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The functions defined have domain R...

The functions defined have domain R

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The correct Answer is:
(A) `rarr` (r ); (B) `rarr` (q); (C ) `rarr` (p); (D) `rarr` (s)
`y=7x+1`
`f(x)=7x+1`

Let `x_(1), x_(2) in D_(f)`,
then `f(x_(1))=f(x_(2))`
`implies 7x_(1)+1=7x_(2)+1impliesx_(1)=x_(2)`
f is one-one, `AAx inR`
Now, `y=7x+1impliesx=(y-1)/(7)`
for each `y in R`, we get `x in R`
f is into function
(B) y = cos x
for `x in[0, pi], y in [-1,1]`
`therefore` f is one-one on `[0 pi]`, `AAx inR,yin[-1,1]`
y is not onto R.
(C ) `y = sin x or f(x) = sin x`
for `x in[0, pi], y in[0, 1]`
`f((pi)/(3))=(sqrt(3))/(2)andf((2pi)/(3))=(sqrt(3))/(2)`
`therefore` f is not one-one on `(0, pi)`, `AAx in R and y in [-1, 1]`
`therefore` f is onto [-1, 1].
(D) y = 1 + In x and f(x) = 1 + In x
y is defined for `x in (0, oo)`
Let `x_(1),x_(2)inD`
then `f(x_(1))=f(x_(2))`
`implies 1+Inx_(1)=1+Inx_(2)`
`implies x_(1)=x_(2)`
`therefore` f is one-one, `AAx in(0, oo)`
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