Odd and even function MCQ Quiz in मल्याळम - Objective Question with Answer for Odd and even function - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 22, 2025

നേടുക Odd and even function ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Odd and even function MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Odd and even function MCQ Objective Questions

Top Odd and even function MCQ Objective Questions

Odd and even function Question 1:

22(sinx+cosx)dx

  1. 2sin2
  2. 2cos2
  3. 2
  4. 1
  5. 0

Answer (Detailed Solution Below)

Option 1 : 2sin2

Odd and even function Question 1 Detailed Solution

Concept used:

Integral property

aaf(x)dx={20af(x)dx,Iff(x)=f(x)0,Iff(x)=f(x)

Calculation:

22(sinx+cosx)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

⇒ 202cosxdx

⇒ 2 [sinx]02

⇒ 2[sin2 - sin0]

⇒ 2sin2

Odd and even function Question 2:

π2π2(x13+xcosx+tan15x+1)dx = ________.

  1. 1
  2. 2
  3. π
  4. 0

Answer (Detailed Solution Below)

Option 3 : π

Odd and even function Question 2 Detailed Solution

Concept Used:

If f(x) is an odd function, then aaf(x)dx=0.
If f(x) is an even function, then aaf(x)dx=20af(x)dx.

Calculation:

Given:

π2π2(x13+xcosx+tan15x+1)dx

x13 is an odd function.

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

tan15x is an odd function.

1 is an even function.

π2π2(x13+xcosx+tan15x+1)dx=π2π2x13dx+π2π2xcosxdx+π2π2tan15xdx+π2π21dx

π2π21dx=20π21dx=2[x]0π2=2(π20)=π

Hence option 3 is correct

Odd and even function Question 3:

Let f(x)=0xg(t)loge(1t1+t)dt, where g is a continuous odd function.  

If π/2π/2(f(x)+x2cosx1+ex)dx=(πα)2α, then α is equal to ____

Answer (Detailed Solution Below) 2

Odd and even function Question 3 Detailed Solution

Calculation

f(x)=0xg(t)ln(1t1+t)dt

⇒ f(x)=0xg(t)ln(1t1+t)dt

 f(x)=0xg(y)ln(1+y1y)dy

 0xg(y)ln(1y1+y)dy (g is odd) 

⇒ f(-x) = - f(x) ⇒ f is also odd

Now, 

I=π/2π/2(f(x)+x2cosx1+ex)dx...(1)

⇒ I=π/2π/2(f(x)+x2excosx1+ex)dx...(2)

Adding (1) and (2)

⇒ 2I=π/2π/2x2cosxdx=20π/2x2cosxdx

⇒ I=(x2sinx)0π/20π/22xsinxdx

 π242(xcosx+cosxdx)0π/2

 π242(0+1)=π242(π2)22

∴ α = 2

Odd and even function Question 4:

Let f(x) and ϕ(x) be two continuous functions on R satisfying ϕ(x)=axf(t)dt,a0 and another continuous function g(x) satisfying g(x + α) + g(x) = 0, ∀ x ∈ R, α > 0 and b2kg(t)dt is independent of b, then

I. If f(x) is an even function, then ϕ(x) is also even.

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

III. f(x) and ϕ(x) are independent.

Which of the above statement(s) is/are correct?

  1. Only I
  2. Only II
  3. Only III
  4. None of these

Answer (Detailed Solution Below)

Option 3 : Only III

Odd and even function Question 4 Detailed Solution

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 

ϕ(x)=axf(t)dt,a0

Let ∫ f(x) dx = F(x)

⇒ ϕ(x)=F(t)|ax

⇒ ϕ(x)=F(x)F(a)             ____(i)

If f(x) is an even function 

⇒ f(-x) = f(x)

⇒ F(x) is an odd function

⇒ F(-x) = - F(x)

Now, ϕ(x)=axf(t)dt

⇒ ϕ(x)=F(t)|ax

⇒ ϕ(x)=F(x)F(a)

⇒ ϕ(x)=F(x)F(a)

⇒ ϕ(x)=[F(x)+F(a)]     ___(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).

Odd and even function Question 5:

π/2  π/2esin2x.sin2n+1x dx = ?

  1. 1
  2. 0
  3. -1
  4. π

Answer (Detailed Solution Below)

Option 2 : 0

Odd and even function Question 5 Detailed Solution

Concept:

  • If f(x)is an even function, then a  af(x) dx=20af(x) dx.
  • If f(x)is an odd function, then a  af(x) dx=0.

 

Calculation:

Let f(x) = esin2x.sin2n+1x. It can also be written as f(x) = e(sin1x)2.(sinx)2n+1.

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

f(-x) = e[sin1(x)]2.[sin(x)]2n+1

Using the fact that sin(θ)=sinθ and sin1(x)=sin1x, we get:

⇒ f(-x) = e[sin1x]2.[sinx]2n+1

⇒ f(-x) = e(1)2(sin1x)2.(1)2n+1(sinx)2n+1

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

⇒ f(-x) = e(sin1x)2.(1)(sinx)2n+1

⇒ f(-x) = -f(x)

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

π/2  π/2esin2x.sin2n+1x dx = 0 is the answer.

Odd and even function Question 6:

What is aa(x2+sin x)dx equal to?

  1. a
  2. 0
  3. 2a23
  4. 2a33

Answer (Detailed Solution Below)

Option 4 : 2a33

Odd and even function Question 6 Detailed Solution

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 aaf(x)dx=20af(x)dx

If f(x)  is odd function then aaf(x)dx=0

 

Calculation:

Let I = aa(x2+sin x)dx

=aax2dx+aasin xdx

= I1 + I2

Now, 

I1 =aax2dx

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 aaf(x)dx=20af(x)dx

Therefore, I1 = 20ax2dx

=2×[x33]0a

=2×[a330]=2a33

Now, 

I2 = aasin xdx

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 aaf(x)dx=0

 

I = I1 + I2 = 2a33 + 0 = 2a33

Odd and even function Question 7:

22x4(4x2)72dx=

  1. 4π
  2. π16
  3. 28π
  4. 3π128

Answer (Detailed Solution Below)

Option 3 : 28π

Odd and even function Question 7 Detailed Solution

Concept

Wallis formula for definite integrals:

0π/2sinmxcosnxdx=(m1)(m3)...(n1)(n3)...(m+n)(m+n2)...(2 or 1)π2

If m, n are both positive integers.

Calculation:

Given:

I = 22x4(4x2)72dx.

Let x=2sinθ, so dx=2cosθdθ.

When x=2, sinθ=1, so θ=π2.

When x=2, sinθ=1, so θ=π2.

Then the integral becomes:

π2π2(2sinθ)4(4(2sinθ)2)722cosθdθ

=π2π216sin4θ(44sin2θ)722cosθdθ

=π2π216sin4θ(4(1sin2θ))722cosθdθ

=π2π216sin4θ(4cos2θ)722cosθdθ

=π2π216sin4θ(22cos2θ)722cosθdθ

=π2π216sin4θ27cos7θ2cosθdθ

=212π2π2sin4θcos8θdθ

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

=2130π2sin4θcos8θdθ

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

2130π2sin4θcos8θdθ=21331753112108642π2 = 28π 

Hence option 3 is correct.

Odd and even function Question 8:

Let f(x)=0xg(t)dt, where g is a non-zero even function. If f(x+5)=g(x), then 0xf(t)dt equals -

  1. x+55g(t)dt
  2. 5x+55g(t)dt
  3. 5x+5g(t)dt
  4. 25x+5g(t)dt

Answer (Detailed Solution Below)

Option 1 : x+55g(t)dt

Odd and even function Question 8 Detailed Solution

f(x)=0xg(t)dt

f(x)=0xg(t)dt

put t=u

=0xg(u)du

=0xg(u)d(u)=f(x)

f(x)=f(x)

f(x) is an odd function

Also f(5+x)=g(x)

f(5x)=g(x)=g(x)=f(5+x)

f(5x)=f(5+x)

Now

I=0xf(t)dt

t=u+5

I=5x5f(u+5)du

=5x5g(u)du

=5x5f(u)du

=f(x5)f(5)

=f(5x)+f(5)

=f(5)f(5+x)

=5+x5f(t)dt=5+x5g(t)dt

Odd and even function Question 9:

Let f(x)=ax(a>0) be written as f(x)=f1(x)+f2(x), where f1(x) is an even function of f2(x) is an odd function. Then f1(x+y)+f1(xy) equals

  1. 2f1(x)f1(y)
  2. 2f1(x)f2(y)
  3. 2f1(x+y)f2(xy)
  4. 2f1(x+y)f1(xy)

Answer (Detailed Solution Below)

Option 1 : 2f1(x)f1(y)

Odd and even function Question 9 Detailed Solution

f(x)=ax,a>0

f(x)=ax+ax+axax2

f1(x)=ax+ax2

f2(x)=axax2

ax+y+axy2+axy+ax+y2

=f1(x)×2f1(y)

=2f1(x)f1(y)

Odd and even function Question 10:

Let f : R → R and g : R → R be continuous functions. Then the value of the integral π2π2[f(x)+f(x)][g(x)g(x)]dx is

  1. π 
  2. 1
  3. -1
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Odd and even function Question 10 Detailed Solution

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:

π2π2[f(x)+f(x)][g(x)g(x)]dx

f and g are continuous functions.

Let I=π2π2[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=π2π2F(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.

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