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

$fog(x)=\dfrac{3x^{2}-4x+2}{(x-1)^{2}}$, $gof(x)=1+\dfrac{1}{x^{2}+1}$; $fog(2)=6$, $gof(-3)=\dfrac{11}{10}$

CBSE Class 12 · 2014 · Relations And Functions XIILong Answer

If the function

f : R \to R

is given by

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

and

g : R \to R

is given by

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

x \neq = 1

, find

f o g

and

g o f

, and hence find

f o g (2)

and

g o f (- 3)

Answer: $f o g (x) = \frac{3 x ^{2} - 4 x + 2}{( x - 1 ) ^{2}}$ , $g o f (x) = 1 + \frac{1}{x ^{2} + 1}$ ; $f o g (2) = 6$ , $g o f (- 3) = \frac{11}{10}$

Solution

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

. Put over a common denominator:

f o g (x) = \frac{x ^{2} + 2 ( x - 1 ) ^{2}}{( x - 1 ) ^{2}} = \frac{3 x ^{2} - 4 x + 2}{( x - 1 ) ^{2}}

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

.
Now evaluate.

f o g (2) = \frac{3 ( 4 ) - 8 + 2}{( 2 - 1 ) ^{2}} = \frac{6}{1} = 6

g o f (- 3) = 1 + \frac{1}{( - 3 ) ^{2} + 1} = 1 + \frac{1}{10} = \frac{11}{10}

f o g (x) = \frac{3 x ^{2} - 4 x + 2}{( x - 1 ) ^{2}}, g o f (x) = 1 + \frac{1}{x ^{2} + 1}, f o g (2) = 6, g o f (- 3) = \frac{11}{10}

CBSE Class 12 Board · 2014 · Subjective

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CBSE Class 12 · 2014 · Relations And Functions XIILong Answer

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