Q.5. Let upto 40 terms. If is a root of the equation , then is equal to :

Sol. $a = (3 + 8 + 13 ... \text{ upto 20 terms}) + (4 + 9 + 14 + ... \text{ upto 20 terms})$ $a = \frac{20}{2}[6+19\times 5] + \frac{20}{2}[8+19\times 5]$ $= 2040$ $x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=1 \text{ or } x=-2$ $(\tan\beta)^{1020} = 1 \Rightarrow \tan\beta = 1 \text{ (since } \beta \in (0, \frac{\pi}{2}) \text{, } \tan\beta \text{ must be positive)}$ $\Rightarrow \sin^2\beta = \frac{1}{2} \text{ & } \cos^2\beta = \frac{1}{2}$ $\sin^2\beta + 3\cos^2\beta = \frac{1}{2} + 3(\frac{1}{2}) = \frac{4}{2} = 2$

What is the correct answer for JEE Main Mathematics Q5?

The correct answer is A. Sol. $a = (3 + 8 + 13 ... \text{ upto 20 terms}) + (4 + 9 + 14 + ... \text{ upto 20 terms})$ $a = \frac{20}{2}[6+19\times 5] + \frac{20}{2}[8+19\times 5]$ $= 2040$ $x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=1 \text{ or } x=-2$ $(\tan\beta)^{1020} = 1 \Rightarrow \tan\beta = 1 \text{ (since } \beta \in (0, \frac{\pi}{2}) \text{, } \tan\beta \text{ must be positive)}$ $\Rightarrow \sin^

Q.5. Let upto 40 terms. If is a root of the equation , then is equal to :

The correct answer is A. Sol. $a = (3 + 8 + 13 ... \text{ upto 20 terms}) + (4 + 9 + 14 + ... \text{ upto 20 terms})$ $a = \frac{20}{2}[6+19\times 5] + \frac{20}{2}[8+19\times 5]$ $= 2040$ $x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=1 \text{ or } x=-2$ $(\tan\beta)^{1020} = 1 \Rightarrow \tan\beta = 1 \text{ (since } \beta \in (0, \frac{\pi}{2}) \text{, } \tan\beta \text{ must be positive)}$ $\Rightarrow \sin^

What are the options for this JEE Main Mathematics question?

A. 2 · B. 7/4 · C. 5/2 · D. 3/2

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Q.5. Let $a = 3+4+8+9+13+14+...$ upto 40 terms. If $(\tan\beta)^{1020}$ is a root of the equation $x^2 + x - 2 = 0, \beta \in (0, \frac{\pi}{2})$, then $\sin^2\beta + 3\cos^2\beta$ is equal to :

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