Odd and even function MCQ Quiz - Objective Question with Answer for Odd and even function - Download Free PDF

Concept Used:

If f(x) is an odd function, then \(\int_{-a}^{a} f(x) dx = 0\).
If f(x) is an even function, then \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\).

Calculation:

Given:

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x + 1) dx\)

\(x^{13}\) is an odd function.

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

\(\tan^{15} x\) is an odd function.

1 is an even function.

⇒ \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x + 1) dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{13} dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^{15} x dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 dx\)

⇒ \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 dx = 2 \int_{0}^{\frac{\pi}{2}} 1 dx = 2[x]_0^{\frac{\pi}{2}} = 2(\frac{\pi}{2} - 0) = \pi\)

Hence option 3 is correct

Concept Used:

\(\sin^2 x = \frac{1 - \cos 2x}{2}\)
The function \(\sin^2 x\) is an even function.

Calculation:

\(\int_{-\pi/4}^{\pi/4} \sin^2 x \, dx = 2 \int_{0}^{\pi/4} \sin^2 x \, dx\) (since \(\sin^2 x\) is even)

⇒ \( 2 \int_{0}^{\pi/4} \frac{1 - \cos 2x}{2} \, dx\)

⇒ \( \int_{0}^{\pi/4} (1 - \cos 2x) \, dx\)

⇒ \( \left[ x - \frac{\sin 2x}{2} \right]_0^{\pi/4}\)

⇒ \( \left( \frac{\pi}{4} - \frac{\sin (\pi/2)}{2} \right) - \left( 0 - \frac{\sin 0}{2} \right)\)

⇒ \( \frac{\pi}{4} - \frac{1}{2}\)

Hence option 2 is correct

Concept

Wallis formula for definite integrals:

\(\int_{0}^{π/2} \sin^m x \cos^n x dx = \frac{(m-1)(m-3)...(n-1)(n-3)...}{(m+n)(m+n-2)...(2~or~1)} \cdot \frac{π}{2}\)

If m, n are both positive integers.

Calculation:

Given:

I = \(\int_{-2}^{2} x^4(4-x^2)^{\frac{7}{2}} dx\).

Let \(x = 2\sin\theta\), so \(dx = 2\cos\theta d\theta\).

When \(x = -2\), \(\sin\theta = -1\), so \(\theta = -\frac{π}{2}\).

When \(x = 2\), \(\sin\theta = 1\), so \(\theta = \frac{π}{2}\).

Then the integral becomes:

\(\int_{-\frac{π}{2}}^{\frac{π}{2}} (2\sin\theta)^4 (4 - (2\sin\theta)^2)^{\frac{7}{2}} 2\cos\theta d\theta\)

\(= \int_{-\frac{π}{2}}^{\frac{π}{2}} 16\sin^4\theta (4 - 4\sin^2\theta)^{\frac{7}{2}} 2\cos\theta d\theta\)

\(= \int_{-\frac{π}{2}}^{\frac{π}{2}} 16\sin^4\theta (4(1 - \sin^2\theta))^{\frac{7}{2}} 2\cos\theta d\theta\)

\(= \int_{-\frac{π}{2}}^{\frac{π}{2}} 16\sin^4\theta (4\cos^2\theta)^{\frac{7}{2}} 2\cos\theta d\theta\)

\(= \int_{-\frac{π}{2}}^{\frac{π}{2}} 16\sin^4\theta (2^2\cos^2\theta)^{\frac{7}{2}} 2\cos\theta d\theta\)

\(= \int_{-\frac{π}{2}}^{\frac{π}{2}} 16\sin^4\theta \cdot 2^7 \cos^7\theta \cdot 2\cos\theta d\theta\)

\(= 2^{12} \int_{-\frac{π}{2}}^{\frac{π}{2}} \sin^4\theta \cos^8\theta d\theta\)

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

\(= 2^{13} \int_{0}^{\frac{π}{2}} \sin^4\theta \cos^8\theta d\theta\)

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

\(2^{13} \int_{0}^{\frac{π}{2}} \sin^4\theta \cos^8\theta d\theta = 2^{13} \cdot \frac{3 \cdot 1\cdot 7\cdot5\cdot3\cdot1}{12 \cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{π}{2}\) = 28π 

Hence option 3 is correct.

\(I=\int^{\pi/2}_{-\pi/2}\frac{\cos\,x}{1+e^x}dx\) ...(i)

\(I=\int^{\pi/2}_{-\pi/2}\frac{\cos(\pi/2-\pi/2-x)}{1+e^{(\pi/2-\pi/2-x)}}dx\)

\(=\int^{\pi/2}_{-\pi/2}\frac{\cos(-x)}{1+e^{-x}}dx\)

\(I=\int^{\pi/2}_{-\pi/2}\frac{\cos\,x}{1+e^{-x}}dx\)

\(=\int^{\pi/2}_{-\pi/2}\frac{e^x\cos\,x}{1+e^x}dx\) ...(ii)

On adding Eqs. (i) and (ii), we get

\(2I=\int^{\pi/2}_{-\pi/2}\frac{(1+e^x)\cos\,x}{(1+e^x)}dx\)

\(=\int^{\pi/2}_{-\pi/2} \cos\,x\,dx\)

\(=2\int^{\pi/2}_{0}\cos\,x\,dx\)

[Since, \(\cos x\) is an even function.]

\(\therefore 2I=2[\sin\,x]^{\pi/2}_0=2(1-0)=2\)

\(\Rightarrow I=1\)

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:

\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x) + f(-x)][g(x) - g(-x)] dx\)

f and g are continuous functions.

Let \(I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x) + f(-x)][g(x) - g(-x)] dx\)

⇒ Let \(F(x) = f(x) + f(-x)\) and \(G(x) = g(x) - g(-x)\)

 \(F(-x) = f(-x) + f(-(-x)) = f(-x) + f(x) = F(x)\)

⇒ F(x) is an even function.

\(G(-x) = g(-x) - g(-(-x)) = g(-x) - g(x) = -[g(x) - g(-x)] = -G(x)\)

⇒ G(x) is an odd function.

\(I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} F(x) G(x) dx\)

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.

⇒ \(I = 0\)

Hence, option 4) 0 is the correct answer.

Concept:

Integral property:

\(\rm \mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},\rm If\;f\left( { - {\rm{x}}} \right) = f\left( x \right)}\\ {0,\;\rm If\;f\left( { - {\rm{x}}} \right) = - f\left( x \right)} \end{array}} \right.\)

Calculation:

I = \(\rm \int_{-2}^2\) (sin x + cos x)dx​

​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

I  = \(\rm 2\mathop \smallint \nolimits_0^2 cosx\;dx\)

= 2 

= 2[sin 2 - sin 0]

= 2sin 2

Concept:

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

Properties of definite integral

If f(x)  is even function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 2\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

If f(x)  is odd function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} =0\)

Calculation:

Let I = \(\displaystyle\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \rm \dfrac{sin^5 \ x \ cos^3 \ x}{x^4}dx\)

Let f(x) = \(\rm \dfrac{sin^5 \ x \ cos^3 \ x}{x^4}\)

Replaced x by -x, 

⇒ f(-x) = \(\rm \dfrac{sin^5 \ (-x) \ cos^3 \ (-x)}{(-x)^4}\)

As we know sin (-θ) = - sin θ and cos (-θ) = cos θ

= \(\rm \dfrac{-sin^5 \ x \ cos^3 \ x}{x^4}\)

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

Concept:

Definite Integrals:

For an odd function f(x), \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).

Calculation:

Let I = \(\rm\int_{-2}^{\ \ 2}\left(ax^3+bx+c\right)dx\).

⇒ I = \(\rm\int_{-2}^{\ \ 2}\left(ax^3+bx\right)dx\) + \(\rm\int_{-2}^{\ \ 2}(c)dx\)

Now, ax³ + bx is an odd function because if we substitute -x in place of x, we get -(ax³ + bx).

Using the property of definite integrals for odd functions, we get:

⇒ I = 0 + \(\rm\int_{-2}^{\ \ 2}(c)dx\)

⇒ I = 0 + \(\rm c\left[x\right]_{-2}^{\ \ \ 2}\)

⇒ I = 0 + c[ 2 - (-2)]

⇒ I = 4c.

Additional Information:

• \(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).
• If f(x) = f(2a - x), then \(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

A function f(x) is:

• Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
• Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.

Concept:

A function, f(x) is odd if f(-x) = - f(x) and the function is even if f(-x) = f(x)

If the function is even or odd and the interval is [-a, a], we can apply these rules:

When f(x) is even:

\(\rm \int_{-a}^{a} f(x)\ dx = 2\int_{0}^{a} f(x)\ dx\)

When f(x) is odd:

\(\rm \int_{-a}^{a} f(x)\ dx =0\)

Calculation:

The given function is, f(x) = sin⁹x

f(-x) = sin⁹(-x) 

= - sin⁹x

= - f(x)

Since the function is odd the value of \(\rm \int_{-\pi/2}^{\pi/2} sin^9x\ dx\) = 0

Concept:

Odd function: f(-x) = -f(x)

Even the function: f(-x) = f(x)

sin (-x) = -sin x

cos (-x) = cos x

tan (-x) = -tan x

Property of definte integral:

\(\rm\int_{-a}^{a}f(x)dx=\left\{\begin{matrix} \rm2\int_{0}^{a} f(x)dx&, \rm \text{ Even function }\\ 0&, \text{Odd function} \end{matrix}\right.\)

Calculation:

\(\rm f(x)=sinx+tanx \)

\(\rm f(-x)=sin(-x)+tan(-x) \)

\(\rm f(-x)=-[sin(x)+tan(x)] \)

Hence f(x) is odd function

\(\rm \therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin x+\tan x)dx=0 \)

Hence , option 3 is correct.

Concept: 

The rules for integrating even and odd functions ,

If the function is even or odd and the interval is [-a, a], we can apply these rules:

• When f(x) is even ⇔ \(\rm\int_{-a}^{a}f(x)dx= 2 \int_{0}^{a}f(x)dx\)
• When f(x) is odd ⇔ \(\rm\int_{-a}^{a}f(x)dx= 0\) 

​Calculation: 

Let f (x) = x⁵ sin⁴ x 

f (-x) = (-x)⁵ sin⁴ (-x) = -x⁵ (-sin x )⁴ = -x⁵ sin⁴ x .

⇒ f(-x) = - f(x) 

So, f(x) is an odd function . 

We knoiw that, when f(x) is odd ⇔ \(\rm\int_{-a}^{a}f(x)dx= 0\) 

Hence , \(\rm \int_{-\pi/2}^{\pi/2} x^{5}\sin^{4}x dx\) = 0 .  

The correct option is 4 .

Concept:

Odd and even function:

If a function f(x) such that f(-x) = f(x) then f(x) is an even function and if f(-x) = -f(x) then f(x) is an odd function.

Integration of odd and even function:

The following two cases hold for a differentiable function f(x).

If f(x) is an even function then \( \rm \displaystyle\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx\).

If f(x) is an odd function then \(\rm \displaystyle\int_{-a}^{a}f(x)dx = 0\).

Calculation:

Let \(\rm f(x) = x\sin x\), put -x instead of x then we get \(\rm f(-x) = (-x)\sin(-x) = x\sin x\).

Therefore, f(-x) = f(x) implies that f(x) is an even function.

Thus, the integral is given by,

\(\rm \begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\sin x dx &= 2\int_{0}^{\frac{\pi}{2}}x\sin x dx\\ &= 2\left[x(-\cos x) - \int(-\cos xdx)\right]_{0}^{\frac{\pi}{2}}\\ &= 2\left[-x\cos x + \sin x\right]_{0}^{\frac{\pi}{2}}\\ &= 2\left[\left(-\dfrac{\pi}{2}\cos\left(\dfrac{\pi}{2}\right) + \sin\left(\dfrac{\pi}{2}\right)\right)-\left(-0\cos 0+\sin 0\right)\right]\\ &= 2 \end{align*}\)

Hence, the value of the given integral is 2.

Concept:

A function is said to even if  f(-x) = f(x)

A function is said to be odd if  f(-x) = -f(x)

Calculation:

For an even function

\(\rm \Rightarrow \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\)

If function is odd then 

\(\rm \Rightarrow \int_{-a}^{a}f(x)dx=0\)

Hence , 4 is correct

Concept:

\(\displaystyle \int_{-a}^{a}f(x)dx=\left\{\begin{matrix} 0; if\hspace{0.2cm} f(-x)=-f(x)\\ 2\displaystyle\int_0^af(x)dx; if \hspace{0.2cm}f(-x)=f(x) \end{matrix}\right.\)

Calculation:

Given, I = \(\rm\displaystyle\int_{−1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1}dx\)

⇒ I = \(\rm\displaystyle\int_{−1}^1 \frac{x^3}{x^2+2|x|+1}dx+\int_{−1}^1 \frac{|x|+1}{x^2+2|x|+1}dx\)

⇒ I = 0 + \(2\rm\displaystyle\int_{0}^1 \frac{|x|+1}{x^2+2|x|+1}dx\) [∵ For first integral f(-x) = -f(x), and for the second integral f(-x) = f(x)]

⇒ I = \(2\rm\displaystyle\int_{0}^1 \frac{x+1}{x^2+2x+1}dx\) [∵ |x| = x for 0 ≤ x ≤ 1]

⇒ I = \(2\rm\displaystyle\int_{0}^1 \frac{x+1}{(x+1)^2}dx\)

⇒ I = \(2\rm\displaystyle\int_{0}^1 \frac{1}{x+1}dx\)

⇒ I = \(\displaystyle2\left[\log |x+1|\right]_0^1\)

⇒ I = 2[log 2 - log 1]

⇒ I = 2 log 2

∴ The value of the integral is 2 log 2 

Concept:

Odd function: f(-x) = -f(x)

Even the function: f(-x) = f(x)

sin (-x) = -sin x

cos (-x) = cos x

tan (-x) = -tan x

Property of definte integral:

\(\rm\int_{-a}^{a}f(x)dx=\left\{\begin{matrix} \rm2\int_{0}^{a} f(x)dx&, \rm \text{ Even function }\\ 0&, \text{Odd function} \end{matrix}\right.\)

Calculation:

\(\rm f(x)=sinx+tanx\)

\(\rm f(-x)=sin(-x)+tan(-x)\)

\(\rm f(-x)=-(sin(x)+tan(x))\)

Hence f(x) is odd function

\(\rm \therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin x+\tan x)dx=0\)

Hence , option 3 is correct.