(C) The given equation is $x^2 - 2mx + m^2 - 1 = 0$.This can be rewritten as $(x - m)^2 - 1 = 0$.$(x - m)^2 = 1$.Taking the square root on both sides,

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(C) The given equation is $x^2 - 2mx + m^2 - 1 = 0$.This can be rewritten as $(x - m)^2 - 1 = 0$.$(x - m)^2 = 1$.Taking the square root on both sides,we get $x - m = \pm 1$.Thus,the roots are $x_1 = m - 1$ and $x_2 = m + 1$.We are given that both roots are greater than $-2$ and less than $4$,so $-2 From $-2 -1$.From $m + 1 Combining these,we get $-1 Therefore,the interval is $(-1, 3)$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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All the values of $m$ for which both roots of the equation $x^2 - 2mx + m^2 - 1 = 0$ are greater than $-2$ but less than $4$ lie in the interval

AIEEE 2006 All AIEEE

A
$(-2, 0)$
B
$(3, \infty)$
C
$(-1, 3)$
D
$(1, 4)$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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All the values of $m$ for which both roots of the equation $x^2 - 2mx + m^2 - 1 = 0$ are greater than $-2$ but less than $4$ lie in the interval

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