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To determine the correct statement about the function $f(x) = -2x^8$, we can analyze its first derivative to find critical points and intervals of increase/decrease. 1. **Calculate the first derivative:** $$f'(x) = \frac{d}{dx}(-2x^8) = -16x^7$$ 2. **Find critical points:** Set $f'(x) = 0$: $$-16x^7 = 0 \implies x^7 = 0 \implies x = 0$$ The only critical point is $x=0$. 3. **Analyze the sign of $f'(x)$:** * For $x < 0$, $x^7 < 0$, so $f'(x) = -16(\text{negative number}) > 0$. This means $f(x)$ is increasing for $x < 0$. * For $x > 0$, $x^7 > 0$, so $f'(x) = -16(\text{positive number}) < 0$. This means $f(x)$ is decreasing for $x > 0$. 4. **Identify the extremum:** Since $f'(x)$ changes from positive to negative at $x=0$, there is a local maximum at $x=0$. The value of the function at this point is $f(0) = -2(0)^8 = 0$. Furthermore, for any $x \neq 0$, $x^8 > 0$, so $-2x^8 < 0$. Since $f(0) = 0$ and $f(x) < 0$ for all other $x$, $(0,0)$ is a global maximum. The correct statement is: The function $f(x)$ has a global maximum at $x=0$.

Every differentiable function is continuous

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for x=1, LHL=1 and RHL=5

Given \(y=\sin^{-1}x\) with \(-1 \le x \le 0\). For the principal branch of the inverse sine function, \[ -\frac{\pi}{2} \le \sin^{-1}x \le \frac{\pi}{2}. \] At the endpoints: \[ x=-1 \;\Rightarrow\; y=\sin^{-1}(-1)=-\frac{\pi}{2}, \] \[ x=0 \;\Rightarrow\; y=\sin^{-1}(0)=0. \] Since \(\sin^{-1}x\) is increasing on \([-1,1]\), the range is \[ \boxed{-\frac{\pi}{2} \le y \le 0}. \]

Question: If f : N → W is defined by f(n) = { n/2, if n is even; 0, if n is odd }, then f is:
A. injective only   B. surjective only   C. a bijection   D. neither surjective nor injective

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Solution:
Injective (one–one): By definition, f is injective if f(x₁)=f(x₂) ⇒ x₁=x₂. Here, f(1)=0, f(3)=0, f(5)=0. Different natural numbers have the same image, hence f is not injective.

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Surjective (onto): A function f : N → W is surjective if for every y ∈ W, there exists n ∈ N such that f(n)=y. For y=0, choose any odd n. For y≥1, choose n=2y ⇒ f(2y)=y. Thus every element of W has a preimage, so f is surjective.

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Conclusion: The function is surjective but not injective. Correct option: B

Question: A function f : R₊ → R (where R₊ is the set of all non-negative real numbers) is defined by f(x) = 4x + 3. Then f is:
(A) one–one but not onto (B) onto but not one–one (C) both one–one and onto (D) neither one–one nor onto

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Solution:
One–one: Let f(x₁) = f(x₂). Then 4x₁ + 3 = 4x₂ + 3 ⇒ x₁ = x₂. Hence, f is one–one.

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Onto: For f to be onto, for every y ∈ R there must exist x ∈ R₊ such that y = 4x + 3 ⇒ x = (y − 3)/4. Since x ≥ 0, we must have y ≥ 3. Thus, values y < 3 are not obtained. Hence, f is not onto R.

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Conclusion: The function is one–one but not onto.
Correct option: (A)