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Byju's Answer

Standard XII

Mathematics

Binary Operation

If fx=x+1 a...

Question

If $f (x) = x + 1$ and $g (x) = x^{2}$ Find
$F o g (1)$
$G o f (- 1)$
$F o g (2)$
$G o f (3)$

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Solution

$f (x) = x + 1$
$g (x) = x^{2}$
Now,
$F o g (x) = F (x^{2}) = x^{2} + 1$
Therefore,
$F o g (1) = 1^{2} + 1 = 2$
$F o g (2) = 2^{2} + 1 = 5$
Also,
$G o f (x) = G {(x + 1)}^{2} = {(x + 1)}^{2}$
Therefore,
$G o f (- 1) = {(- 1 + 1)}^{2} = 0$
$G o f (- 1) = {(3 + 1)}^{2} = 4^{2} = 16$

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Similar questions

Q. If the function

f : R \to R

be given by

f (x) = x^{2} + 2

and

g : R \to R

be given by

g (x) = \frac{x}{x - 1}, x \neq 1

, find fog and gof and hence find

f o g (2)

and

g o f (- 3)

If the function $f : R \to R$ be given by $f (x) = x^{2} + 2$ and $g : R \to R$ be given by $g (x) = \frac{x}{x - 1}, x \neq 1$ ,
find fog and gof and hence fing fog(2) and gof(-3).

Q. If

f (x) = 3 x - 2

and

g (x) = \frac{x + 1}{2}

, then

g o f (3) =

Q. Find fog and gof if
(i)

f (x) = e^{x}, g (x) = \log_{e} x

(ii)

f (x) = x^{2}, g (x) = \cos x

(iii)

f (x) = | x |, g (x) = \sin x

(iv)

f (x) = x + 1, g (x) = e^{x}

(v)

f (x) = \sin^{- 1} x, g (x) = x^{2}

(vi)

f (x) = x + 1, g (x) = \sin x

(vii)

f (x) = x + 1, g (x) = 2 x + 3

(viii)

f (x) = c, c \in R, g (x) = \sin x^{2}

(ix)

f (x) = x^{2} + 2, g (x) = 1 - \frac{1}{1 - x}

Q. Let f(x)= [x] and g(x)=|x| then find (gof) (-1/3) - (fog) (1/3) .

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If f(x)=x+1 and g(x)=x2 Find Fog (1)Gof (−1)Fog (2)Gof (3)

f(x)=x+1g(x)=x2Now,Fog(x)=F(x2)=x2+1Therefore,Fog(1)=12+1=2Fog(2)=22+1=5Also,Gof(x)=G(x+1)2=(x+1)2Therefore,Gof(−1)=(−1+1)2=0Gof(−1)=(3+1)2=42=16

If $f (x) = x + 1$ and $g (x) = x^{2}$ Find
$F o g (1)$
$G o f (- 1)$
$F o g (2)$
$G o f (3)$

$f (x) = x + 1$
$g (x) = x^{2}$
Now,
$F o g (x) = F (x^{2}) = x^{2} + 1$
Therefore,
$F o g (1) = 1^{2} + 1 = 2$
$F o g (2) = 2^{2} + 1 = 5$
Also,
$G o f (x) = G {(x + 1)}^{2} = {(x + 1)}^{2}$
Therefore,
$G o f (- 1) = {(- 1 + 1)}^{2} = 0$
$G o f (- 1) = {(3 + 1)}^{2} = 4^{2} = 16$