JEE Main 2026 — Mathematics Algebra

The given series is $\alpha = 3+4+8+9+13+14+\ldots$α=3+4+8+9+13+14+… up to $40$40 terms.

We can split the series into two arithmetic progressions, each containing $20$20 terms:

$S_1 = 3 + 8 + 13 + \ldots$S1​=3+8+13+… up to $20$20 terms

$S_2 = 4 + 9 + 14 + \ldots$S2​=4+9+14+… up to $20$20 terms

Using the sum formula for an AP, $S_n = \dfrac{n}{2}[2a + (n-1)d]$Sn​=2n​[2a+(n−1)d]:

$$S_1 = \dfrac{20}{2}[2(3) + (20-1)5] = 10[6 + 95] = 1010$$S1​=220​[2(3)+(20−1)5]=10[6+95]=1010

$$S_2 = \dfrac{20}{2}[2(4) + (20-1)5] = 10[8 + 95] = 1030$$S2​=220​[2(4)+(20−1)5]=10[8+95]=1030

Therefore, $\alpha = S_1 + S_2 = 1010 + 1030 = 2040$α=S1​+S2​=1010+1030=2040.

The exponent in the given expression is $\dfrac{\alpha}{1020} = \dfrac{2040}{1020} = 2$1020α​=10202040​=2.

Thus, $(\tan\beta)^2 = \tan^2\beta$(tanβ)2=tan2β is a root of the equation $x^2+x-2=0$x2+x−2=0.

Solving $x^2+x-2=0$x2+x−2=0 gives $(x+2)(x-1)=0 \Rightarrow x = -2$(x+2)(x−1)=0⇒x=−2 or $x = 1$x=1.

Since $\tan^2\beta \ge 0$tan2β≥0, we must have $\tan^2\beta = 1$tan2β=1.

Given $\beta \in \left(0, \dfrac{\pi}{2}\right)$β∈(0,2π​), $\tan\beta = 1 \Rightarrow \beta = \dfrac{\pi}{4}$tanβ=1⇒β=4π​.

We need to find the value of $\sin^2\beta + 3\cos^2\beta$sin2β+3cos2β.

Substituting $\beta = \dfrac{\pi}{4}$β=4π​:

$\sin^2\left(\dfrac{\pi}{4}\right) + 3\cos^2\left(\dfrac{\pi}{4}\right) = \left(\dfrac{1}{\sqrt{2}}\right)^2 + 3\left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2} + \dfrac{3}{2} = 2$sin2(4π​)+3cos2(4π​)=(2​1​)2+3(2​1​)2=21​+23​=2.

Answer: $2$2