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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Functions

NDA 2019 · Mathematics · Functions

If f(x)=3^ 1+x , then f(x) f(y) f(z) is equal to

If f(x)=31+xf(x)=3^{1+x}, then f(x)f(y)f(z)f(x) f(y) f(z) is equal to

Options

  1. Af(x+y+z)f(x+y+z)
  2. Bf(x+y+z+1)f(x+y+z+1)
  3. Cf(x+y+z+2)f(x+y+z+2)
  4. Df(x+y+z+3)f(x+y+z+3)

Correct answer

C. f(x+y+z+2)f(x+y+z+2)

Step-by-step solution

f(x)=31+xf(x)=3^{1+x}f(x)⋅f(y)⋅f(z)=31+x⋅31+y⋅31+z=31+x+1+y+1+z=33+x+y+z=31+x+y+z+2=f(x+y+z+2)\begin{aligned} & \mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \cdot \mathrm{f}(\mathrm{z})=3^{1+\mathrm{x}} \cdot 3^{1+\mathrm{y}} \cdot 3^{1+z} \\ & =3^{1+x+1+y+1+z} \\ & =3^{3+x+y+z} \\ & =3^{1+x+y+z+2} \\ & =\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z}+2)\end{aligned}$$
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