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  •  Find the values of m, for which exactly one root of the equation x2 - 2mx + m2 -1 = 0  lies in the interval (-2, 4).Op

 Find the values of m, for which exactly one root of the equation x2 - 2mx + m2 -1 = 0  lies in the interval (-2, 4).

Option: 1

m \in (-3, -1) U (3,5)


Option: 2

m\in(-1,3)


Option: 3

m\in(-\infty,-3)\cup (5,\infty)


Option: 4

m\in(-\infty,-3)\cup (-1,3)\cup (5,\infty)


Answers (1)

best_answer

As exactly one root lies in (-2,4), so one of af(-2) and af(4) will be positive and the other will be negative. Hence their product will be -ve, so\\\mathrm{f(-2)f(4) < 0 \Rightarrow \left ( (-2)^2 - 2m(-2) + m^2 -1 \right )\left ( (4)^2 - 2m(4) + m^2 -1 \right )}

\\\mathrm{=\left ( m^2 + 4m + 3 \right )\left ( m^2 - 8m + 15 \right ) <0}

Factorizing them further

(m+3)(m+1)(m-3)(m-5) < 0

So we have :

m \in (-3, -1) ? (3,5)

Also, D > 0, solving for this condition we get m in real.

  So m \in??????? (-3, -1) ? (3,5) this is the final solution.

Correct option is (a)

Posted by

manish

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