JEE MAIN - Mathematics (2009 - No. 4)
Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge - 1$$
Statement - 1 : The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.
Statement - 2 : $$f$$ is a bijection.
Statement - 1 : The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.
Statement - 2 : $$f$$ is a bijection.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
Statement - 1 is true, Statement - 2 is false
Statement - 1 is false, Statement - 2 is true
Explanation
Given that $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,\,x \ge - 1$$
Clearly $${D_f} = \left[ { -1 ,\infty } \right)$$ but co-domain is not given
$$\therefore$$ $$f(x)$$ need not be necessarily onto.
But if $$f(x)$$ is onto then as $$f\left( x \right)$$ is one one also,
$$(x+1)$$ being something $$+ve,$$ $${f^{ - 1}}\left( x \right)$$ will exist where
$${\left( {x + 1} \right)^2} - 1 = y$$
$$ \Rightarrow x + 1 = \sqrt {y + 1} $$
$$\,\,\,\,\,$$ $$\left( { + ve} \right.$$ square root as $$x + 1 \ge 0$$$$\left. {} \right)$$
$$ \Rightarrow x = - 1 + \sqrt {y + 1} $$
$$ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1$$
Then $$\,\,\,\,\,\,\,\,$$ $$f\left( x \right) = {f^{ - 1}}\left( x \right)$$
$$ \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1$$
$$ \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} $$
$$ \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right)$$
$$ \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0$$
$$ \Rightarrow x = - 1,0$$
$$\therefore$$ The statement -$$1$$ is correct but statement- $$2$$ is false.
Clearly $${D_f} = \left[ { -1 ,\infty } \right)$$ but co-domain is not given
$$\therefore$$ $$f(x)$$ need not be necessarily onto.
But if $$f(x)$$ is onto then as $$f\left( x \right)$$ is one one also,
$$(x+1)$$ being something $$+ve,$$ $${f^{ - 1}}\left( x \right)$$ will exist where
$${\left( {x + 1} \right)^2} - 1 = y$$
$$ \Rightarrow x + 1 = \sqrt {y + 1} $$
$$\,\,\,\,\,$$ $$\left( { + ve} \right.$$ square root as $$x + 1 \ge 0$$$$\left. {} \right)$$
$$ \Rightarrow x = - 1 + \sqrt {y + 1} $$
$$ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1$$
Then $$\,\,\,\,\,\,\,\,$$ $$f\left( x \right) = {f^{ - 1}}\left( x \right)$$
$$ \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1$$
$$ \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} $$
$$ \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right)$$
$$ \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0$$
$$ \Rightarrow x = - 1,0$$
$$\therefore$$ The statement -$$1$$ is correct but statement- $$2$$ is false.
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