Functions (VSAQs)

Maths-1A | 1. Functions – VSAQs:
Welcome to VSAQs in Chapter 1: Functions. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : If f : R → R, g : R → R are defined by f(x) = 3x – 1 and g(x) = x² + 1, then find (fog)(2)

Step-1 : Given Functions: You have two functions,
$$f(x) = 3x – 1$$ $$g(x) = x^2 + 1$$

Step-2 : Compute g(2): To find f[g(2)], you first evaluate g(x) at x=2,
$$g(2) = 2^2 + 1 = 4 + 1 = 5$$

Step-3 : Evaluate f(g(2)): Then, you substitute g(2)=5 into f(x),
$$f(5) = 3(5) – 1 = 15 – 1 = 14$$

Step-4 : Composition of f and g: The notation (f ∘g)(x) represents the composition of f and g, meaning you first apply g to x, and then apply f to the result of g(x). For x=2,
$$(f \circ g)(2) = f[g(2)] = 14$$

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VSAQ-2 : If f(x) = 2x – 1, g(x) = (x+1)/2 for all x ∈ R, find (i) (gof)(x) (ii) (fog)(x)

Composition g ∘ f

Given: $$(g \circ f)(x) = g(f(x))$$

$$f(x) = 2x – 1$$

$$g(x) = \frac{x + 1}{2}$$

Substitute f(x) into g(x):

$$(g \circ f)(x) = g(f(x)) = g(2x-1) = \frac{(2x-1)+1}{2} = \frac{2x}{2} = x$$

Composition f ∘ g

Given: $$(f \circ g)(x) = f(g(x))$$

$$g(x) = \frac{x + 1}{2}$$

$$f(x) = 2x – 1$$

Substitute g(x) into f(x):

$$(f \circ g)(x) = f(g(x)) = f\left(\frac{x+1}{2}\right) = 2\left(\frac{x+1}{2}\right) – 1 = x + 1 – 1 = x$$

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VSAQ-3 : If A = {-2, -1, 0, 1, 2} and f : A → B is a surjection defined by f(x) = x² + x + 1 then find B.

Given $$A = {-2, -1, 0, 1, 2}$$ and the function $$f(x) = x^2 + x + 1$$ applying f(x) to each element in A yields:

$$f(-2) = 3$$

$$f(-1) = 1$$

$$f(0) = 1$$

$$f(1) = 3$$

$$f(2) = 7$$

Thus, the resulting set $$B = f(A) = {3, 1, 1, 3, 7}$$ which simplifies to $$B = {3, 1, 7}$$ after removing duplicates.

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VSAQ-4 : If A = {0, π/6, π/4, π/3, π/2} and f : A → B is a surjection defined by f(x) = cosx then find B.

Given A = {0, π/6, π/4, π/3, π/2}
with the function $$f(x) = \cos(x)$$ the function applied to each element in A yields:

$$f(0) = \cos(0) = 1$$

$$f\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

$$f\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$

$$f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

$$f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$$

Thus, applying $$f(x) = \cos(x)$$ to set A results in a new set of values without needing to remove any duplicates since all values are unique.

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VSAQ-5 : If f : Q→Q is defined by f(x) = 5x + 4, find f^-1.

To find the inverse of the function $$f(x) = 5x + 4$$ we set $$f(x) = y$$ and solve for x in terms of y, which leads to finding $$f^{-1}(y)$$ the inverse function. The process involves reversing the operations applied to x in f(x):

Step-1 : Start with the equation: $$5x + 4 = y$$

Step-2 : Subtract 4 from both sides: $$5x = y – 4$$

Step-3 : Divide both sides by 5: $$x = \frac{y – 4}{5}$$

Thus, the inverse function $$f^{-1}(y)$$ is expressed as $$f^{-1}(y) = \frac{y – 4}{5}$$ which means the inverse of f(x) when replacing y with x is $$f^{-1}(x) = \frac{x – 4}{5}$$

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VSAQ-6 : Find the inverse of the real function f(x) = ax + b, a ≠ 0.

To find the inverse of the function $$f(x) = ax + b$$ where a and b are constants, we express f(x) as y and then solve for x in terms of y, leading to the derivation of $$f^{-1}(y)$$ the inverse function. Here’s the step-by-step process:

Step-1 : Start with the equation: $$ax + b = y$$

Step-2 : Subtract b from both sides to isolate the term with x: $$ax = y – b$$

Step-3 : Divide both sides by a to solve for x: $$x = \frac{y – b}{a}$$

Thus, the inverse function $$f^{-1}(y)$$ is expressed as $$f^{-1}(y) = \frac{y – b}{a}$$ When we replace y with x to find the expression in terms of x, we get $$f^{-1}(x) = \frac{x – b}{a}$$

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VSAQ-7 : Find the inverse function of f(x) = 5^x

To find the inverse of the exponential function $$f(x) = 5^x$$ we express f(x) as y and solve for x in terms of y, which leads us to the inverse function $$f^{-1}(y)$$ Here’s how it’s done:

Step-1 : Start with $$5^x = y$$

Step-2 : To solve for x, take the logarithm base 5 of both sides: $$\log_5 5^x = \log_5 y$$

Step-3 : Using the logarithmic identity $$\log_b b^x = x$$ we find $$x = \log_5 y$$

Therefore, the inverse function $$f^{-1}(y)$$ is given by $$f^{-1}(y) = \log_5(y)$$ which means the inverse of f(x) when expressed in terms of x is $$f^{-1}(x) = \log_5(x)$$

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VSAQ-8 : Find the inverse function of f(x) = log2^x

To find the inverse of the logarithmic function $$f(x) = \log_2{x}$$ we set $$f(x) = y$$ and then solve for x in terms of y, which leads to the derivation of $$f^{-1}(y)$$ the inverse function.

Step-1 : Start with the equation: $$\log_2{x} = y$$

Step-2 : Using the property of logarithms that if $$\log_b{a} = c$$ then $$a = b^c$$ we rewrite the equation to solve for x: $$x = 2^y$$

Thus, the inverse function $$f^{-1}(y)$$ is expressed as $$f^{-1}(y) = 2^y$$ When we replace y with x to find the expression in terms of x, we get $$f^{-1}(x) = 2^x$$

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VSAQ-9 : If f = {(1,2), (2,-3), (3,-1)} then find (i) 2+f (ii) √f

Given a function f represented as a set of ordered pairs $${(1, 2), (2, -3), (3, -1)}$$ we perform operations on the second element of each pair:

Step-1 : Adding 2 to f, denoted as (2+f): This operation adds 2 to the second element of each pair in f.

• For (1,2), adding 2 gives (1,4).

• For (2,−3), adding 2 gives (2,−1).

• For (3,−1), adding 2 gives (3,1).

Thus, $$(2 + f) = {(1, 4), (2, -1), (3, 1)}$$

Step-2 : Taking the square root of f, denoted as f​: This operation applies the square root to the second element of each pair in f. Note, however, that square roots of negative numbers are not real numbers (without considering complex numbers).

• For (1,2), the square root is real, so 2​ is valid.

• For (2,−3) and (3,−1) the square roots would be of negative numbers, which are not real, so these pairs are not included when considering only real outputs.

Therefore, with only real number outputs considered, $$\sqrt{f} = {(1, \sqrt{2})}$$

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VSAQ-10 : If f = {(1,2), (2,-3), (3,-1)} then find (i) 2f (ii) f²

Operation 2f:

Step-1 : What it does: Multiplies the output values (y-values) of the function by 2.

Step-2 : Effect: Doubles the size of the output values without changing the input values (x-values), scaling the function’s outputs vertically.

Step-3 : Example: If $$f = {(1, 2), (2, -3), (3, -1)}$$ then $$2f = {(1, 4), (2, -6), (3, -2)}$$

Operation f²:

Step-1 : What it does: Squares the output values (�y-values) of the function.

Step-2 : Effect: Changes the magnitude of the output values and makes all outputs non-negative, affecting both the distribution and the sign of the outputs.

Step-3 : Example: If $$f = {(1, 2), (2, -3), (3, -1)}$$ then $$f^2 = {(1, 4), (2, 9), (3, 1)}$$

In essence, 2f scales the outputs linearly, while f² transforms the outputs non-linearly by squaring them, impacting both their magnitude and positivity.

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VSAQ-11 : Find the domain of the real function √(9- x² )

Step-1 : Condition Transformation: Move terms to get $$x^2 – 9 \leq 0$$

Step-2 : Factoring: Factor to $$(x – 3)(x + 3) \leq 0$$ identifying critical points at $$x = 3$$ and $$x = -3$$

Step-3 : Domain Determination: Solve the inequality to find $$x \in [-3, 3]$$

Thus, the domain of f(x) is $$[-3, 3]$$ indicating f(x) is defined for x values from −3 to 3, inclusive.

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VSAQ-12 : Find the domain of the real function √(16 – x² )

To determine the domain of f(x) given $$16 – x^2 \geq 0$$

Step-1 : Transform to $$x^2 – 16 \leq 0$$

Step-2 : Factor to $$(x – 4)(x + 4) \leq 0$$

Step-3 : Solve to find $$x \in [-4, 4]$$

Thus, the domain of f(x) is [−4,4], meaning f(x) is defined for x values from −4 to 4, inclusive.

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VSAQ-13 : Find the domain of the real function √(x² – 25)

To find the domain of f(x) given $$x^2 – 25 \geq 0$$

Step-1 : Factor to $$(x – 5)(x + 5) \geq 0$$

Step-2 : Solve to find $$x \in (-\infty, -5] \cup [5, \infty)$$

Thus, the domain of f(x) is $$x \in (-\infty, -5] \cup [5, \infty)$$ meaning f(x) is defined for x values less than or equal to −5 and greater than or equal to 5.

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VSAQ-14 : Find the domain of √(x² – 3x + 2)

For $$f(x) = \sqrt{x^2 – 3x + 2}$$ f(x) is defined when the expression under the square root, $$x^2 – 3x + 2$$ is non-negative:

Step-1 : Factor the quadratic: $$x^2 – 3x + 2 = (x – 1)(x – 2)$$

Step-2 : Determine the domain: $$x \in (-\infty, 1] \cup [2, \infty)$$

Thus, the domain of f(x) is $$x \in (-\infty, 1] \cup [2, \infty)$$ indicating f(x) is defined for x values less than or equal to 1 and greater than or equal to 2.

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VSAQ-15 : Find the domain of the real function log(x² – 4x + 3)

For f(x) defined when $$x^2 – 4x + 3 > 0$$

Step-1 : Factor the quadratic: $$x^2 – 4x + 3 = (x – 1)(x – 3) > 0$$

Step-2 : Determine the domain: $$x \in (-\infty, 1) \cup (3, \infty)$$

Given the inequality is strict (>), the correct domain should not include the endpoints. Therefore, the accurate domain of f(x) is indeed $$x \in (-\infty, 1) \cup (3, \infty)$$ not including 1 or 3 as part of the domain.

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VSAQ-16 : Find the domain of the real function 1/√(x² – a² )

For f(x) defined when $$x^2 – a^2 > 0$$

Step-1 : Factor the quadratic: $$x^2 – a^2 = (x – a)(x + a) > 0$$

Step-2 : Determine the domain: $$x \in (-\infty, -a) \cup (a, \infty)$$

Thus, the correct domain of f(x) is $$x \in (-\infty, -a) \cup (a, \infty)$$ indicating f(x) is defined for x values less than −a and greater than a, excluding −a and a themselves due to the strict inequality (>).

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VSAQ-17 : Find the domain of the real function f(x) = √(4x-x²)

For the function given by $$\sqrt{4x – x^2}$$ it’s defined when the expression under the square root, $$4x – x^2$$ is non-negative:

Step-1 : Rearrange and factor the inequality: $$x^2 – 4x \leq 0$$ translates to $$x(x – 4) \leq 0$$

Step-2 : Factor further to show the roots explicitly: $$(x – 0)(x – 4) \leq 0$$

Hence, the domain of f(x) is [0,4], meaning f(x) is defined for x values within and including the endpoints 0 and 4.

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VSAQ-18 : Find the domain of the real function 1/√(1-x²)

For the function f(x) defined when $$1 – x^2 > 0$$

Step-1 : Rearrange the inequality to $$x^2 – 1 < 0$$

Step-2 : Factor the expression to $$(x + 1)(x – 1) < 0$$

Thus, the domain of f(x) is (−1,1), indicating f(x) is defined for x values strictly between -1 and 1, excluding the endpoints -1 and 1 due to the strict inequality (<).

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VSAQ-19 : Find the domain of the real function 1/((x² – 1)(x + 3))

For the function f(x) defined when $$(x^2 – 1)(x + 3) \neq 0$$

Factor the expression: $$(x – 1)(x + 1)(x + 3) \neq 0$$

Thus, the domain of f(x) is $$\mathbb{R} – {1, -1, -3}$$ meaning f(x) is defined for all real numbers except for 11, −1−1, and −3−3.

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VSAQ-20 : Find the domain of the real function (2x² – 5x + 7)/((x-1)(x-2)(x-3))

For the function $$f(x) = \frac{2x^2 – 5x + 7}{(x – 1)(x – 2)(x – 3)}$$ it’s not defined when its denominator equals zero:

The denominator $$(x – 1)(x – 2)(x – 3) = 0$$ gives roots $$x = 1, 2, 3$$

Therefore, the domain of f(x) is $$\mathbb{R} – {1, 2, 3}$$ indicating f(x) is defined for all real numbers except for 1, 2, and 3.