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Functions: Let Denote Greatest Integer Fractional Part Real Valued

JEE Maths question with a full step-by-step solution.

Question

Let

[k]

and

\{k\}

denote the greatest integer and fractional part of

k

. A real-valued function

f

is defined for all real

x \neq -1

f(x) = \left(\left[\log_e(1 + \{x\})\right]\int_{0}^{x^2+1}\sin(e^t)\,dt + \frac{2026 - x^5}{x^5 + 1}\right)^{1/5}.

Find the value of

\frac{f^{-1}(200) - f(200) + f(f(2026))}{f(f(2))}.

Solution

Answer: 1013

Step 1: Evaluate the floor coefficient. For every real

x

0 \le \{x\} < 1

, so

1 \le 1 + \{x\} < 2 \Rightarrow 0 \le \log_e(1 + \{x\}) < \log_e 2 < 1 \Rightarrow \left[\log_e(1 + \{x\})\right] = 0.

The integral is multiplied by

0

, so that term vanishes for every

x

and need not be evaluated. Step 2: Reduced form.

f(x) = \left(\frac{2026 - x^5}{x^5 + 1}\right)^{1/5}.

Step 3: Show

f

is self-inverse. Put

y = f(x)

and raise to the fifth power.

y^5 = \frac{2026 - x^5}{x^5 + 1} \Rightarrow y^5(x^5 + 1) = 2026 - x^5 \Rightarrow x^5(y^5 + 1) = 2026 - y^5,

\Rightarrow x = \left(\frac{2026 - y^5}{y^5 + 1}\right)^{1/5} = f(y).

Hence

f^{-1} = f

, giving

f^{-1}(x) = f(x)

and

f(f(x)) = x

for all

x

in the domain. Step 4: Evaluate. The first two terms cancel since

f^{-1}(200) = f(200)

, and each composite collapses by

f(f(x)) = x

\frac{f^{-1}(200) - f(200) + f(f(2026))}{f(f(2))} = \frac{0 + 2026}{2} = 1013.

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