Composition of Functions MCQ Quiz - Objective Question with Answer for Composition of Functions - Download Free PDF

Concept:

Composite Functions and Derivatives:

• Composite Function: A composite function is a function formed by applying one function to the results of another. For two functions \( f \) and \( g \), the composite function \( f \circ g \) is defined as \( (f \circ g)(x) = f(g(x)) \).
• Derivative of a Composite Function: The derivative of a composite function is found using the chain rule. The chain rule states that if we have two functions \( f(x) \) and \( g(x) \), then the derivative of \( f(g(x)) \) is:
  • \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
• Inverse Functions: If \( f \) and \( g \) are inverses of each other, then \( f(g(x)) = x \) and \( g(f(x)) = x \).

Calculation:

We are given the following functions:

• \( f(x) = \log(x^2 + 2x + 4) \)
• \( g(x) = \frac{4}{1 + e^{-2x}} \)

The task is to find the derivative of the composite function \( f \circ g^{-1}(x) \) at \( x = 2 \), where \( g^{-1} \) is the inverse of the function \( g(x) \).

Step 1: Find the inverse of the function \( g(x) \)

We need to solve for \( g^{-1}(x) \). Start with the equation for \( g(x) \):

\( g(x) = \frac{4}{1 + e^{-2x}} \)

We can express \( g^{-1}(x) \) by swapping \( x \) and \( y \) and solving for \( y \):

\( x = \frac{4}{1 + e^{-2y}} \)

Rearranging this equation gives:

\( 1 + e^{-2y} = \frac{4}{x} \)

\( e^{-2y} = \frac{4}{x} - 1 \)

\( e^{-2y} = \frac{4 - x}{x} \)

Taking the natural logarithm of both sides:

\( -2y = \ln\left( \frac{4 - x}{x} \right) \)

\( y = -\frac{1}{2} \ln\left( \frac{4 - x}{x} \right) \)

Thus, the inverse function is:

\( g^{-1}(x) = -\frac{1}{2} \ln\left( \frac{4 - x}{x} \right) \)

Step 2: Apply the chain rule to compute the derivative of the composite function

Now, we will compute the derivative of the composite function \( f(g^{-1}(x)) \).

The chain rule states that:

\( \frac{d}{dx} [ f(g^{-1}(x)) ] = f'(g^{-1}(x)) \cdot \frac{d}{dx} [ g^{-1}(x) ] \)

We already have \( f(x) = \log(x^2 + 2x + 4) \). The derivative of \( f(x) \) is:

\( f'(x) = \frac{d}{dx} \log(x^2 + 2x + 4) = \frac{2x + 2}{x^2 + 2x + 4} \)

Next, we need to differentiate \( g^{-1}(x) \):

\( \frac{d}{dx} [ g^{-1}(x) ] = -\frac{1}{2} \cdot \frac{d}{dx} \ln\left( \frac{4 - x}{x} \right) \)

The derivative of the logarithm is:

\( \frac{d}{dx} \ln\left( \frac{4 - x}{x} \right) = \frac{-1}{(4 - x) / x} \cdot \frac{d}{dx} \left( \frac{4 - x}{x} \right) \)

After calculating the derivative of the quotient, we get:

\( \frac{d}{dx} [ g^{-1}(x) ] = \frac{1}{x} \)

Step 3: Evaluate at \( x = 2 \)

We now substitute \( x = 2 \) into the formula for the derivative of the composite function:

\( f'(g^{-1}(x)) = \frac{2x + 2}{x^2 + 2x + 4} \)

Substituting \( x = 2 \) gives:

\( f'(g^{-1}(2)) = \frac{2(2) + 2}{(2)^2 + 2(2) + 4} = \frac{6}{12} = \frac{1}{2} \)

\( \frac{d}{dx} [ g^{-1}(x) ] = \frac{1}{x} = \frac{1}{2} \)

Hence, the derivative at \( x = 2 \) is:

\( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)

Conclusion:

Hence, the value of the derivative of the composite function at \( x = 2 \) is \( 0.25 \).

Calculation: 

Let fog(x) = h(x)

⇒ \(\mathrm{h}^{-1}(\mathrm{x})=\left(\frac{\mathrm{x}-7}{2}\right)^{\frac{1}{3}}\)

⇒ h(x) = fog(x) = 2x³ + 7

fog(x) = a(x^b + c) – 3 

⇒ a = 2, b = 3, c = 5 

⇒ fog(ac) = fog(10) = 2007

g(f(x) = (2x – 3)³ + 5 

⇒ gof(b) = gof(3) = 32 

⇒ sum = 2039 

Hence, the correct answer is 2039. 

Concept:

Properties of Composition of Functions:

• Function Composition: For two functions f: A → B and g: B → C, the composite function g ∘ f is defined as (g ∘ f)(x) = g(f(x)).
• Injective (One-One): A function is one-one if every element of the codomain is mapped by at most one element of the domain.
• Surjective (Onto): A function is onto if every element of the codomain has a pre-image in the domain.
• Natural Numbers: The set of natural numbers N = {1, 2, 3, ...}
• Integers: The set of integers Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}

Calculation:

Given,

Function f: N → Z

f(n) = \(\left\{\begin{array}{ll} (n+1) / 2 & \text { if } n \text { is odd ,} \\ (4-n) / 2 & \text { if } n \text { is even, } \end{array}\right.\)   

⇒ f(1) = 1, f(2) = −1, f(3) = 2, f(4) = −2, f(5) = 3, ...

⇒ The image of f contains all integers (both positive and negative)

⇒ f is onto.

Now checking if f is one-one:

f(2) = −1 and f(4) = −2, f(6) = −3, ... (even numbers give negative integers)

f(1) = 1, f(3) = 2, f(5) = 3, ... (odd numbers give positive integers)

But f(−1) and f(−3) do not exist in N ⇒ f is not one-one.

Now g: Z → N

g(n) = \(\left\{\begin{array}{cc} 3+2 n & \text { if } n \geq 0, \\ -2 n & \text { if } n<0. \end{array}\right.\)   

⇒ g(0) = 3, g(1) = 5, g(2) = 7, ... (odd numbers ≥ 3)

g(−1) = 2, g(−2) = 4, g(−3) = 6, ... (even numbers ≥ 2)

⇒ g(Z) = {2, 3, 4, 5, 6, ...} (excluding 1)

⇒ g is not onto N

Now, gof (n) = g(f(n))

If n is odd, f(n) = (n+1)/2 ≥ 1

⇒ f(n) ∈ Z, f(n) ≥ 1

⇒ use g(n ≥ 0)

⇒ g(f(n)) = 3 + 2 × (n+1)/2 = 3 + (n+1) = n + 4

If n is even, f(n) = −n/2

⇒ f(n) < 0

⇒ use g(n < 0)

⇒ g(f(n)) = −2 × (−n/2) = n

⇒ gof(n) = \(\left\{\begin{array}{cc} n+4 & \text { if } \text{n is odd} , \\ n & \text { if } \text{n is even} . \end{array}\right.\)

Let n₁ = 2, n₂ = 6

⇒ gof(2) = 2, gof(6) = 6

But gof(1) = 5, gof(3) = 7 ⇒ all values distinct

Still, g(f(n)) does not cover all of N

⇒ gof is not onto

Also, for gof to be one-one, g(f(n)) should be unique for all n

g(f(1)) = 5, g(f(2)) = 2, g(f(3)) = 7, g(f(4)) = 4

⇒ All values unique

⇒ gof is one-one

Now fog(n) = f(g(n))

If n ≥ 0

⇒ g(n) = 3 + 2n (odd number)

⇒ f(g(n)) = (g(n) + 1)/2

⇒ f(g(n)) = (3 + 2n + 1)/2 = (4 + 2n)/2 = 2 + n ∈ N

If n < 0

⇒ g(n) = −2n (even)

⇒ f(g(n)) = −g(n)/2 = n ∈ Z

⇒ f(g(n)) = \(\left\{\begin{array}{cc} n+2 & \text { if } n\geq 0 , \\ n & \text { if } n<0 . \end{array}\right.\)

So, fog(n) maps Z to N and is one-one

fog is one-one, but image is not all of N

⇒ fog is not onto

∴ Correct options are 1 and 4. 

Explanation -

g(x) = x2 + x – 1

gof(x) = 4x2 – 10x + 5

g(f(x) = 4x2 – 10x + 5

f2(x) + f(x) – 1 = 4x2 – 10x + 5

Putting x = 5/4 and f(5/4) = t

\(t^2 + t + \frac{1}{4} = 0\)

t = -1/2 or f(5/4) = -1/2

Hence Option (2) is correct.

Explanation:

\(f_{1} = f_{0}(f_{0}(x))\)

\(f_{1} = \dfrac{1}{1 - f_{0}(x)}\)  

\(f_{2} = f_{0}(f_{1}(x)) = \dfrac{1}{1 - f_{1}(x)}\)

Substitute the value of \(f_{1}\) to get the following expression:

\(f_{2} = \dfrac{f_{0}(x) - 1}{f_{0}(x)}\)  

\(f_{3} = f_{0}(x)\) (3)

\(f_{4}(x)\) will be the same as \(f_{1}(x)\)

\(f_{100}(x)\) will be the same as \(f_{1}(x)\)

\(f_{100}(3) + f_{1}\left( \dfrac{2}{3} \right) + f_{2}\left( \dfrac{3}{2} \right) = \dfrac{2}{3} - \dfrac{1}{2} + \dfrac{3}{2} = \dfrac{5}{3}\)

Concept:

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if the co-domain of g = domain of f i.e D = A and (gof) (x) will exist if and only if co-domain of f = domain of g i.e B = C.

In our case, the co-domain of g is R and the domain of f is also R  so (fog) (x) is defined.

Composition of function:

(fog) (x) = f[g(x)]

Calculations:

Given: f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| - x ∀ x ∈ R

Given, x < 0

If(x) = |x| + x and g(x) = |x| - x 

Now, (fog) (x) = f[g(x)]

= |g(x)| + g(x)

For x < 0, g(x) = - x + x = 0

⇒ (fog) (x) = f[g(x)] = |g(x)| + g(x)

= 0 + 0 = 0

Concept:

• For two functions f(x) and g(x), (f o g)(x) is defined as f[g(x)].

Calculation:

f(x) = sin x and g(x) = x2.

∴ (f o g)(x) = f[g(x)] = sin [g(x)] = sin (x²) = sin x².

Concept:

For any two functions f and g, f o g is defined as f[g(x)].

Calculation:

Given that,

f(x) = 4x + 3   

Using the above concept

⇒ fof(x) = 4(4x + 3) + 3

⇒ fof(x) = 16x + 15 

Again using the same concept

⇒ fofof(x) = 16(4x + 3) + 15

⇒ fofof(x) = 64x + 63

Put x = -1 in the above function

⇒ fofof(-1) = 64(-1) + 63 = -1

∴  f o f o f(-1) equal to -1.

Concept: 

Composition of function: Let f, and g be any functions then, f ∘ g(x) = f[g(x)]

Calculation:

Here, f(x) = \(\rm \frac{x+1}{x-1}\)

f{f(x)} = \(\rm \frac{f(x)+1}{f(x)-1}\)

= \(\rm \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}\)

= \(\rm \frac{2x}{2}\)

= x

Hence, option (3) is correct.

Concept: 

Composition of function: Let f, and g be any functions then, f ∘ g(x) = f[g(x)]

Calculation:  

Given f (x) = 16x⁴ and g (x) = x^1/4 

gof (x) = g[f(x)] = \(\rm [16x^{4}]^{\frac{1}{4}}\) 

⇒ gof (x) = \(\rm [(2x)^{4}]^{\frac{1}{4}}\) 

⇒ gof (x) = 2x . 

The correct option is 4. 

Concept: 

Composition of function: Let f, and g be any functions then, f ∘ g(x) = f[g(x)]

Calculation:  

Given f (x) = 8x3 and g (x) = x1/3 

gof (x) = g[f(x)] = \(\rm [8x^{3}]^{\frac{1}{3}}\) 

⇒ gof (x) = \(\rm [(2x)^{3}]^{\frac{1}{3}}\) 

⇒ gof (x) = 2x . 

Now put x = 2

So, gof (2) = 4 . 

The correct option is 2. 

Concept: 

Composition of function: Let f, and g be any functions then, f ∘ g(x) = f[g(x)]

Calculations:

Given, f(x) = e^x and g(x) = loge x

Now, f ∘ g(x) = f[g(x)]

= e^g(x)

= e^loge x

= x                (∵ e^loge x = x)

f ∘ g(x) = x

Put x = 1, we get

fog(1) = 1

Concept:

Composition of function: fog(x) = f(g(x)), it means that the value of x for the function f(x) is g(x).

Example: Suppose f(x ) = x + 2 and g(x) = 2x. Then f(g(x)) = g(x) + 2 = 2x + 2

\(\log {{\bf{a}}^{\bf{n}}} = {\bf{n}}\log {\bf{a}}\)

Calculation:

Given that, f(x) = \(\log \frac{{{\rm{x}} + 4}}{{{\rm{x}} + 1}}\) and g(x) = \(2{\rm{x}} + {{\rm{x}}^2}\)

f[g(x)] = \(\log \frac{{{\rm{g}}\left( {\rm{x}} \right) + 4}}{{{\rm{g}}\left( {\rm{x}} \right) + 1}}\)

\( = \log \left( {\frac{{2{\rm{x}} + {{\rm{x}}^2} + 4{\rm{\;}}}}{{2{\rm{x}} + {{\rm{x}}^2} + 1}}} \right)\)

Hence, option (4) is correct.

Mistake Points
For option C, the Numerator should be (4x + x2 + 4) but it is given as (2x + x2 + 4). Therefore, option 4 is correct answer.

Calculation:

Given: f(x) = px + q and g(x) = mx + n

f (g(x)) = g (f(x))

⇒ f (mx + n) = g (px + q)

⇒ p (mx + n) + q = m (px + q) + n

⇒ pmx + pn + q = pmx + mq + n

⇒ pn + q = mq + n

⇒ f (n) = g (q)

∴ Option 3 is correct answer.

Calculation:

Here, f(x) + 2f(\(\rm \frac 1 x\)) = \(\rm \frac 1 x\)                ....(1)

Replace x by \(\rm \frac 1 x\), we get 

f(\(\rm \frac 1 x\)) + 2f(x) = x                          ....(2)

Adding (1) and (2), we get  

 3f(x) + 3f(\(\rm \frac 1 x\))) = x + \(\rm \frac 1 x\)               ....(3)

Now, subtracting (1) from (2), we get

f(x) - f(\(\rm \frac 1 x\)) = x - \(\rm \frac 1 x\)

Multiplying by 3,

3f(x) - 3f(\(\rm \frac 1 x\)) = 3x - \(\rm \frac 3 x\)                  ....(4)

Now, adding (3) and (4) we get 

6f(x) = 4x - \(\rm \frac 2 x\)

f(x) = \(\rm \frac{1}{3} (2x-\frac 1 x)\)

Hence, option (4) is correct.