BMTC-131 Solved Assignment 2024

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BMTC-131 Solved Assignment 2024 Sample Solution
bmtc-131-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

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

BMTC-131 Solved Assignment 2024
  1. 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 f fff, defined by f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x)=\cos x+\sin xf(x)=cos⁡x+sin⁡x, is an odd function.
Answer:
To determine whether the statement "The function f f fff, defined by f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x) = \cos x + \sin xf(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 ) f ( x ) f(x)f(x)f(x).

Definition of an Odd Function

A function f ( x ) f ( x ) f(x)f(x)f(x) is said to be odd if it satisfies the condition f ( − x ) = − f ( x ) f ( − x ) = − f ( x ) f(-x)=-f(x)f(-x) = -f(x)f(−x)=−f(x) for all x x xxx in its domain. This means that the function is symmetric with respect to the origin.

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

The given function is f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x) = \cos x + \sin xf(x)=cos⁡x+sin⁡x. To check if it’s odd, we compute f ( − x ) f ( − x ) f(-x)f(-x)f(−x) and compare it with − f ( x ) − f ( x ) -f(x)-f(x)−f(x).
  1. Compute f ( − x ) f ( − x ) f(-x)f(-x)f(−x):
    f ( − x ) = cos ⁡ ( − x ) + sin ⁡ ( − x ) f ( − x ) = cos ⁡ ( − x ) + sin ⁡ ( − x ) f(-x)=cos(-x)+sin(-x)f(-x) = \cos(-x) + \sin(-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 f ( − x ) = cos ⁡ x − sin ⁡ x f(-x)=cos x-sin xf(-x) = \cos x – \sin xf(−x)=cos⁡x−sin⁡x
  2. Compare f ( − x ) f ( − x ) f(-x)f(-x)f(−x) with − f ( x ) − f ( x ) -f(x)-f(x)−f(x):
    − f ( x ) = − ( cos ⁡ x + sin ⁡ x ) = − cos ⁡ x − sin ⁡ x − f ( x ) = − ( cos ⁡ x + sin ⁡ x ) = − cos ⁡ x − sin ⁡ x -f(x)=-(cos x+sin x)=-cos x-sin x-f(x) = -(\cos x + \sin x) = -\cos x – \sin x−f(x)=−(cos⁡x+sin⁡x)=−cos⁡x−sin⁡x
    Clearly, f ( − x ) = cos ⁡ x − sin ⁡ x f ( − x ) = cos ⁡ x − sin ⁡ x f(-x)=cos x-sin xf(-x) = \cos x – \sin xf(−x)=cos⁡x−sin⁡x is not equal to − f ( x ) = − cos ⁡ x − sin ⁡ x − f ( x ) = − cos ⁡ x − sin ⁡ x -f(x)=-cos x-sin x-f(x) = -\cos x – \sin x−f(x)=−cos⁡x−sin⁡x.

Conclusion

Since f ( − x ) ≠ − f ( x ) f ( − x ) ≠ − f ( x ) f(-x)!=-f(x)f(-x) \neq -f(x)f(−x)≠−f(x) for the function f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x) = \cos x + \sin xf(x)=cos⁡x+sin⁡x, the function is not an odd function. Therefore, the statement is false.
b) d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2 d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2 (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2\frac{\mathrm{d}}{\mathrm{dx}}\left[\int_2^{\mathrm{e}^{\mathrm{x}}} \ln \mathrm{tdt}\right]=\mathrm{x}-\ln 2ddx[∫2exln⁡tdt]=x−ln⁡2.
Answer:
To determine the truth of the statement
d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2 , d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2 , (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2,\frac{d}{dx}\left[\int_2^{e^x} \ln t \, dt\right] = x – \ln 2,ddx[∫2exln⁡tdt]=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 ) F ( x ) F(x)F(x)F(x) is an antiderivative of f ( x ) f ( x ) f(x)f(x)f(x), then
d d x [ ∫ a g ( x ) f ( t ) d t ] = f ( g ( x ) ) ⋅ g ′ ( x ) . d d x ∫ a g ( x )   f ( t ) d t = f ( g ( x ) ) ⋅ g ′ ( x ) . (d)/(dx)[int_(a)^(g(x))f(t)dt]=f(g(x))*g^(‘)(x).\frac{d}{dx}\left[\int_a^{g(x)} f(t) \, dt\right] = f(g(x)) \cdot g'(x).ddx[∫ag(x)f(t)dt]=f(g(x))⋅g′(x).
In this case, f ( t ) = ln ⁡ t f ( t ) = ln ⁡ t f(t)=ln tf(t) = \ln tf(t)=ln⁡t and g ( x ) = e x g ( x ) = e x g(x)=e^(x)g(x) = e^xg(x)=ex. The derivative of g ( x ) = e x g ( x ) = e x g(x)=e^(x)g(x) = e^xg(x)=ex is g ′ ( x ) = e x g ′ ( x ) = e x g^(‘)(x)=e^(x)g'(x) = e^xg′(x)=ex.

Applying the Theorem

Applying the theorem to the given integral, we get:
d d x [ ∫ 2 e x ln ⁡ t d t ] = ln ⁡ ( e x ) ⋅ e x . d d x ∫ 2 e x   ln ⁡ t d t = ln ⁡ ( e x ) ⋅ e x . (d)/(dx)[int_(2)^(e^(x))ln tdt]=ln(e^(x))*e^(x).\frac{d}{dx}\left[\int_2^{e^x} \ln t \, dt\right] = \ln(e^x) \cdot e^x.ddx[∫2exln⁡tdt]=ln⁡(ex)⋅ex.
Since ln ⁡ ( e x ) = x ln ⁡ ( e x ) = x ln(e^(x))=x\ln(e^x) = xln⁡(ex)=x, this simplifies to:
x ⋅ e x . x ⋅ e x . x*e^(x).x \cdot e^x.x⋅ex.

Comparing with the Given Statement

The given statement is:
d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2. d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2. (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2.\frac{d}{dx}\left[\int_2^{e^x} \ln t \, dt\right] = x – \ln 2.ddx[∫2exln⁡tdt]=x−ln⁡2.
However, from our calculation, we found that the derivative is actually x ⋅ e x x ⋅ e x x*e^(x)x \cdot e^xx⋅ex, not x − ln ⁡ 2 x − ln ⁡ 2 x-ln 2x – \ln 2x−ln⁡2.

Conclusion

The statement is false. The correct derivative of the integral ∫ 2 e x ln ⁡ t d t ∫ 2 e x   ln ⁡ t d t int_(2)^(e^(x))ln tdt\int_2^{e^x} \ln t \, dt∫2exln⁡tdt with respect to x x xxx is x ⋅ e x x ⋅ e x x*e^(x)x \cdot e^xx⋅ex, not x − ln ⁡ 2 x − ln ⁡ 2 x-ln 2x – \ln 2x−ln⁡2.
c) The function f f fff, defined by f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x)=|x-2|f(x)=|x−2|, is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
Answer:
To determine whether the statement "The function f f fff, defined by f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2|, is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]" is true, we need to examine the differentiability of the function f ( x ) f ( x ) f(x)f(x)f(x) within the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][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 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2|

The function f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2| can be expressed as:
  • f ( x ) = 2 − x f ( x ) = 2 − x f(x)=2-xf(x) = 2 – xf(x)=2−x for x ≤ 2 x ≤ 2 x <= 2x \leq 2x≤2
  • f ( x ) = x − 2 f ( x ) = x − 2 f(x)=x-2f(x) = x – 2f(x)=x−2 for x > 2 x > 2 x > 2x > 2x>2
In the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1], f ( x ) = 2 − x f ( x ) = 2 − x f(x)=2-xf(x) = 2 – xf(x)=2−x, since all values of x x xxx in this interval are less than 2.

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

For x ∈ [ 0 , 1 ] x ∈ [ 0 , 1 ] x in[0,1]x \in [0, 1]x∈[0,1], f ( x ) = 2 − x f ( x ) = 2 − x f(x)=2-xf(x) = 2 – xf(x)=2−x, which is a linear function. The derivative of f ( x ) f ( x ) f(x)f(x)f(x) in this interval is f ′ ( x ) = − 1 f ′ ( x ) = − 1 f^(‘)(x)=-1f'(x) = -1f′(x)=−1, which exists for all x x xxx in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1].

Conclusion

Since the derivative of f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2| exists and is constant ( − 1 − 1 -1-1−1) throughout the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1], the function f f fff is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]. Therefore, the statement is true.
d) y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y=x^2-3 x^3y=x2−3x3 has no points of inflection.
Answer:
To determine whether the function y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y = x^2 – 3x^3y=x2−3x3 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 x = c x=cx = cx=c if and only if the second derivative y ″ ( c ) y ″ ( c ) y^(″)(c)y”(c)y″(c) changes sign at that point. Specifically, y ″ ( c ) = 0 y ″ ( c ) = 0 y^(″)(c)=0y”(c) = 0y″(c)=0 and the sign of y ″ ( x ) y ″ ( x ) y^(″)(x)y”(x)y″(x) changes as x x xxx crosses c c ccc.
Let’s find the first and second derivatives of the function y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y = x^2 – 3x^3y=x2−3x3:
  1. First derivative:
    y ′ = 2 x − 9 x 2 y ′ = 2 x − 9 x 2 y^(‘)=2x-9x^(2)y’ = 2x – 9x^2y′=2x−9x2
  2. Second derivative:
    y ″ = 2 − 18 x y ″ = 2 − 18 x y^(″)=2-18 xy” = 2 – 18xy″=2−18x
Now, we need to find the values of x x xxx where y ″ ( x ) = 0 y ″ ( x ) = 0 y^(″)(x)=0y”(x) = 0y″(x)=0:
2 − 18 x = 0 2 − 18 x = 0 2-18 x=02 – 18x = 02−18x=0
Solving for x x xxx:
18 x = 2 ⟹ x = 2 18 = 1 9 18 x = 2 ⟹ x = 2 18 = 1 9 18 x=2Longrightarrowx=(2)/(18)=(1)/(9)18x = 2 \implies x = \frac{2}{18} = \frac{1}{9}18x=2⟹x=218=19
So, y ″ ( x ) = 0 y ″ ( x ) = 0 y^(″)(x)=0y”(x) = 0y″(x)=0 at x = 1 9 x = 1 9 x=(1)/(9)x = \frac{1}{9}x=19.
Now, we need to determine the sign of y ″ ( x ) y ″ ( x ) y^(″)(x)y”(x)y″(x) in the intervals around x = 1 9 x = 1 9 x=(1)/(9)x = \frac{1}{9}x=19:
  • For x < 1 9 x < 1 9 x < (1)/(9)x < \frac{1}{9}x<19, y ″ ( x ) = 2 − 18 x > 0 y ″ ( x ) = 2 − 18 x > 0 y^(″)(x)=2-18 x > 0y”(x) = 2 – 18x > 0y″(x)=2−18x>0 (since x x xxx is positive).
  • For x > 1 9 x > 1 9 x > (1)/(9)x > \frac{1}{9}x>19, y ″ ( x ) = 2 − 18 x < 0 y ″ ( x ) = 2 − 18 x < 0 y^(″)(x)=2-18 x < 0y”(x) = 2 – 18x < 0y″(x)=2−18x<0 (since x x xxx is positive).
Since the sign of y ″ ( x ) y ″ ( x ) y^(″)(x)y”(x)y″(x) changes from positive to negative as x x xxx crosses x = 1 9 x = 1 9 x=(1)/(9)x = \frac{1}{9}x=19, we have a change in concavity at that point.
Therefore, the function y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y = x^2 – 3x^3y=x2−3x3 does have a point of inflection at x = 1 9 x = 1 9 x=(1)/(9)x = \frac{1}{9}x=19.
So, the statement " y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y=x^2-3 x^3y=x2−3x3 has no points of inflection" is false.
e) y = − x 2 y = − x 2 y=-x^(2)y=-x^2y=−x2 is increasing in [ − 5 , − 3 ] [ − 5 , − 3 ] [-5,-3][-5,-3][−5,−3].
Answer:
To determine whether the function y = − x 2 y = − x 2 y=-x^(2)y = -x^2y=−x2 is increasing in the interval [ − 5 , − 3 ] [ − 5 , − 3 ] [-5,-3][-5, -3][−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 a aaa and b b bbb in the interval where a < b a < b a < ba < ba<b, the function values satisfy f ( a ) ≤ f ( b ) f ( a ) ≤ f ( b ) f(a) <= f(b)f(a) \leq f(b)f(a)≤f(b).
Let’s evaluate the function y = − x 2 y = − x 2 y=-x^(2)y = -x^2y=−x2 within the interval [ − 5 , − 3 ] [ − 5 , − 3 ] [-5,-3][-5, -3][−5,−3]:
  1. At x = − 5 x = − 5 x=-5x = -5x=−5: y ( − 5 ) = − ( − 5 ) 2 = − 25 y ( − 5 ) = − ( − 5 ) 2 = − 25 y(-5)=-(-5)^(2)=-25y(-5) = -(-5)^2 = -25y(−5)=−(−5)2=−25
  2. At x = − 3 x = − 3 x=-3x = -3x=−3: y ( − 3 ) = − ( − 3 ) 2 = − 9 y ( − 3 ) = − ( − 3 ) 2 = − 9 y(-3)=-(-3)^(2)=-9y(-3) = -(-3)^2 = -9y(−3)=−(−3)2=−9
Since y ( − 5 ) = − 25 y ( − 5 ) = − 25 y(-5)=-25y(-5) = -25y(−5)=−25 is less than y ( − 3 ) = − 9 y ( − 3 ) = − 9 y(-3)=-9y(-3) = -9y(−3)=−9, we have f ( a ) < f ( b ) f ( a ) < f ( b ) f(a) < f(b)f(a) < f(b)f(a)<f(b) for a = − 5 a = − 5 a=-5a = -5a=−5 and b = − 3 b = − 3 b=-3b = -3b=−3, which means the function is decreasing in this interval.
Therefore, the statement " y = − x 2 y = − x 2 y=-x^(2)y=-x^2y=−x2 is increasing in [ − 5 , − 3 ] [ − 5 , − 3 ] [-5,-3][-5,-3][−5,−3]" is false. The function y = − x 2 y = − x 2 y=-x^(2)y = -x^2y=−x2 is actually decreasing in the interval [ − 5 , − 3 ] [ − 5 , − 3 ] [-5,-3][-5, -3][−5,−3].
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