The interval of a for which the equation tan2 x - (a - 4) tan x + 4 - 2a = 0 has at least one solution ∀x∈[0,π/4]

tan⁡x=a−4±(a−4)2−4(4−2a)2=a−4±a2=a−2,−2∴ tan⁡x=a−2 (∵tan⁡x≠−2)∵ x∈0,π4∴ 0≤a−2≤1⇒ 2≤a≤3

The interval of a for which the equation tan2 x - (a - 4) tan x + 4 - 2a = 0 has at least one solution ∀x∈[0,π/4]

tan⁡x=a−4±(a−4)2−4(4−2a)2=a−4±a2=a−2,−2∴ tan⁡x=a−2 (∵tan⁡x≠−2)∵ x∈0,π4∴ 0≤a−2≤1⇒ 2≤a≤3

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The interval of a for which the equation tan² x - (a - 4) tan x + 4 - 2a = 0 has at least one solution $\forall x \in [0, π / 4]$

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$a \in (2, 3)$

$a \in [2, 3]$

$a \in (1, 4)$

$a \in [1, 4]$

answer is B.

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Detailed Solution

$\begin{matrix} \tan x = \frac{a - 4 \pm \sqrt{(a - 4)^{2} - 4 (4 - 2 a)}}{2} \\ = \frac{a - 4 \pm a}{2} = a - 2, - 2 \end{matrix}$

$∴ \tan x = a - 2 (∵ \tan x \neq - 2)$

$\begin{array}{ll} ∵ x \in [0, \frac{π}{4}] \\ ∴ 0 \leq a - 2 \leq 1 \\ \Rightarrow 2 \leq a \leq 3 \end{array}$

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The interval of a for which the equation tan2 x - (a - 4) tan x + 4 - 2a = 0 has at least one solution ∀x∈[0,π/4]

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The interval of a for which the equation tan² x - (a - 4) tan x + 4 - 2a = 0 has at least one solution $\forall x \in [0, π / 4]$