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If both roots of the quadratic equation $x^2 - 2ax + a^2 - 1$ lie in $(-2, 2)$ then which of the following can be $[a]$ ?

$[a]$ denotes greatest Integer function of $a$

$$A. -1$$

$$B. 1$$

$$C. 2$$

$$D. 3$$

I have solve it using graphs:

The graph will intersect the x-axis somewhere between $-2$ and $2$. Hence we can conclude that $f(2)$ and $f(-2)$ will be greater than zero.

Now there will be two quadratic equation in $a$ and we will get $4$ values of $a$ I.e. $-3, -1 , 3 ,1$. From these values the answer should be opton $D.$ but the answer is given as option $A.$

Kindly help.

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  • $\begingroup$ HINT: write the function as $(x-a)^2-1$, so that its shape is easy to recognize. $\endgroup$ – b00n heT Aug 25 '16 at 8:09
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    $\begingroup$ What does "$[a]$" mean? $\endgroup$ – user228113 Aug 25 '16 at 8:09
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    $\begingroup$ @G.Sassatelli Edited. $\endgroup$ – Saksham Aug 25 '16 at 8:13
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Observe that $$x^2 - 2ax + a^2 - 1=0$$ $$\implies x^2 - 2ax + a^2 = 1$$ $$\implies (x-a)^2=1$$ $$\implies x-a=\pm1$$ $$\implies x=a+1,a-1$$

Now it is given that both the roots lie between $-2$ and $2$.

Therefore, we have that $$-2 < a+1< 2$$ $$\implies -3 < a < 1\tag1$$ and $$-2 < a-1< 2$$ $$\implies -1 < a< 3 \tag2$$ This means from $(1)$ and $(2)$ that $-1 < a < 1$.
So $[a]$ should be either $0$ or $-1$.
Since $0$ is not in the option but $-1$ is, so the answer should be $(A)$.

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  • $\begingroup$ Is there a way to conclude answer from the method that I've used? $\endgroup$ – Saksham Aug 25 '16 at 8:19
  • $\begingroup$ @user109256 Your conclusions are wrong. The graph actually cuts the x-axis twice between $-2$ and $2$ since it is a quadratic equation. So $f(2)$ and $f(-2)$ are either both positive or both negative. Still the method doesn't seem to help anyway. $\endgroup$ – SchrodingersCat Aug 25 '16 at 8:22
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    $\begingroup$ Suppose if the question was a bit different I mean if we were not able to for something symmetrical to $(x-a)$ there it would have been tedious to solve it. What about that type? $\endgroup$ – Saksham Aug 25 '16 at 8:25
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    $\begingroup$ Depends upon the equation. You don't have any general procedure for such questions. $\endgroup$ – SchrodingersCat Aug 25 '16 at 8:37
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    $\begingroup$ More and more practice will help. Thank you $\endgroup$ – Saksham Aug 25 '16 at 8:38

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