Comprehensive IGNOU BMTC-131 Solved Assignment 2024 for B.Sc (G) CBCS Students

IGNOU BMTC-131 Solved Assignment 2024 | B.Sc (G) CBCS

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Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTC-131 Assignment Question Paper 2024

bmtc-131-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-131 Solved Assignment 2024

State whether the following statements are True or False? Justify your answers with the help of a short proof or a counter example:

a) The function

f

, defined by

f (x) = \cos x + \sin x

, is an odd function.

\frac{d}{d x} [\int_{2}^{e^{x}} \ln t d t] = x - \ln 2

c) The function

f

, defined by

f (x) = | x - 2 |

, is differentiable in

[0, 1]

y = x^{2} - 3 x^{3}

has no points of inflection.

y = - x^{2}

is increasing in

[- 5, - 3]

a) Find $\frac{d y}{d x}$ , if $y = x \sin^{- 1} x + \sqrt{1 - x^{2}}$ .

b) Evaluate

\int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x

c) Find

lim_{x \to 1} \frac{x^{3} - 3 x + 2}{x^{2} - 5 x + 4}

Trace the curve $y^{2} = x^{2} (x + 1)$ by the stating all the properties used to trace it.
a) Find the length of the curve given by $x = t^{2}, y = 2 t^{2}$ in $0 \leq t \leq 2$ .

b) Find the angle between the curves

y^{2} = a x

and

a y^{2} = x^{3} (a > 0)

, at the points of intersection other than the origin.

a) Evaluate

\int \frac{x^{2} d x}{(x - 3) (x - 5) (x - 7)}

b) Use Simpson’s method to approximate

\int_{0}^{8} (x^{2} - x + 3) d x

with 8 sub-intervals.

c) Find the derivatives of

\ln (1 + x^{2})

w.r.t.

\tan^{- 1} x

a) The curve $a y^{2} = x (x - a)^{2}, a > 0$ has a loop between $x = 0$ and $x = a$ . Find the area of this loop.

b) Obtain the largest possible domain, and corresponding range, of the function

f

, defined by

f (x) = \frac{x - 2}{3 - x}

c) Expand

e^{2 x}

in powers of

(x - 1)

, up to four terms.

a) Verify Rolle’s theorem for the function $f$ , defined by $f (x) = x (x - 2) e^{- x}$ , on the interval $[0, 2]$ .

b) Is the function

f

, defined by

f (x) = \frac{x^{2} - 5 x + 4}{x^{2} - 16}, x \neq 4

f (4) = 0

, continuous at

x = 4

? Give reasons for your answer.

c) Evaluate

\int_{0}^{1} x^{2} e^{3 x} d x

a) Using Trapezoidal rule, calculate $\int_{0}^{1} \frac{d x}{1 + x^{2}}$ by dividing the interval $[0, 1]$ in 5 equal subintervals. Hence evaluate $π$ .

b) Find

\frac{d y}{d x}

, if

y = x^{\sin x} + (\sin x)^{x}

c) Find the perimeter of the cardioid

r = a (1 - \cos θ)

a) If the first three non-zero terms of Maclaurin’s series for $\sin x$ are used to approximate $\sin π / 2$ , show that the error is less than $1 / 50$ .

b) Find the least value of

a^{2} \sec^{2} x + b^{2} {cosec}^{2} x

, where

a > 0, b > 0

c) Evaluate

lim_{x \to 0} (1 + x)^{1 / x}

a) Find the slope of the normal to the curve $y = x^{3}$ at $(\frac{1}{2}, \frac{1}{8})$ .

b) Find the points of inflexion of the curve

y = \frac{a^{2} x}{x^{2} + a^{2}}

. Also, show that they lie on a straight line.

c) Evaluate

\int e^{x} \cdot \frac{x^{2} - x + 1}{{(1 + x^{2})}^{3 / 2}} d x

BMTC-131 Sample Solution 2024

bmtc-131-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

BMTC-131 Solved Assignment 2024

State whether the following statements are True or False? Justify your answers with the help of a short proof or a counter example:

a) The function

f

, defined by

f (x) = \cos x + \sin x

, is an odd function.

Answer:

To determine whether the statement "The function

f

, defined by

f (x) = \cos x + \sin x

, is an odd function" is true, we need to understand the definition of an odd function and then apply it to

f (x)

Definition of an Odd Function

A function

f (x)

is said to be odd if it satisfies the condition

f (- x) = - f (x)

for all

x

in its domain. This means that the function is symmetric with respect to the origin.

Applying the Definition to $f (x)$

The given function is

f (x) = \cos x + \sin x

. To check if it’s odd, we compute

f (- x)

and compare it with

- f (x)

Compute $f (- x)$ :

$f (- x) = \cos (- x) + \sin (- x)$

Using the even property of cosine and the odd property of sine:

$f (- x) = \cos x - \sin x$
Compare $f (- x)$ with $- f (x)$ :

$- f (x) = - (\cos x + \sin x) = - \cos x - \sin x$

Clearly, $f (- x) = \cos x - \sin x$ is not equal to $- f (x) = - \cos x - \sin x$ .

Conclusion

Since

f (- x) \neq - f (x)

for the function

f (x) = \cos x + \sin x

, the function is not an odd function. Therefore, the statement is false.

\frac{d}{d x} [\int_{2}^{e^{x}} \ln t d t] = x - \ln 2

Answer:

To determine the truth of the statement

\frac{d}{d x} [\int_{2}^{e^{x}} \ln t d t] = x - \ln 2,

we need to use the Fundamental Theorem of Calculus and the chain rule.

Fundamental Theorem of Calculus and Chain Rule

The Fundamental Theorem of Calculus states that if

F (x)

is an antiderivative of

f (x)

, then

\frac{d}{d x} [\int_{a}^{g (x)} f (t) d t] = f (g (x)) \cdot g^{'} (x) .

In this case,

f (t) = \ln t

and

g (x) = e^{x}

. The derivative of

g (x) = e^{x}

g^{'} (x) = e^{x}

Applying the Theorem

Applying the theorem to the given integral, we get:

\frac{d}{d x} [\int_{2}^{e^{x}} \ln t d t] = \ln (e^{x}) \cdot e^{x} .

Since

\ln (e^{x}) = x

, this simplifies to:

x \cdot e^{x} .

Comparing with the Given Statement

The given statement is:

\frac{d}{d x} [\int_{2}^{e^{x}} \ln t d t] = x - \ln 2.

However, from our calculation, we found that the derivative is actually

x \cdot e^{x}

, not

x - \ln 2

Conclusion

The statement is false. The correct derivative of the integral

\int_{2}^{e^{x}} \ln t d t

with respect to

x

x \cdot e^{x}

, not

x - \ln 2

c) The function

f

, defined by

f (x) = | x - 2 |

, is differentiable in

[0, 1]

Answer:

To determine whether the statement "The function

f

, defined by

f (x) = | x - 2 |

, is differentiable in

[0, 1]

" is true, we need to examine the differentiability of the function

f (x)

within the interval

[0, 1]

Definition of Differentiability

A function is differentiable at a point if its derivative exists at that point. For a function to be differentiable on an interval, it must be differentiable at every point in that interval.

Analyzing the Function $f (x) = | x - 2 |$

The function

f (x) = | x - 2 |

can be expressed as:

$f (x) = 2 - x$ for $x \leq 2$
$f (x) = x - 2$ for $x > 2$

In the interval

[0, 1]

f (x) = 2 - x

, since all values of

x

in this interval are less than 2.

Differentiability in $[0, 1]$

For

x \in [0, 1]

f (x) = 2 - x

, which is a linear function. The derivative of

f (x)

in this interval is

f^{'} (x) = - 1

, which exists for all

x

[0, 1]

Conclusion

Since the derivative of

f (x) = | x - 2 |

exists and is constant (

- 1

) throughout the interval

[0, 1]

, the function

f

is differentiable in

[0, 1]

. Therefore, the statement is true.

y = x^{2} - 3 x^{3}

has no points of inflection.

Answer:

To determine whether the function

y = x^{2} - 3 x^{3}

has any points of inflection, we need to check the concavity of the function. A point of inflection occurs when the concavity of the function changes from concave up to concave down or vice versa. This change in concavity is characterized by the second derivative of the function.

A point of inflection occurs at

x = c

if and only if the second derivative

y^{″} (c)

changes sign at that point. Specifically,

y^{″} (c) = 0

and the sign of

y^{″} (x)

changes as

x

crosses

c

Let’s find the first and second derivatives of the function

y = x^{2} - 3 x^{3}

First derivative:

$y^{'} = 2 x - 9 x^{2}$
Second derivative:

$y^{″} = 2 - 18 x$

Now, we need to find the values of

x

where

y^{″} (x) = 0

2 - 18 x = 0

Solving for

x

18 x = 2 ⟹ x = \frac{2}{18} = \frac{1}{9}

So,

y^{″} (x) = 0

x = \frac{1}{9}

Now, we need to determine the sign of

y^{″} (x)

in the intervals around

x = \frac{1}{9}

For $x < \frac{1}{9}$ , $y^{″} (x) = 2 - 18 x > 0$ (since $x$ is positive).
For $x > \frac{1}{9}$ , $y^{″} (x) = 2 - 18 x < 0$ (since $x$ is positive).

Since the sign of

y^{″} (x)

changes from positive to negative as

x

crosses

x = \frac{1}{9}

, we have a change in concavity at that point.

Therefore, the function

y = x^{2} - 3 x^{3}

does have a point of inflection at

x = \frac{1}{9}

So, the statement "

y = x^{2} - 3 x^{3}

has no points of inflection" is false.

y = - x^{2}

is increasing in

[- 5, - 3]

Answer:

To determine whether the function

y = - x^{2}

is increasing in the interval

[- 5, - 3]

, we need to analyze the behavior of the function within this interval.

A function is considered increasing on an interval if, for any two points

a

and

b

in the interval where

a < b

, the function values satisfy

f (a) \leq f (b)

Let’s evaluate the function

y = - x^{2}

within the interval

[- 5, - 3]

At $x = - 5$ : $y (- 5) = - (- 5)^{2} = - 25$
At $x = - 3$ : $y (- 3) = - (- 3)^{2} = - 9$

Since

y (- 5) = - 25

is less than

y (- 3) = - 9

, we have

f (a) < f (b)

for

a = - 5

and

b = - 3

, which means the function is decreasing in this interval.

Therefore, the statement "

y = - x^{2}

is increasing in

[- 5, - 3]

" is false. The function

y = - x^{2}

is actually decreasing in the interval

[- 5, - 3]

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How do I access the Complete Solution?

You can access the Complete Solution through our app, which can be downloaded using this link:

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Simply click “Install” to download and install the app, and then follow the instructions to purchase the required assignment solution. Currently, the app is only available for Android devices. We are working on making the app available for iOS in the future, but it is not currently available for iOS devices.

Is this Complete Solution for IGNOU Assignments?

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. Valid from January 1, 2023 to December 31, 2023.

Is the Complete Solution aligned with the IGNOU requirements?

Yes, the Complete Solution is aligned with the IGNOU requirements and has been solved accordingly.

Is the Complete Solution guaranteed error-free?

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

Can I access the Complete Solution anytime?

As of now, you have access to the Complete Solution for a period of 6 months after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

What if I need help with a specific assignment question?

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

Can I access the educational materials on multiple devices?

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

How do I make a payment for accessing the solutions?

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

What are the instructions for formatting my assignments?

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

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IGNOU BMTC-131 Solved Assignment 2024 | B.Sc (G) CBCS

IGNOU BMTC-131 Assignment Question Paper 2024

bmtc-131-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-131 Sample Solution 2024

bmtc-131-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

Definition of an Odd Function

Applying the Definition to $f (x)$

Conclusion

Fundamental Theorem of Calculus and Chain Rule

Applying the Theorem

Comparing with the Given Statement

Conclusion

Definition of Differentiability

Analyzing the Function $f (x) = | x - 2 |$

Differentiability in $[0, 1]$

Conclusion

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IGNOU BMTC-131 Solved Assignment 2024 | B.Sc (G) CBCS

IGNOU BMTC-131 Assignment Question Paper 2024

bmtc-131-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-131 Sample Solution 2024

bmtc-131-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

Definition of an Odd Function

Applying the Definition to f ( x ) f ( x ) f(x)f(x)f(x)

Conclusion

Fundamental Theorem of Calculus and Chain Rule

Applying the Theorem

Comparing with the Given Statement

Conclusion

Definition of Differentiability

Analyzing the Function f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2|

Differentiability in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]

Conclusion

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Applying the Definition to $f (x)$

Analyzing the Function $f (x) = | x - 2 |$

Differentiability in $[0, 1]$