Question

Find the values of 't' in the equation $x^2$ - 2tx + $t^2$ - 1 = 0 such that exactly one root lies in between the numbers 2 and 4, and no root of the equation is either 2 (or) 4.

• A. 1 < t < 5
• B. 1 ≤ t ≤ 5
• C. 1 < t < 3 and 3 < t < 5
• D. 1 ≤ t ≤ 3 and 4 < t < 5

Answer 1

The correct option is C

1 < t < 3 and 3 < t < 5

Comparing f(x) = $x^2$ - 2tx + $t^2$ - 1 to a$x^2$ + bx + c,

a = 1, b = -2t, c = $t^2$ - 1

As a > 0, graph of f(x) is upward(cup shaped) parabola

For roots to be real, $b^2$ - 4ac > 0

$\Rightarrow$ 4$t^2$ - 4($t^2$ - 1) > 0

4 > 0 [Always true]

Observe from the graph that f(2) > 0 and f(4) < 0.

Hence f(2) f(4) < 0 ------------- (1)

Also, for root $\ne$ +2 (or) 4, f(2) $\ne$ 0 and f(4) $\ne$ 0 ---------- (2)

(1) $\Rightarrow$ f(2) f(4) < 0

[4 - 4t + ($t^2$ - 1)][16 - 8t + $t^2$ - 1] < 0

$\Rightarrow$ ($t^2$ - 4t + 3) ($t^2$ - 8t + 15) < 0

$\Rightarrow$ (t - 1) (t - 3) (t - 3) (t - 5) < 0

$\Rightarrow$ ${(t-3)}^2$ (t - 1) (t - 5) < 0

$\Rightarrow$ t $ϵ$ (1, 5)

(2) $\Rightarrow$ f(2) $\ne$ 0

$\Rightarrow$ t $\ne$ 1, 3

(3) $\Rightarrow$f(4) $\ne$ 0 $\Rightarrow$ t $\ne$ 3, 5

So, t $ϵ$ (1, 5) - {3}