Mathematics

Functions and Laplace Transforms

59 Questions

Functions and Laplace transforms involve mapping mathematical domains and ranges to solve complex differential equations. Questions require finding the Laplace transform of exponential, hyperbolic, and trigonometric functions. These advanced topics are typically assessed in engineering and civil services aptitude tests.

Laplace transform exponentialFunction domain rangeHyperbolic function transformsLinear transformation dimensionOptimization functionals

Functions and Laplace Transforms Questions

Multiple choice general knowledge math & puzzles

Let the closed interval [a , b] be the domain of function f. The domain of f(x - 3) is given by

  1. the open interval (a , b)

  2. the closed interval [a , b]

  3. the closed interval [a - 3 , b - 3]

  4. the closed interval [a + 3 , b + 3]
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The function $f(x-3)$ is a horizontal shift of $f(x)$ to the right by 3 units. If the original domain is $[a, b]$, the new domain is found by solving $a \le x - 3 \le b$, which simplifies to $a + 3 \le x \le b + 3$.

Multiple choice general knowledge math & puzzles

The range of f(x) = x2 -2x + 1 is given by the interval

  1. [0 , +infinity)

  2. (-infinity , 1]

  3. (0 , + infinity)

  4. (1 , + infinity)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function f(x) = x² - 2x + 1 can be written as f(x) = (x-1)² by completing the square. Since any squared quantity is always non-negative, (x-1)² ≥ 0 for all real x. Therefore, the range of f(x) is [0, +∞), meaning all values from 0 to positive infinity, including 0 (when x = 1). Option C is incorrect because it excludes 0, which is achieved at x = 1. Option B incorrectly suggests the function is bounded above, but parabolas opening upward are unbounded. Option D excludes 0 and incorrectly shifts the starting point.

Multiple choice

Which one of the following functions is strictly bounded?
  1. $\frac{1}{x^2}$

  2. ex

  3. x2
  4. $e^{-x^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{We have} \hspace{1cm} lim_{x \rightarrow 0} \frac{1}{x^2} = \infty \\ \hspace{2.7cm} lim_{x \rightarrow \infty} x^2 = \infty \\ \hspace{2.7cm} lim_{x \rightarrow \infty} e^{-x} = \infty \\ \hspace{2.7cm} lim_{x \rightarrow \infty} e^{-x^2} = 0 \\ \hspace{2.7cm} lim_{x \rightarrow 0} e^{-x^2} = 1 \hspace{1cm} \text{Thus $e^{-x^2}$ is strictly bounded.}$

Multiple choice

Laplace transform for the function f(x) = cosh(ax) is
  1. $\frac{a}{s^2 - a^2}$
  2. $\frac{s}{s^2 - a^2}$
  3. $\frac{a}{s^2 + a^2}$
  4. $\frac{s}{s^2 + a^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Laplace transform of cosh(at) is derived from its exponential definition, yielding s / (s^2 - a^2). The transform with a plus sign in the denominator corresponds to cos(at).

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

If $f(x)=\left | \sin x \right |$, then domain of $f$ for the existence of inverse is  

  1. $[0,\pi ]$
  2. $\left [ 0,\dfrac{\pi }{2} \right ]$
  3. $\left [ -\dfrac{\pi }{4},\dfrac{\pi }{4} \right ]$
  4. $\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $-1 \leq \sin x \leq 1$ for $x \in \left [ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] $. 

For $| \sin x |$ to be invertible, the function has to be one-to-one. 
Thus, we need unique values of $x$ that give unique values of $f$ and vice versa.
For $x \in \left [0, \dfrac{\pi}{2} \right]$, $0 \leq \sin x \leq 1 \Rightarrow  0 \leq | \sin x \leq 1$
For $x \in \left [-\dfrac{\pi}{2},0 \right]$, $-1 \leq \sin x \leq 0 \Rightarrow  0 \leq | \sin x \leq 1$.
So, we have
$ \left [0, \dfrac{\pi}{2} \right] \rightarrow\left [0, 1 \right]$
$ \left [- \dfrac{\pi}{2},0 \right] \rightarrow\left [0, 1 \right]$
Since both the domains of $|\sin x|$ map to$\left [0, 1 \right]$, we consider only one of them for $x$ to be unique. 
Here, according to the options, the domain of $f$ must be$\left [0, \dfrac{\pi}{2} \right]$.

Multiple choice reciprocal equations theory of equations maths

The domain of reciprocal equation is :

  1. $R$
  2. $R-\{0\}$
  3. $Q$
  4. $R^+$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that in reciprocal equation, if one value of $x$ is $a$ then the other value of $x$  will be $\dfrac{1}{a}$

But, if $a = 0$ then $\dfrac{1}{a}$ does not exist  

Therefore, $0$ can not be in the domain of a reciprocal equation.
So, domain of reciprocal equation will be set of all real numbers excluding $0$.
Hence, option B is correct.

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

Let $f$ be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false :
$f (x) = 1, f (y) \sqrt 1, f (z) \sqrt 2$. The value of $f^{-1} (1)$ is

  1. x

  2. y

  3. z

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$f(x)=1,\quad f(y)\neq 1,\quad f(z)\neq 1$

Case 1:
$f(x)=1\ f(z)=2\ f(y)=1$
$\therefore f $ is not injective
Case 2: $f(y)\neq 1,\quad f(z)=2,\quad f(x)=1$
Case 3:
$f(z)\neq 2\quad \quad \quad f(z)=3\ f(x)\neq 1\quad \quad \quad f(x)=2\ f(y)=1\quad \quad \quad f(y)=1\ f(x)=2,f(y)=1,f(z)=3\ f^{ -1 }\left( 2 \right) =x,f^{ -1 }\left( 1 \right) =y,f^{ -1 }\left( 3 \right) =z$ 

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

The domain of the derivative of the function
$\displaystyle f\left ( x \right )=\begin{cases}
\tan^{-1}x & \text{ if } \left | x \right |\leq 1 \
\frac{1}{2}\left ( \left | x \right |-1 \right ) & \text{ if } \left | x \right |> 1
\end{cases}$

  1. $R\sim \left \{ 0 \right \}$
  2. $R\sim \left \{ 1 \right \}$
  3. $R\sim \left \{ -1 \right \}$
  4. $R\sim \left \{ -1, 1 \right \}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have
$f\left ( x \right )=\begin{cases}
\left ( 1/2 \right )\left ( -x-1 \right ) & \text{ if } x< -1 \
\tan^{-1}x & \text{ if } -1\leq x\leq 1 \
\left ( 1/2 \right )\left ( x-1 \right ) & \text{ if } x> 1
\end{cases}$
Since $\displaystyle f\left ( -1 \right )=-\frac{\pi }{4}$, $\displaystyle f\left ( 1 \right )=\frac{\pi }{4}$ and $\lim _{x\rightarrow 1-}f\left ( x \right )=-1$, $\lim _{x\rightarrow 1+}f\left ( x \right )=0$
so f is not continuous at -1, 1, hence not differentiable at -1, 1. Also
$\displaystyle {f}'\left ( x \right )=\begin{cases}
-1/2 & \text{ if } x< -1 \
\frac{1}{1+x^{2}} & \text{ if } -1< x< 1 \
1/2 & \text{ if } x> 1
\end{cases}$
Thus the domain of ${f}'$ is $R - \left { -1, 1 \right }$.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $f(x) = a{x^7} + b{x^3} + cx - 5 \,\,\,\,\,a,b,c$ are real constants and $f( - 7) = 7$ then the range of $f(7) + 17\cos x$ is

  1. $\left[ { - 34,0} \right]$
  2. $\left[ {0,34} \right]$
  3. $\left[ { - 34,34} \right]$
  4. $\left[ {34,\infty } \right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$f(x)=ax^7+6x^3+cx-5$

$f(-7)=7$

$f(7)+f(-7)=-10$

$f(7)+7=-10$

$f(7)=-17$

Range of $-17+17\cos x$ is form $[-34,\ 0]$