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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][k] and {k}\{k\} denote the greatest integer and fractional part of kk. A real-valued function ff is defined for all real x≠−1x \neq -1 by
f(x)=([log⁡e(1+{x})]∫0x2+1sin⁡(et) dt+2026−x5x5+1)1/5.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
f−1(200)−f(200)+f(f(2026))f(f(2)).\frac{f^{-1}(200) - f(200) + f(f(2026))}{f(f(2))}.
Solution
Answer: 1013
Step 1: Evaluate the floor coefficient. For every real xx, 0≤{x}<10 \le \{x\} < 1, so
1≤1+{x}<2⇒0≤log⁡e(1+{x})<log⁡e2<1⇒[log⁡e(1+{x})]=0.1 \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 00, so that term vanishes for every xx and need not be evaluated. Step 2: Reduced form.
f(x)=(2026−x5x5+1)1/5.f(x) = \left(\frac{2026 - x^5}{x^5 + 1}\right)^{1/5}.
 Step 3: Show ff is self-inverse. Put y=f(x)y = f(x) and raise to the fifth power.
y5=2026−x5x5+1⇒y5(x5+1)=2026−x5⇒x5(y5+1)=2026−y5,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,
⇒x=(2026−y5y5+1)1/5=f(y).\Rightarrow x = \left(\frac{2026 - y^5}{y^5 + 1}\right)^{1/5} = f(y).
 Hence f−1=ff^{-1} = f, giving f−1(x)=f(x)f^{-1}(x) = f(x) and f(f(x))=xf(f(x)) = x for all xx in the domain. Step 4: Evaluate. The first two terms cancel since f−1(200)=f(200)f^{-1}(200) = f(200), and each composite collapses by f(f(x))=xf(f(x)) = x:
f−1(200)−f(200)+f(f(2026))f(f(2))=0+20262=1013.\frac{f^{-1}(200) - f(200) + f(f(2026))}{f(f(2))} = \frac{0 + 2026}{2} = 1013.
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