JEE Main 2026 — Mathematics Calculus

Since $(2^{1-a} + 2^{1+a})$(21−a+21+a), $f(a)$f(a), and $(3^a + 3^{-a})$(3a+3−a) are in A.P., we have:

$$2f(a) = 2^{1-a} + 2^{1+a} + 3^a + 3^{-a}$$2f(a)=21−a+21+a+3a+3−a

$$f(a) = (2^{-a} + 2^a) + \dfrac{1}{2}(3^a + 3^{-a})$$f(a)=(2−a+2a)+21​(3a+3−a)

Using the AM-GM inequality, $x + \dfrac{1}{x} \ge 2$x+x1​≥2 for $x > 0$x>0.

$2^a + 2^{-a} \ge 2$2a+2−a≥2 and $3^a + 3^{-a} \ge 2$3a+3−a≥2.

The minimum value of $f(a)$f(a) occurs when $a = 0$a=0.

$\alpha = f(0) = (2^0 + 2^0) + \dfrac{1}{2}(3^0 + 3^0) = 2 + 1 = 3$α=f(0)=(20+20)+21​(30+30)=2+1=3.

Now, we evaluate the integral:

$$I = \int_{\log_e 2}^{\log_e 3} \dfrac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e 2}^{\log_e 3} \dfrac{e^{2x}}{e^{4x} - 1} dx$$I=∫loge​2loge​3​e2x−e−2xdx​=∫loge​2loge​3​e4x−1e2x​dx

Let $e^{2x} = t \Rightarrow 2e^{2x} dx = dt$e2x=t⇒2e2xdx=dt.

When $x = \log_e 2$x=loge​2, $t = e^{2\log_e 2} = 4$t=e2loge​2=4.

When $x = \log_e 3$x=loge​3, $t = e^{2\log_e 3} = 9$t=e2loge​3=9.

$$I = \dfrac{1}{2} \int_{4}^{9} \dfrac{dt}{t^2 - 1}$$I=21​∫49​t2−1dt​

$$I = \dfrac{1}{2} \left[ \dfrac{1}{2} \log_e \left| \dfrac{t-1}{t+1} \right| \right]_{4}^{9}$$I=21​[21​loge​​t+1t−1​​]49​

$$I = \dfrac{1}{4} \left( \log_e \dfrac{8}{10} - \log_e \dfrac{3}{5} \right)$$I=41​(loge​108​−loge​53​)

$$I = \dfrac{1}{4} \left( \log_e \dfrac{4}{5} - \log_e \dfrac{3}{5} \right) = \dfrac{1}{4} \log_e \left( \dfrac{4/5}{3/5} \right) = \dfrac{1}{4} \log_e \left( \dfrac{4}{3} \right)$$I=41​(loge​54​−loge​53​)=41​loge​(3/54/5​)=41​loge​(34​)

Answer: $\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)$41​loge​(34​)