Which of the following are true? (A) A function f(x)=x^2 is increasin…

The correct answer is Option (2) → (B), (C) and (D) only

For \(f(x)=x^2\):

    \(f'(x)=2x \lt 0 \text{ for } -1 \le x \lt 0\)
    \(f'(x)=2x \gt 0 \text{ for } 0 \lt x \le 1\)

Differentiating \(x^n\) repeatedly, the power drops by 1 each time:

    \(f'(x)=nx^{n-1},\quad f''(x)=n(n-1)x^{n-2}\)
    \(f^{(n)}(x)=n(n-1)(n-2)\cdots 2\times 1=n!\)

Completing the square:

    \(x^2-2x+4=(x-1)^2+3 \ge 3\)

The value 3 is reached only at x = 1, and the function grows without bound on both sides, so it has no maximum.

For \(y=x^2\):

    \( \displaystyle \frac{dy}{dx}=2x=0 \text{ at } x=0 \)

So the tangent at (0, 0) is the horizontal line y = 0.

So \(x^2\) decreases on [−1, 0], the n^th derivative of \(x^n\) is n!, \(x^2-2x+4\) has exactly one global extreme value (the minimum 3 at x = 1), and the tangent to \(y=x^2\) at the origin is y = 0.

Checking the statements

(A) A function \(f(x)=x^2\) is increasing in \([-1, 1]\) — Incorrect. f’(x) = 2x < 0 on [−1, 0), e.g. f(−1) = 1 > f(0) = 0.

(B) If \(f(x)=x^n\), then n^th derivative is equal to n! — Correct. \(f^{(n)}(x)=n(n-1)\cdots 1=n!\).

(C) \(f(x)=x^2-2x+4\) has a single global maxima/minima — Correct. Its only global extreme value is the global minimum 3 at x = 1.

(D) x-axis is tangent on \(y=x^2\) — Correct. The slope at the origin is 0, so the tangent there is y = 0.