Question

Let $f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \geqslant - 1$
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.
$\left( a \right)$ Statement 1 is true, statement 2 is true. Statement 2 is a correct explanation for statement 1.
$\left( b \right)$ Statement 1 is true, statement 2 is true. Statement 2 is not a correct explanation for statement 1.
$\left( c \right)$ Statement 1 is true, statement 2 is false.
$\left( d \right)$ Statement 1 is false, statement 2 is true.

Answer 1

Hint: In this particular question use the concept that ${f^{ - 1}}\left( x \right)$ only exist if the function is bijection, and use the concept to find the inverse of the function find the value of x in terms of f (x) then replace x to ${f^{ - 1}}\left( x \right)$ and f (x) to x, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given function:
$f\left( x \right) = {\left( {x + 1} \right)^2} - 1$................. (1), $x \geqslant - 1$
Range of equation (1) is, $\left[ { - 1,\infty } \right)$
Now find the inverse of the function.
So first find out the value of x in terms of f (x) we have,
$ \Rightarrow {\left( {x + 1} \right)^2} = f\left( x \right) + 1$
Now take square root on both sides we have,
$ \Rightarrow \left( {x + 1} \right) = \sqrt {f\left( x \right) + 1} $
$ \Rightarrow x = \sqrt {f\left( x \right) + 1} - 1$
Now replace x with ${f^{ - 1}}\left( x \right)$ and f (x) with x we have,
$ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1$................. (2)
So range of equation (2) is $\left[ { - 1,\infty } \right)$
Statement – 1: The set $\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$
So equate equation (1) and (2) we have,
$ \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1$
Now simplify we have,
$ \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} $
Now squatting on both sides we have,
$ \Rightarrow {\left( {x + 1} \right)^4} = x + 1$
$ \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0$
\[ \Rightarrow x = - 1\]
And
$ \Rightarrow {\left( {x + 1} \right)^3} - 1 = 0$
Now as we know that $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ so we have,
$ \Rightarrow \left( {x + 1 - 1} \right)\left[ {{{\left( {x + 1} \right)}^2} + 1 + 1\left( {x + 1} \right)} \right] = 0$
$ \Rightarrow x\left[ {{x^2} + 1 + 2x + x + 2} \right] = 0$
$ \Rightarrow x\left[ {{x^2} + 3x + 3} \right] = 0$
$ \Rightarrow x = 0$, ${x^2} + 3x + 3 = 0$
The determinant of the quadratic equation is $D = \sqrt {{b^2} - 4ac} $, where, a = 1, b = 3, c = 3.
Therefore, $D = \sqrt {9 - 12} = \sqrt { - 3} $ = complex, so the roots of the quadratic equation are complex.
Hence, $x \in \left\{ {0, - 1} \right\}$
Therefore, $\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$
So statement 1 is true.
Statement – 2: f is a bijection.
As we see that the range of both the equations (1) and (2) are same and the inverse of $f\left( x \right)$ exists so $f\left( x \right)$ is said to be a bijection function.
And because of this statement 1 exists.
So statement 2 is correct and also a correct explanation of statement 1.
So this is the required answer.
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that a quadratic equation has no real roots if the discriminant D of the quadratic equation is less than zero, and always recall the basic standard identity i.e. $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$.