Calculus – Continuity, Differentiation & Integration Quiz - ICSE Class 12 Mathematics
Free practice quiz: Calculus – Continuity, Differentiation & Integration (ICSE Class 12 Mathematics). 26 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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If $f(x) = \frac{\sin(x)}{x}$ for $x \neq 0$ and $f(0) = k$, for what value of $k$ is $f(x)$ continuous at $x=0$?
- 1
- 0
- undefined
- $\cos(0)$
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Answer: A. 1
For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x \to 0} f(x) = f(0)$. We know that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. Therefore, $k$ must be 1.
For the function $f(x) = \begin{cases} x+1 & \text{if } x \le 1 \ 3-ax^2 & \text{if } x > 1 \end{cases}$, find the value of $a$ that makes $f(x)$ continuous at $x=1$.
- 2
- 1/2
- 1
- 0
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Answer: C. 1
For continuity at $x=1$, we need $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$. $f(1) = 1+1=2$. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+1) = 1+1=2$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3-ax^2) = 3-a(1)^2 = 3-a$. For continuity, $2 = 3-a$, which gives $a=1$.
Consider the function $f(x) = \frac{x^2-9}{x-3}$ for $x \neq 3$. Is $f(x)$ continuous at $x=3$ if we define $f(3)=6$?
- Yes
- No
- Depends on the definition of f(3)
- Cannot be determined
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Answer: A. Yes
For continuity at $x=3$, we need $\lim_{x \to 3} f(x) = f(3)$. $\lim_{x \to 3} \frac{x^2-9}{x-3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 3+3=6$. Since $f(3)$ is defined as 6, which is equal to the limit, the function is continuous at $x=3$.
For the function $f(x) = x^n$, when is it continuous?
- Always
- Only for positive integers n
- Only for negative integers n
- Never
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Answer: A. Always
Polynomials are continuous everywhere. $f(x) = x^n$ is a polynomial for any real number $n$ (though the domain might be restricted for non-integer $n$). For the context of calculus up to Grade 12, $x^n$ is considered continuous over its domain.
Find the derivative of $f(x) = \sinh(x)$.
- $\cosh(x)$
- $\sinh(x)$
- $-\cosh(x)$
- $-\sinh(x)$
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Answer: A. $\cosh(x)$
The derivative of the hyperbolic sine function, $\sinh(x)$, is the hyperbolic cosine function, $\cosh(x)$.