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a x = a - a y ,
Now , a y > 0 for every real " y " and for every a > 1 .
Hence ,
0 < a x ≤ 1
So , - ∞ < x ≤ 0 .
So the answer given is wrong actually .
*edited*sorry- title is a misnomer
1.The domain of defination of function y(x) given by the equation ax+ay=a(a>1)is..?
Ans.[0. 1)
2.g(x)=1+x1/3..then a function such that f(g(x))=3-3√x+x is ..?
Ans.x3-3x2+2x+3
3.The function f(x)=x/(ex-1)+x/2+1 is..
a.even
b.odd
c.periodic
d.neither even nor odd
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9 Answers
1
39But what if we do it like this Ricky..
ay > 0
=> -ay< 0
=> ax - a < 0
=> ax < a
Now on both sides, take log to the base a.
=> x < 1
There must be some other restriction which brings about x ≥ 0, though I have no idea.
And for Q3, neither even nor odd? It can't be periodic, xex - 1 is not a periodic function, is it..
11y=3-\sqrt[3]x+x\Rightarrow g(x)=4-y+x
Thus we have f\left\{g(x) \right\}=-g(x)+4+\left\{g(x)-1 \right\}^3
Replace g(x) by x to have f(x)=x^3-3x^2+2x+3.
62\\f(x)=\frac{x}{e^x-1}+\frac{x}{2}+1 \\f(-x)=\frac{-x}{e^{-x}-1}+\frac{-x}{2}+1 \\f(-x)=\frac{-x}{e^{-x}-1}+\frac{x}{2}-x+1 \\f(-x)=\frac{-xe^x}{1-e^x}+\frac{x}{2}-x+1 \\f(-x)=\frac{-xe^x}{1-e^x}-x+\frac{x}{2}+1 \\f(-x)=-x\left[\frac{-e^x}{1-e^x}-1\right]+\frac{x}{2}+1 \\f(-x)=-x\left[\frac{-1}{1-e^x}\right]+\frac{x}{2}+1 \\f(x)=\frac{x}{e^x-1}+\frac{x}{2}+1
Hence even
62\\f(g(x))=3-\sqrt[3]{x}+x=3-\sqrt[3]{x}-1+\left[\sqrt[3]{x}\right]^3=3-g(x)+\left[\sqrt[3]{x}+1-1\right]^3\\=3-g(x)+\left[g(x)-1\right]^3=g^3(x)-3g^2(x)+2g(x)+2\\f(x)=x^3-3x^2+2x+2