Let = 3 + 4 + 8 + 9 + 13 + 14 + ... uto 40 terms. If is a root of the equation , , then is equal to: [2026]

Question

Let $α$ = 3 + 4 + 8 + 9 + 13 + 14 + ... uto 40 terms. If ${(\tan β)}^{\frac{α}{1020}}$ is a root of the equation $x^{2} + x - 2 = 0$ , $β \in (0, \frac{π}{2})$ , then $\sin^{2} β + 3 \cos^{2} β$ is equal to: [2026]

Smart Achievers · Accepted Answer

(1)

$α = 3 + 4 + 8 + 9 + 13 + 14 + . . . (40 terms)$

$α = (3 + 8 + 13 + . . . (20 terms) + (4 + 9 + 14 + . . . (20 terms)$

$\Rightarrow α = \frac{20}{2} [6 + 19 \times 5] + \frac{20}{2} [8 + 19 \times 5]$

$\Rightarrow α = 10 [101 + 103] = 2040$

Now, is a root of the equation.

$\Rightarrow x^{2} + x - 2 = 0$

$\Rightarrow x^{2} + 2 x - x - 2 = 0$

$\Rightarrow x = - 2, 1$

$∴ \tan^{2} β = 1 \Rightarrow β = \frac{π}{4}$

So, $\sin^{2} β + 3 \cos^{2} β = \frac{1}{2} + \frac{3}{2} = 2$

Solution

Q.
Let $α$ = 3 + 4 + 8 + 9 + 13 + 14 + ... uto 40 terms. If ${(\tan β)}^{\frac{α}{1020}}$ is a root of the equation $x^{2} + x - 2 = 0$ , $β \in (0, \frac{π}{2})$ , then $\sin^{2} β + 3 \cos^{2} β$ is equal to: [2026]

1 2

2 $\frac{7}{4}$

3 $\frac{5}{2}$

4 $\frac{3}{2}$

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Solution

Q. Let α = 3 + 4 + 8 + 9 + 13 + 14 + ... uto 40 terms. If (tanβ)α1020 is a root of the equation x2+x–2=0, β∈(0,π2), then sin2β+3cos2β is equal to: [2026]

1 2

2 74

3 52

4 32

Ans. (1) α=3+4+8+9+13+14+...(40 terms) α=(3+8+13+...(20 terms)+(4+9+14+...(20 terms) ⇒ α=202[6+19×5]+202[8+19×5] ⇒ α=10[101+103]=2040 Now, is a root of the equation. ⇒ x2+x–2=0 ⇒ x2+2x–x–2=0 ⇒ x=–2,1 ∴ tan2β=1 ⇒β=π4 So, sin2β+3cos2β=12+32=2

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Q.
Let $α$ = 3 + 4 + 8 + 9 + 13 + 14 + ... uto 40 terms. If ${(\tan β)}^{\frac{α}{1020}}$ is a root of the equation $x^{2} + x - 2 = 0$ , $β \in (0, \frac{π}{2})$ , then $\sin^{2} β + 3 \cos^{2} β$ is equal to: [2026]

2 $\frac{7}{4}$

3 $\frac{5}{2}$

4 $\frac{3}{2}$