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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Functions

NDA 2016 · Mathematics · Functions

Let f(x) be the greatest integer function and g(x) be the modulus...

Let f(x)f(x) be the greatest integer function and g(x)g(x) be the modulus function. What is (f∘f)(−95)+(g∘g)(−2)(f {}^{\circ} f)\left(-\frac{9}{5}\right)+(g {}^{\circ} g)(-2) equal to?

Options

  1. A–1
  2. B0
  3. C1
  4. D2

Correct answer

B. 0

Step-by-step solution

f(x)→f(x) \rightarrow greatest integer function f(x)=[x]g(x)→f(x)=[x]g(x) \rightarrow modulus function g(x)=∣x∣g(x)=|x|f∘f(x)=f(f(x))=f([x])f∘f(−95)=f(−2)∵[−95]=−2\begin{aligned} & f {}^{\circ} f(x)=f(f(x))=f([x]) \\ & f {}^{\circ} f\left(-\frac{9}{5}\right)=f(-2) \quad \because\left[-\frac{9}{5}\right]=-2 \end{aligned}=[−2]=−2gog⁡(x)=g(∣x∣)=∥x∥=∣x∣=[-2]=-2\operatorname{gog}(x)=g(|x|)=\|x\|=|x|gog⁡(−2)=∣−2∣=2(f∘f)(−95)+gog⁡(−2)=−2+2=0\begin{aligned} & \operatorname{gog}(-2)=|-2|=2 \\ & (f {}^{\circ} f)\left(-\frac{9}{5}\right)+\operatorname{gog}(-2)=-2+2=0\end{aligned}$$
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