"Let m and M be respectively the minimum and maximum values of $$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$ Then the ordered pair (m, M) is equal to :"

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"$$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$

R₁ $\to$ R₁ – R₂, R₂ $\to$ R₂ – R₃

$$\left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$

= –1(sin² x) – 1(1 + sin2x + cos² x)

= - sin2x - 2

$\therefore$ minimum value when sin2x = 1

m = - 2 - 1 = -3

$\therefore$ Maximum value when sin2x = –1

M = -2 + 1 = -1

$\therefore$ (m, M) = (–3, –1)"

Let m and M be respectively the minimum and maximum values of

$$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$

Then the ordered pair (m, M) is equal to :

Solution

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Let m and M be respectively the minimum and maximum values of $$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$ Then the ordered pair (m, M) is equal to :

Solution

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

Let m and M be respectively the minimum and maximum values of

$$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$

Then the ordered pair (m, M) is equal to :