Find all the values of $^{'} a^{'}$ for which the roots of the equation $x^{2} - 2 a x + a^{2} - 1 = 0$ lies between 2 & 4. Also find the values of $^{'} a^{'}$ for which exactly one root lie between 2&4.

Open in App

Solution

Given equation $x^{2} - 2 a x + a^{2} - 1 = 0$
First we can say all cases. If coefficient of $x^{2}$ is then graph of parabola is up word.
$\begin{matrix} D > 0 & [B e c a u s e t w o r o o t s] f (2) f (4) < 0 D = 4 a^{2} - 4 (1) (a^{2} - 1) \Rightarrow 0 = 4 a^{2} - 4 a^{2} + 4 \Rightarrow D > 0 a \in R . N o w, f (2) = (4 - 4 a + a^{2} - 1) & f (4) = (16 - 8 a + a^{2} - 1) f (2) f (4) < 0 \Rightarrow (a^{2} - 4 a + 3) (a^{2} - 3 a + 15) < 0 \Rightarrow (a - 1) (a - 3) (a - 3) (a - 5) < 0 \Rightarrow (a - 1) {(a - 3)}^{2} (a - 5) < 0 \end{matrix}$
Now, ${(a - 3)}^{2}$ is always +ve so $(a - 1)$ & $(a - 5)$ are
effective So, by number line.
But
$\begin{matrix} {(a - 3)}^{2} \geq 0 I f a = 3, t h e n i t i s 0. S o, w e r e m o v e 3. S o \Rightarrow a \in (1, 3) \cup (3, 5) \end{matrix}$

Suggest Corrections

Similar questions

For the equation $x^{2}$ - (a - 3) x + a = 0 (a ∈ R), find the values of 'a' such that exactly one root lies in between 1 and 2.

For the equation $x^{2}$ - 2ax + $a^{2}$ - 1 = 0, the values of 'a' for which 3 lies in between the roots of given equation is

a

(1 - a^{2}) x^{2} + 2 a x - 1 = 0

0

1

x^{2} - (a + 1) x + 2 a = 0

(0, 3)

The values of ‘a’ for which exactly one root of the equation $e^{a} x^{2} - e^{2 a} x + e^{a} - 1 = 0$ lies between 1 and 2 are given by

Join BYJU'S Learning Program