisfies this. So we need to make sure that the upper bound is positive.

Thus, $\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2>0\iff \sqrt{(a+3)^2-20}>a+1$

If $a+1$ negative then the inequality is true, and if $a+1\geq0$, we can square both sides and simplify to obtain $a>3$ which is consistent with $a+1\geq0$. Also, we need $(a+3)^2-20$ to be non-negative, so $(a+3)^2\geq20\iff a\geq\sqrt{20}-3 \text{ or } a\leq-\sqrt{20}-3$.

Thus, $ a>3 \text{ or } a\leq-\sqrt{20}-3$.

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thanks. A very very nice solution. – SHOBHIT GAUTAM Aug 6 '13 at 3:58

To get you started, try letting $(\tan x) = y$, and determine first the values of $a$ and $y = \tan x$ at which the corresponding quadratic equation is equal to zero:

$$\tan^2x + (a+1)\tan x-(a-3)<0\tag{given}$$

$$y^2 + (a+1) y - (a - 3) = 0\tag{2}$$

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Is there something wrong in ur answer, as per the comment above? or should i proceed with the hint u provided. Thanks anyway. – SHOBHIT GAUTAM Aug 5 '13 at 23:48
Shobbit: no, nothing is wrong; the edit I was referring to was changing the formatting of $tan x$ to $\tan x$. You're welcome, and let me know how you progress with this. I'll help if needed. – amWhy Aug 5 '13 at 23:58
@amWhy: Deserves another TU! +1 – Amzoti Aug 6 '13 at 0:33

Hint :

put: $u=\tan x$ and you need the $\Delta >0$ of the quadratic equation $u^2+(a+1)u-(a-3)$=0 by take the numbers $\sqsupset -\infty,c\sqsubset$ and $\sqsupset d,\infty\sqsubset$ where c and d is the solution of $0=(a+1)^2+4(a-3)$ and $d>c$

then to be $u^2+(a+1)u-(a-3)<0$ we take the numbers from t to m where t and m is the solution of $u^2+(a+1)u-(a-3)=0$ and $m>t$

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Why should the discriminant be negative? A quadratic can be negative at some points even if it has $2$ real roots. – Alraxite Aug 6 '13 at 0:10
@Alraxite look at the answer after edit – mhd.math Aug 6 '13 at 7:07

We are really looking at $t^2+(a+1)t-(a-3)=0$, and want at least one positive root.

First observe that if $a\gt 3$ then the constant term is negative, so there is a positive root and a negative root.

For $a=3$, the larger root $0$ barely misses our interval.

For $a \lt 3$, the real roots, if any, have the same sign, since their product is positive.

For the rest, we use the fact that the negative of the coefficient of $t$ is the sum of the (possibly non-real) roots.

As $a$ decreases from $3$ for a while, the coefficient of $t$ is positive. So the real roots, if any, are both negative.

Below $a=-1$, the real roots, if any, are positive. This is because the coefficient of $t$ is the negative of the sum of the roots.

But we need to make sure that the roots are real. For that, the discriminant $(a+1)^2+4(a-3)$ must be $\ge 0$. For negative $a$ this happens at $a=-3-\sqrt{20}$ and below.

Conclusion: There is a positive root if $a\gt 3$ or if $a\le -(3+\sqrt{20})$.

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$a<-1 \text{ and } (a+1)^2+4(a-3)\geq0\iff a\leq-\sqrt{20}-3$. So the solution should be $a>3 \text{ or } a\leq-\sqrt{20}-3$. Nice answer by the way! – Alraxite Aug 24 '13 at 19:19
Thank you for telling me about the missing minus signs. – André Nicolas Aug 24 '13 at 19:47

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