Let {f}(x)=(x+1)²-1, x ≥slant-1, then the set ≤ft\{x / f(x)=f^{-1}(x)\} is

Correct Answer is : {0,−1}\{0,-1\}{0,−1}

≤ft\{0,-1, {-3+{i} {3}}{2}, {-3-{i} {3}}{2}., where ≤ft.{i}={-1}\}

Correct Answer is : {0,−1}\{0,-1\}{0,−1}

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MHTCET 15 May 2024 Shift 1 Paper

Question : 126 of 150

Marks: +1, -0

Let

\mathrm{f}(x)=(x+1)^{2}-1, x \geqslant-1

\left\{x / f(x)=f^{-1}(x)\right\}

\{0, 1, -1\}
$\{0,-1\}$
$\left\{0,-1, \frac{-3+\mathrm{i} \sqrt{3}}{2}, \frac{-3-\mathrm{i} \sqrt{3}}{2}\right.$ , where $\left.\mathrm{i}=\sqrt{-1}\right\}$
$\phi$

Solution:

Concept:

For an invertible function,

f(x) = f^{-1}(x)

implies

f(f(x)) = x

So we solve

f(f(x)) = x

for real

x

in the given domain.

Explanation:

Given

f(x) = (x+1)^2 - 1

for

x \geq -1

Since

f

is strictly increasing on this domain,

f^{-1}

exists.

Now apply

f

on both sides of

f(x) = f^{-1}(x)

f(f(x)) = x

Compute

f(f(x))

f(f(x)) = (f(x)+1)^2 - 1

Since

f(x)+1 = (x+1)^2

, we get:

f(f(x)) = (x+1)^4 - 1

So,

(x+1)^4 - 1 = x

Rearrange:

(x+1)^4 - (x+1) = 0

Factor:

(x+1)\left[(x+1)^3 - 1\right] = 0

Thus,

x+1 = 0

(x+1)^3 = 1

x = -1

x+1 = 1 \Rightarrow x = 0

Both values satisfy

x \geq -1

, and complex roots are not considered since

x

is real.

Answer:

\{0, -1\}

Correct option: B.

\{0, -1\}

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