JEE Main 2026 — Mathematics Algebra

Given $f: \{1, 2, 3, 4\} \rightarrow \{1, e, e^2, e^3\}$f:{1,2,3,4}→{1,e,e2,e3} is a strictly decreasing bijective function.

Arranging the domain and codomain in increasing order, we get $f(1) = e^3$f(1)=e3, $f(2) = e^2$f(2)=e2, $f(3) = e$f(3)=e, and $f(4) = 1$f(4)=1.

Given $g: \{1, e, e^2, e^3\} \rightarrow \left\{1, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}\right\}$g:{1,e,e2,e3}→{1,21​,31​,41​} is a strictly increasing bijective function.

Arranging the domain and codomain in increasing order, we get $g(1) = \dfrac{1}{4}$g(1)=41​, $g(e) = \dfrac{1}{3}$g(e)=31​, $g(e^2) = \dfrac{1}{2}$g(e2)=21​, and $g(e^3) = 1$g(e3)=1.

Evaluating $\phi(x) = \left[f^{-1}\left\{g^{-1}\left(\dfrac{1}{2}\right)\right\}\right]^x$ϕ(x)=[f−1{g−1(21​)}]x:

From the mapping of $g$g, $g^{-1}\left(\dfrac{1}{2}\right) = e^2$g−1(21​)=e2.

From the mapping of $f$f, $f^{-1}(e^2) = 2$f−1(e2)=2.

Substituting these values, we get $\phi(x) = 2^x$ϕ(x)=2x.

The region $R$R is given by $x^2 \leq y \leq 2^x$x2≤y≤2x for $0 \leq x \leq 1$0≤x≤1.

The area of the region $R$R is:

$$\int_{0}^{1} (2^x - x^2) dx = \left[ \dfrac{2^x}{\log_e(2)} - \dfrac{x^3}{3} \right]_{0}^{1}$$∫01​(2x−x2)dx=[loge​(2)2x​−3x3​]01​

$$= \left( \dfrac{2}{\log_e(2)} - \dfrac{1}{3} \right) - \left( \dfrac{1}{\log_e(2)} - 0 \right)$$=(loge​(2)2​−31​)−(loge​(2)1​−0)

$$= \dfrac{1}{\log_e(2)} - \dfrac{1}{3}$$=loge​(2)1​−31​

$$= \dfrac{3 - \log_e(2)}{3\log_e(2)}$$=3loge​(2)3−loge​(2)​

Answer: $\dfrac{3 - \log_e(2)}{3\log_e(2)}$3loge​(2)3−loge​(2)​