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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Definite Integrals

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)^{\frac{1}{3}} d x

is ____.

Answer: 5

Solution

Let

f (x) = lo g_{2} (x^{3} + 1)

. Then

y = lo g_{2} (x^{3} + 1)

implies

x^{3} + 1 = 2^{y}

, so

x = (2^{y} - 1)^{1/3} = f^{- 1} (y)

.

Using the property

\int_{a}^{b} f (x) d x + \int_{f (a)}^{f (b)} f^{- 1} (x) d x = b f (b) - a f (a)

:

Here

f (1) = lo g_{2} (2) = 1

and

f (2) = lo g_{2} (9)

.

So the sum equals:

2 lo g_{2} 9 - 1 \cdot 1 = 2 lo g_{2} 9 - 1

Since

8 < 9 < 2^{7/2}

, we have

3 < lo g_{2} 9 < 3.5

, giving

5 < 2 lo g_{2} 9 - 1 < 6

.

Therefore,

[2 lo g_{2} 9 - 1] = 5

.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer

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