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:
  • $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• 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:

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

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

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

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:

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

Substituting $x=2$ gives:

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

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

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

⇒ 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) = $\{\begin{array}{cc}3+2n& \text{ if }n\ge 0,\\ -2n& \text{ if }n<0.\end{array}$   

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

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)) = $\{\begin{array}{cc}n+2& \text{ if }n\ge 0,\\ n& \text{ if }n<0.\end{array}$

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:

$f_1=f_0(f_0(x))$

$f_1={\displaystyle \frac{1}{1-f_0(x)}}$  

$f_2=f_0(f_1(x))={\displaystyle \frac{1}{1-f_1(x)}}$

Substitute the value of $f_1$ to get the following expression:

$f_2={\displaystyle \frac{f_0(x)-1}{f_0(x)}}$  

$f_3=f_0(x)$ (3)

$f_4(x)$ will be the same as $f_1(x)$

$f_{100}(x)$ will be the same as $f_1(x)$

$f_{100}(3)+f_1\left({\displaystyle \frac{2}{3}}\right)+f_2\left({\displaystyle \frac{3}{2}}\right)={\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{2}}={\displaystyle \frac{5}{3}}$

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) = $\frac{\mathrm{x}+1}{\mathrm{x}-1}$

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

= $\frac{\frac{\mathrm{x}+1}{\mathrm{x}-1}+1}{\frac{\mathrm{x}+1}{\mathrm{x}-1}-1}$

= $\frac{2\mathrm{x}}{2}$

= 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)] = $[16{\mathrm{x}}^4{]}^{\frac{1}{4}}$ 

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

⇒ 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)] = $[8{\mathrm{x}}^3{]}^{\frac{1}{3}}$ 

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

⇒ 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

$\log {\mathbf{a}}^{\mathbf{n}}=\mathbf{n}\log \mathbf{a}$

Calculation:

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

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

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

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($\frac{1}{\mathrm{x}}$) = $\frac{1}{\mathrm{x}}$                ....(1)

Replace x by $\frac{1}{\mathrm{x}}$, we get 

f($\frac{1}{\mathrm{x}}$) + 2f(x) = x                          ....(2)

Adding (1) and (2), we get  

 3f(x) + 3f($\frac{1}{\mathrm{x}}$)) = x + $\frac{1}{\mathrm{x}}$               ....(3)

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

f(x) - f($\frac{1}{\mathrm{x}}$) = x - $\frac{1}{\mathrm{x}}$

Multiplying by 3,

3f(x) - 3f($\frac{1}{\mathrm{x}}$) = 3x - $\frac{3}{\mathrm{x}}$                  ....(4)

Now, adding (3) and (4) we get 

6f(x) = 4x - $\frac{2}{\mathrm{x}}$

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

Hence, option (4) is correct.