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Question

Find the values of 't' in the equation x2 - 2tx + t2 - 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

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B

1 t 5

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C

1 < t < 3 and 3 < t < 5

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D

1 t 3 and 4 < t < 5

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Solution

The correct option is C

1 < t < 3 and 3 < t < 5


Comparing f(x) = x2 - 2tx + t2 - 1 to ax2 + bx + c,

a = 1, b = -2t, c = t2 - 1

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

For roots to be real, b2 - 4ac > 0

4t2 - 4(t2 - 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 +2 (or) 4, f(2) 0 and f(4) 0 ---------- (2)

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

[4 - 4t + (t2 - 1)][16 - 8t + t2 - 1] < 0

(t2 - 4t + 3) (t2 - 8t + 15) < 0

(t - 1) (t - 3) (t - 3) (t - 5) < 0

(t3)2 (t - 1) (t - 5) < 0

t ϵ (1, 5)

(2) f(2) 0

t 1, 3

(3) f(4) 0 t 3, 5

So, t ϵ (1, 5) - {3}


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