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

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

Calculation:

\( \mathop \smallint \nolimits_{ - 2}^2 \left( {sinx + cosx} \right)dx\)

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

⇒ \(2\mathop \smallint \nolimits_0^2 cosx\;dx\)

⇒ 2 \(\rm[sinx]^2_0\)

⇒ 2[sin2 - sin0]

⇒ 2sin2

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

Calculation

\(f(x)=\int_0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t\)

⇒ \(\mathrm{f}(-\mathrm{x})=\int_0^{-\mathrm{x}} \mathrm{g}(\mathrm{t}) \ln \left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right) \mathrm{dt}\)

⇒ \(f(-x)=-\int_0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y\)

⇒ \(-\int_0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y \text { (g is odd) }\)

⇒ f(-x) = - f(x) ⇒ f is also odd

Now, 

\(\rm I=\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x \quad...(1)\)

⇒ \(\rm I=\int_{-\pi / 2}^{\pi / 2}\left(f(-x)+\frac{x^2 e^x \cos x}{1+e^x}\right) d x\quad...(2)\)

Adding (1) and (2)

⇒ \(\rm 2 I=\int_{-\pi / 2}^{\pi / 2} x^2 \cos x d x=2 \int_0^{\pi / 2} x^2 \cos x d x\)

⇒ \(\rm I=\left(x^2 \sin x\right)_0^{\pi / 2}-\int_0^{\pi / 2} 2 x \sin x d x\)

⇒ \(\frac{\pi^2}{4}-2\left(-x \cos x+\int \cos x d x\right)_0^{\pi / 2}\)

⇒ \(\frac{\pi^2}{4}-2(0+1)=\frac{\pi^2}{4}-2 \Rightarrow\left(\frac{\pi}{2}\right)^2-2\)

∴ 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 

\(ϕ (x)=\displaystyle ∫_a^x f(t)\:dt, a \ne 0\)

Let ∫ f(x) dx = F(x)

⇒ \(ϕ (x)=\displaystyle F(t) |_{a}^{x}\)

⇒ \(ϕ (x) = F(x) - F(a)\)             ____(i)

If f(x) is an even function 

⇒ f(-x) = f(x)

⇒ F(x) is an odd function

⇒ F(-x) = - F(x)

Now, \(ϕ (-x)=\displaystyle ∫_a^{-x} f(t)\:dt\)

⇒ \(ϕ (-x)=\displaystyle F(t) |_{a}^{-x}\)

⇒ \(ϕ (-x) = F(-x) - F(a)\)

⇒ \(ϕ (-x) = -F(x) - F(a)\)

⇒ \(ϕ (-x) = -[F(x) + F(a)]\)     ___(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).

Concept:

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

Calculation:

Let f(x) = \(\rm e^{\sin^{-2}x}.\sin^{2n+1}x\). It can also be written as f(x) = \(\rm {{e}^{{{\left( {{\sin }^{-1}}x \right)}^{2}}}}.{{\left( \sin x \right)}^{2n+1}}\).

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

f(-x) = \(\rm {{e}^{{{\left[ {{\sin }^{-1}}(-x) \right]}^{2}}}}.{{\left[ \sin (-x) \right]}^{2n+1}}\)

Using the fact that \(\rm \sin \left( -\theta \right)=-\sin \theta\) and \(\rm {{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x\), we get:

⇒ f(-x) = \(\rm {{e}^{{{\left[ -{{\sin }^{-1}}x \right]}^{2}}}}.{{\left[ -\sin x \right]}^{2n+1}}\)

⇒ f(-x) = \(\rm {{e}^{{{\left( -1 \right)}^{2}}{{\left( {{\sin }^{-1}}x \right)}^{2}}}}.{{\left( -1 \right)}^{2n+1}}{{\left( \sin x \right)}^{2n+1}}\)

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

⇒ f(-x) = \(\rm {{e}^{{{\left( {{\sin }^{-1}}x \right)}^{2}}}}.\left( -1 \right){{\left( \sin x \right)}^{2n+1}}\)

⇒ f(-x) = -f(x)

This means that f(x) is an odd function and therefore \(\rm \int\limits_{-a}^{a}f(x)\ dx\) must be 0, for any real number a.

∴ \(\rm \int_{-\pi /2}^{\ \ \pi /2}e^{\sin^{-2x}}.\sin^{2n+1}x\ dx\) = 0 is the answer.

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 = \(\rm\displaystyle\int_{-a}^a (x^2+sin \ x) dx\)

\(= \rm\displaystyle\int_{-a}^a x^2 \; dx+ \displaystyle\int_{-a}^a sin \ x\; dx\)

= I₁ + I₂

Now, 

I₁ \(= \rm\displaystyle\int_{-a}^a x^2 \; dx \)

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

Therefore, I₁ = \(2\rm\displaystyle\int_{0}^a x^2 \; dx \)

\(= \rm 2\times [\frac{x^{3}}{3}]_0^a\)

\(= \rm 2\times [\frac{a^{3}}{3}-0] = \frac{2a^{3}}{3}\)

Now, 

I₂ = \(\rm \displaystyle\int_{-a}^a sin \ x\; dx\)

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 \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} =0\)

I = I₁ + I₂ = \(\rm \frac{2a^{3}}{3}\) + 0 = \(\rm \frac{2a^{3}}{3}\)

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.

\( f(x) = \int_0^x g(t) dt \)

\( f(-x) = \int_0^{-x} g(t)dt \)

put \( t = -u \)

\( = - \int^x_0 g(-u) du \)

\( = - \int^x_0 g(u) d (u)=-f(x) \)

\( \Rightarrow f(-x) = -f(x) \)

\( \Rightarrow f(x) \) is an odd function

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

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

\( \Rightarrow f(5-x) = f(5+x) \)

Now

\( I = \int_0^x f(t) dt \)

\( t = u + 5 \)

\( I = \int_{-5}^{x-5} f(u +5)du \)

\( =\int_{-5}^{x-5} g(u) du \)

\( =\int_{-5}^{x-5} f'(u) du \)

\( =f (x-5) - f(-5) \)

\( =-f(5-x) + f(5) \)

\( = f(5) - f(5+x) \)

\( =\int^5_{5+x} f'(t) dt = \int _{5+x}^5 g(t) dt \)

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.