If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

\([x + 1] = (x+1)^2 - 1 = x^2 + 2x + 1 -1\)

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

Answer is E.

We can solve it by plugging in numbers too. Say x = 2, then x + 1 = 3

\([3] - [2] = (3^2 - 1) - (2^2 - 1) = 5\)

Put x = 2 in the options. Only (E) satisfies.

[x] = x^2 – 1

=> [x + 1] - [x] = \((x+1)^2 - 1 - ( x^2 - 1)\) = \(x^2 + 2x + 1 -1 - x^2 + 1\) = 2x + 1

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

­Straightforward function- keep each expression in parentheses to be safe:

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Step 1: Use the definition of \([x]\)
The problem defines \([x]\) as:

\[
[x] = x^2 - 1
\]

This tells you that whenever you see \([x]\), you can replace it with \(x^2 - 1\).

Step 2: Apply the function to \([x + 1]\)
Now you need to apply the same definition to \([x + 1]\). This means, wherever you have \(x + 1\), you replace it into the function \(x^2 - 1\):

\[
[x + 1] = (x + 1)^2 - 1
\]

Now, expand \((x + 1)^2\):

\[
(x + 1)^2 = x^2 + 2x + 1
\]

So:

\[
[x + 1] = (x^2 + 2x + 1) - 1 = x^2 + 2x
\]

Step 3: Now subtract \([x]\) from \([x + 1]\)

We already know from the definition that:

\[
[x] = x^2 - 1
\]

So now, subtract \([x]\) from \([x + 1]\):

\[
[x + 1] - [x] = (x^2 + 2x) - (x^2 - 1)
\]

Step 4: Simplify

Now, simplify this expression:

\[
x^2 + 2x - x^2 + 1 = 2x + 1
\]

Final Result:

The result of \([x + 1] - [x]\) is:

\[
2x + 1
\]