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

Concept:

Exponential Generating Function & Series Coefficients:

• The expression uses combinations ⁿC_r and factorials, hinting at exponential and binomial expansion identities.
• Function f(x) = 1 + (1 + x)/1! + (1 + x)²/2! + (1 + x)³/3! + ... is considered to evaluate the coefficient of x².
• This function can be transformed using the identity: e^1+x / (1 + x)
• The coefficient of x² in this expansion corresponds to the RHS of 'a' series.
• The value of b is derived using the identity: 1 + 2/1! + 2²/2! + 2³/3! + ... = e²

Calculation:

Let f(x) = 1 + (1 + x)/1! + (1 + x)²/2! + (1 + x)³/3! + ...

⇒ f(x) = e^(1+x) / (1 + x)

Expand RHS:

= (1 + x + x²/2! + ...) / (1 + x)

⇒ (1 + x + (1 + x)²/2! + (1 + x)³/3! + (1 + x)⁴/4! + ...)

So, coefficient of x² in RHS is:

²C₂/3! + ³C₂/4! + ⁴C₂/5! + ... = a - 1

coefficient of x2 in RHS:

e × (1 + x+ x2/2!) × (1 -x+ x2/2!)

is e- e+ e/2! =a 

Now, expand LHS expression:

e × (1 + x+ x²/2!) × (1 -x+ x²/2!)

⇒ e × (1 - (x⁴/4!)) = e 

So, coefficient of x² = e × e = e²

Thus, b = 1 + 2/1! + 2²/2! + 2³/3! + ... = e²

a = e\2!

⇒ 8b / a² = 2 × e² / (e/2!)² = 32

∴ 8b / a² = 32

Calculation

Taking log to the base 5 on both sides

Let 

OR 

OR 

OR 

So 

The product of all positive real values of x is 1

Explanation:

Graph for exponential growth is

i.e., the curve obtained in exponential growth is J shaped curve.

Option (2) is true.

Concept:

Base Rule

If b raised to the xth power is equal to b raised to the yth power, that implies that x = y.” 

 ⇒ x = y

Calculations:

Given equation is 4x - 3(2x + 3) + 128 = 0

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ x = 4 or x = 3

The roots of the equation 4x - 3(2x + 3) + 128 = 0 are 4 and 3

Its Sum = 4 + 3 = 7

Concept:

Smallest integer function:(Ceiling function)

The function f (x) = [x) is called the smallest integer function and it means that the smallest integer greater than or equal to x i.e [x) ≥ x.

If f :A → B and g : C → D.

Then (fog) (x) will exist if and only if 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.

Calculation:

Given: 

f(x) = ex and g(x) = ⌈x) where ⌈.⌉ denotes smallest integer function

Here, we have to find the value of f o g(9/2)

⇒ f o g(9/2) = f(g(9/2))

Since, g(x) = [x)

⇒ g(9/2) = [4.5) = 5

⇒ f o g(9/2) = f(5)

Since, f(x) = e^x 

⇒ f(5) = e⁵

Hence, f o g(9/2) = e⁵.

Concept:

Base Rule

If b raised to the xth power is equal to b raised to the yth power, that implies that x = y.” 

 ⇒ x = y

Calculations:

Given equation is 4x - 3(2x + 3) + 128 = 0

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ x = 4 or x = 3

The roots of the equation 4x - 3(2x + 3) + 128 = 0 are 4 and 3

Its Sum = 4 + 3 = 7

Mistake PointsPlease Note that, here, ⌈⌉ represents the smallest integer function, not the greatest integer function.

Concept:

Smallest integer function (Ceiling function):

It is a function that takes all the values (−∞,∞) and gives only the integer

part i.e. range of the smallest integer the function is Z (all integers).

For e.g., [5.1] = 6, [- 5.1] = - 5

Composite Function:

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.

Calculation:

Given: f(x) = ⌈x⌉ ,

where [ ] denotes smallest integer function and g(x) = 5x 

Here, we have to find the value of g o f(1/2)

⇒ g o f(1/2) = g( f(1/2))

As, f(x) = ⌈x⌉

⇒ f(1/2) =  ⌈1/2⌉ =   ⌈0.5⌉

⇒ f(1/2) = 1

⇒ g o f(1/2) = g(1)

∵ g(x) = 5^x so, g(0) = 5¹ = 5

Hence, g o f(1/2) = 5

Confusion PointsGreatest integer function (Floor Function):

The greatest integer function is a function that gives the greatest integer

less than or equal to a given number. The greatest integer less than or

equal to a number x is represented as ⌊x⌋.  

For e.g  [5.1] = 5 & [- 5.1] = - 6 

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

Domain of [x] is R and range is I.

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if 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.

Calculation:

Given: f(x) = ex and g(x) = [x] where [.] denotes greatest integer function.

Here, we have to find the value of f o g(- 5/2)

⇒ f o g(- 5/2) = f( g(-5/2))

∵ g(x) = [x] so, g(-5/2) = - 3

⇒ f o g(- 5/2) = f(- 3)

∵ f(x) = ex so, f(- 3) = e^{- 3}

Hence, f o g(- 5/2) = e^{- 3}

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

Domain of [x] is R and range is I.

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if 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.

Calculation:

Given: f(x) = [x] where [.] denotes greatest integer function and g(x) = 2x 

Here, we have to find out the value of g o f(- 3/2)

⇒ g o f(- 3/2) = g( f(- 3/2))

∵ f(x) = [x], so f(- 3/2) = [- 3/2] = - 2

⇒ g o f(- 3/2) = g(- 2)

∵ g(x) = 2^x so, g(- 2) = 1/4

Hence, g o f(- 3/2) = 1/4

CONCEPT:

Exponential functions are functions of the form f(x) = b^x for a fixed base b which could be any positive real number.

The inverse of an exponential function is a logarithmic function.

CALCULATIONS:

Given exponential function is y = q^x, also x = 4 and y = 1296

∴ 1296 = q⁴

⇒ q = 6

Concept: 

Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

Domain of [x] is R and range is I.

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if 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.

Calculation:

Given: f(x) = [x] where [.] denotes greatest integer function and g(x) = 2x 

Here, we have to find out the value of g o f(- 3/2) + g o f(5/2)

First lets find out the value of g o f(- 3/2)

⇒ g o f(- 3/2) = g( f(- 3/2))

∵ f(x) = [x], so f(- 3/2) = [- 3/2] = - 2

⇒ g o f(- 3/2) = g(- 2)

∵ g(x) = 2x so, g(- 2) = 1/4

⇒ g o f(- 3/2) = 1/4----------(1)

Similarly lets ind out the value of g o f(5/2)

⇒ g o f(5/2) = g( f(5/2))

∵ f(x) = [x], so f(5/2) = [5/2] = 2

⇒ g o f(5/2) = g(2)

∵ g(x) = 2x so, g(2) = 4

⇒ g o f(5/2) = 4---------(2)

Now, from equation (1) and (2), we get

⇒ g o f(- 3/2) + g o f(5/2) = 4 + 1/4 = 17/4

Concept:

Smallest integer function:(Ceiling function)

The function f (x) = [x] is called the smallest integer function and it means that smallest integer greater than or equal to x i.e [x] ≥ x.

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if 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.

Calculation:

Given: f(x) = [x], where [.] denotes smallest integer function and g(x) = 5x 

Here, we have to find the value of g o f(1/2) + g o f(- 3/2)

First lets find out the value of g o f(1/2)

⇒ g o f(1/2) = g( f(1/2))

∵ f(x) = [x] so, f(1/2) = [1/2] = 1

⇒ g o f(1/2) = g(1)

∵ g(x) = 5x so, g(1) = 5

So, g o f(1/2) = 5-----------(1)

Similarly, lets find out the value of g o f(- 3/2)

⇒ g o f(- 3/2) = g( f(- 3/2))

∵ f(x) = [x] so, f(- 3/2) = [- 3/2] = - 1

⇒ g o f(- 3/2) = g(- 1)

∵ g(x) = 5x so, g(- 1) = 1/5

So, g o f(1/2) = 1/5-----------(2)

Now, from equation (1) and (2) we get

⇒ g o f(1/2) + g o f(- 3/2) = 5 + 1/5 = 26/5

Concept:

Smallest integer function:(Ceiling function)

The function f (x) = [x) is called the smallest integer function and it means that the smallest integer greater than or equal to x i.e [x) ≥ x.

If f :A → B and g : C → D.

Then (fog) (x) will exist if and only if 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.

Calculation:

Given: 

f(x) = ex and g(x) = ⌈x) where ⌈.⌉ denotes smallest integer function

Here, we have to find the value of f o g(9/2)

⇒ f o g(9/2) = f(g(9/2))

Since, g(x) = [x)

⇒ g(9/2) = [4.5) = 5

⇒ f o g(9/2) = f(5)

Since, f(x) = e^x 

⇒ f(5) = e⁵

Hence, f o g(9/2) = e⁵.

Concept:

Base Rule

If b raised to the xth power is equal to b raised to the yth power, that implies that x = y.” 

 ⇒ x = y

Calculations:

Given equation is 4x - 3(2x + 3) + 128 = 0

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ x = 4 or x = 3

The roots of the equation 4x - 3(2x + 3) + 128 = 0 are 4 and 3

Its Sum = 4 + 3 = 7