Odd and even function MCQ Quiz in తెలుగు - Objective Question with Answer for Odd and even function - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept used:

Integral property

Calculation:

Let f₁(x) = sinx and f₂(x) = cosx

f₁(x) = sinx

f₁(-x) = sin(-x) = -sinx = -f₁(x)

f₂(x) = cosx

f₂(-x) = cox(-x) = cosx = f₂(x)

By integration property f₁(x) = 0

⇒ 

⇒ 2 

⇒ 2[sin2 - sin0]

⇒ 2sin2

Concept Used:

If f(x) is an odd function, then .
If f(x) is an even function, then .

Calculation:

Given:

is an odd function.

is an odd function (since x is odd and cos x is even).

is an odd function.

1 is an even function.

⇒

⇒

Hence option 3 is correct

Calculation

⇒ 

⇒ 

⇒ 

⇒ f(-x) = - f(x) ⇒ f is also odd

Now, 

⇒ 

Adding (1) and (2)

⇒ 

⇒ 

⇒ 

⇒ 

∴ α = 2

Concept:

The indefinite integral of an even function is an odd function plus constant.

That is , If f(x) is an even function, and f(x) dx = F(x) + C 

Then F(x) is an odd function.

Calculation:

Given, f(x) and ϕ(x) be two continuous functions on R satisfying 

Let ∫ f(x) dx = F(x)

⇒ 

⇒              ____(i)

If f(x) is an even function 

⇒ f(-x) = f(x)

⇒ F(x) is an odd function

⇒ F(-x) = - F(x)

Now, 

⇒ 

⇒ 

⇒ 

⇒      ___(ii)

Comparing (i) and (ii), we get

ϕ(x) is neither odd nor even function when f(x) is even.

So, ϕ(x) and f(x) are independent function.

∴ The correct answer is option (3).

Concept:

• If f(x)is an even function, then .
• If f(x)is an odd function, then .

Calculation:

Let f(x) = . It can also be written as f(x) = .

Let us find the expression for f(-x) to compare and check whether it is an even function or odd.

f(-x) = 

Using the fact that  and , we get:

⇒ f(-x) = 

⇒ f(-x) = 

Since 2n+1 is odd for any value of n, we get:

⇒ f(-x) = 

⇒ f(-x) = -f(x)

This means that f(x) is an odd function and therefore  must be 0, for any real number a.

∴ = 0 is the answer.

Concept:

If f(x)  is even function then f(-x) = f(x)

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then 

If f(x)  is odd function then 

Calculation:

Let I = 

= I₁ + I₂

Now, 

I₁ 

Here f(x) = x²

Replace x by -x, we get

⇒ f(-x) = (-x)² = x²

⇒ f(-x) = f(x)

So, f(x) is even function.

As we know, If f(x) even function then 

Therefore, I₁ = 

Now, 

I₂ = 

Here f(x) = sin x

Replace x by -x, we get

⇒ f(-x) = sin (-x) = -sin x                (∵ sin (-θ) = - sin θ)

⇒ f(-x) = -f(x)

So, f(x) is odd function.

As we know, If f(x) even function then 

I = I₁ + I₂ =  + 0 = 

Concept

Wallis formula for definite integrals:

If m, n are both positive integers.

Calculation:

Given:

I = .

Let , so .

When , , so .

When , , so .

Then the integral becomes:

Since the integrand is an even function, we can write:

In our case, m = 4 and n = 8, so applying Wallis formula:

 = 28π 

Hence option 3 is correct.

put

is an odd function

Also

Now

Concept Used:

Properties of definite integrals.

If f(x) is an even function, f(-x) = f(x).

If g(x) is an odd function, g(-x) = -g(x).

Calculation:

Given:

f and g are continuous functions.

Let

⇒ Let and

⇒ F(x) is an even function.

⇒ G(x) is an odd function.

Since F(x) is even and G(x) is odd, F(x)G(x) is an odd function.

The integral of an odd function over a symmetric interval [-a, a] is 0.

⇒

Hence, option 4) 0 is the correct answer.