JEE Advanced 2022 — Paper 2 — MATH-I solution

JEE Advanced2022Paper 2MATH-I

The greatest integer less than or equal to $\int_{1}^{2} lo g_{2} (x^{3} + 1) d x + \int_{1}^{l o g_{2} 9} (2^{x} - 1)^{1/3} d x$ is ______.

Your answer

Solution

The two integrals are inverses of each other. Let $f (x) = lo g_{2} (x^{3} + 1)$ on $[1, 2]$ . Then $f (1) = 1$ and $f (2) = lo g_{2} 9$ .

The inverse $f^{- 1} (y)$ is obtained from $y = lo g_{2} (x^{3} + 1) \Rightarrow x = (2^{y} - 1)^{1/3}$ , defined on $[1, lo g_{2} 9]$ .

By the standard identity for a monotone function and its inverse, $\int_{a}^{b} f (x) d x + \int_{f (a)}^{f (b)} f^{- 1} (y) d y = b \cdot f (b) - a \cdot f (a)$ .

Here $a = 1$ , $b = 2$ , $f (a) = 1$ , $f (b) = lo g_{2} 9$ , so the sum equals $2 lo g_{2} 9 - 1 \cdot 1 = 2 lo g_{2} 9 - 1$ .

Numerically $2 lo g_{2} 9 \approx 6.34$ , so the value is approximately $5.34$ . The greatest integer not exceeding it is $5$ .

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