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

Calculation

Taking log to the base 5 on both sides

\(\left(16\left(\log _{5} x\right)^{3}-68\left(\log _{5} x\right)\right)\left(\log _{5} x\right)=-16\)

Let \(\left(\log _{5} x\right)=t\)

\(16 t^{4}-68 t^{2}+16=0\)

OR \(4 t^{4}-16 t^{2}-t^{2}+4=0\)

OR \(\left(4 t^{2}-1\right)\left(t^{2}-4\right)=0\)

OR \(t= \pm \frac{1}{2}, \pm 2\)

So \(\log _{5} x= \pm \frac{1}{2} \mathrm{OR} \pm 2\)

\(\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^{2}, 5^{-2}\)

The product of all positive real values of x is 1

\( \Rightarrow \log _{5}\left ( 16-x^{2} \right )-\log _{5}\left ( 2x+5 \right )\leq \log _{5}5 \quad [\because \log_{1/a}b=-\log_ab \quad \text& \quad \log_aa=1] \)

\( \therefore \frac{16-x^{2}}{2x+5}\leq 5\:or\:\left ( 16-x^{2} \right )\leq 10x+25 \quad [\because \log a-\log b=\log\dfrac{a}{b}] \)

\( x^{2}+10x+9\geq 0 \)

\( \left ( x+9 \right )\left ( x+1 \right )\geq 0 \)

\( \therefore x\leq -9 \) or \( x\geq -1 \) ...(1)

Now by definition of \( \log \), we must have

\( 2x+5> 0 \), i.e., \( x> -\dfrac52 \)

and \( 16-x^{2}> 0 \) or \( x^{2}-16< 0 \)

or \( \left ( x+4 \right )\left ( x-4 \right )< 0 \)

\( \therefore -4< x< 4 \) and \( x> -\dfrac52 \) ...(2)

Hence from (1) and (2), \( x\geq -1 \) and \( x< 4 \).

\( \therefore x \in \left [ -1,4 \right ) \).

Ans: B

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.” 

\(\rm b^x = b^y \) ⇒ x = y

Calculations:

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

⇒ \(\rm (2^2)^x - 3 (2^x.2^3) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 24 (2^x) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 16 (2^x) - 8(2^x)+ 128 = 0\)

⇒ \(\rm (2^x - 16)(2^x - 8) = 0\)

⇒ \(\rm 2^x = 16 \;\;\text{or}\;\; 2^x = 8\)

⇒ \(\rm 2^x = 2^4 \;\;\text{or}\;\; 2^x = 2^3\)

⇒ 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:

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

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.” 

\(\rm b^x = b^y \) ⇒ x = y

Calculations:

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

⇒ \(\rm (2^2)^x - 3 (2^x.2^3) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 24 (2^x) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 16 (2^x) - 8(2^x)+ 128 = 0\)

⇒ \(\rm (2^x - 16)(2^x - 8) = 0\)

⇒ \(\rm 2^x = 16 \;\;\text{or}\;\; 2^x = 8\)

⇒ \(\rm 2^x = 2^4 \;\;\text{or}\;\; 2^x = 2^3\)

⇒ 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

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.” 

\(\rm b^x = b^y \) ⇒ x = y

Calculations:

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

⇒ \(\rm (2^2)^x - 3 (2^x.2^3) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 24 (2^x) + 128 = 0\)

⇒ \(\rm (2^x)^2 - 16 (2^x) - 8(2^x)+ 128 = 0\)

⇒ \(\rm (2^x - 16)(2^x - 8) = 0\)

⇒ \(\rm 2^x = 16 \;\;\text{or}\;\; 2^x = 8\)

⇒ \(\rm 2^x = 2^4 \;\;\text{or}\;\; 2^x = 2^3\)

⇒ 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