Logarithmic Function MCQ Quiz - Objective Question with Answer for Logarithmic Function - Download Free PDF

Concept:

• We are given:
  • t = e^2x
  • y = log_e(t²) = 2 log_e(t)
• We need to find the second derivative d²y/dx².
• This requires the chain rule and product rule of differentiation.

Calculation:

Step 1: Express y in terms of x

y = 2 log(t), where t = e^2x

Since log(t) = log(e^2x) = 2x

⇒ y = 2 × 2x = 4x

First derivative:

dy/dx = d/dx (4x) = 4

Second derivative:

d²y/dx² = d/dx (4) = 0

∴ The correct answer is: 0.

Let, 

Therefore, 

Therefore, 

Calculation

Given equation is

applying componendo and dividendo gives,

Differentiating with respect to gives,

Hence option 4 is correct

Calculation

y = logn x

x log x log2 x log3 x ..... logn-1 x logn x  = logn x

Hence option 4 is correct 

Answer : 1

Solution :

log(x + y) = 2xy

Differentiating w.r.t. x, we get

For x = 0, log(y) = 0

⇒ y = 1

Concept:

Formula:

log mn = n log m

Calculation:

Let y = x^log x

Taking log both sides, we get

⇒ log y = x^{log x}

⇒ log y = log x log  x            (∵ log mn = n log m)

Differentiating with respect to x, we get

⇒ 

⇒ 

⇒ 

⇒ 

Concept:

Product rule of differentiation:

Let h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x)

Calculation:

Given, y = logy x

⇒ y = 

⇒ y log y = log x

By differentiating both sides w.r.t. x, we get

⇒ 

∴ 

Given:

log 2 =  x, log 3 = y

Concept:

Properties of Logarithms:

Calculation:

log 6 = log (2 x 3)

From 2nd property of logarithms

log 6 = log 2 + log 3

log 6 = x + y

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: 
• Product Rule: 
• log ab = b log a 

Calculation:-  

 , where a is constant 

Taking log both sides, we get

⇒ 

⇒ log x^m + log yⁿ = log x + log y + log a^m+n

⇒ m log x + n log y = log x + log y + (m + n) log a 

Differentiate both sides w.r.t  x, we get

⇒    [∵   ]  

⇒  

⇒   . 

The correct option is 3. 

Concept:

Chain rule: 

If y = log x, then f'(x) = 1/x

And if y = √x, then f'(x) = 

Calculation:

f(x) = log √x

Differentiating with respect to x, we get

⇒ f'(x) = () 

⇒ f'(x) = ⋅ 

⇒ f'(x) = 

Concept:

Chain rule: 

If y = log x, then f'(x) = 1/x

And if y = xⁿ, then f'(x) = nx^{n - 1}

Calculation:

f(x) = log x², x > 1

Differentiating with respect to x, we get

⇒ f'(x) = () 

⇒ f'(x) = 

⇒ f'(x) = 

Concept:

Chain rule: 

If y = log x, then f'(x) = 1/x

And if y = sin x, then f'(x) = cos x

Calculation:

f(x) = log (sin x)

Differentiating with respect to x, we get

⇒ f'(x) = () 

⇒ f'(x) = 

⇒ f'(x) = cot x

Concept:

Chain rule: 

Calculation:

Here, -log (log x), x > 1

Let, log x = y 

Differentiating with respect to x, we get

⇒ dy/dx = 1/x        ....(1)

Now, -log (log x) = -log y

= -1 / (x log x)    ....(from (1))

Hence, option (1) is correct. 

Given the values:

log 2 = 0.2614

log 3 = 0.3521

Formula Used:

Using the logarithm property:

Calculation:

We can write as 6 = 2 x 3

Using the property:

log 6 = log 2 + log 3

Substituting the given values:

⇒ log 6 = 0.6135

Hence the value log 6 is 0.6135.

Therefore, the correct option is (2)

Concept:

Formula:

log mn = n log m

Calculation:

Let y = x^log x

Taking log both sides, we get

⇒ log y = x^{log x}

⇒ log y = log x log  x            (∵ log mn = n log m)

Differentiating with respect to x, we get

⇒ 

⇒ 

⇒ 

⇒