Please clarify my doubts

Respected Members, I have a few doubts in Calculus. I hope if you can spare some amount of your valuable time and solve them and explain their solutions. They are as follows:

1) d20ydx20(2cosxcos3x)\frac { { d }^{ 20 }y }{ d{ x }^{ 20 } } (2\cos { x } \cos { 3x) } is equal to

a) 220(cos2x220cos3x){ 2 }^{ 20 }(\cos { 2x } -{ 2 }^{ 20 }\cos { 3x } )

b) 220(cos2x+220cos4x){ 2 }^{ 20 }(\cos { 2x } +{ 2 }^{ 20 }\cos { 4x } )

c) 220(sin2x+220sin4x){ 2 }^{ 20 }(\sin { 2x } +{ 2 }^{ 20 }\sin { 4x } )

d) 220(sin2x220sin4x){ 2 }^{ 20 }(\sin { 2x } -{ 2 }^{ 20 }\sin { 4x } )

Answer: (b)

2) dndxn(logx)\frac { { d }^{ n } }{ d{ x }^{ n } } (\log { x) } =

a) (n1)!xn\frac { (n-1)! }{ { x }^{ n } }

b) n!xn\frac { n! }{ { x }^{ n } }

c) (n2)!xn\frac { (n-2)! }{ { x }^{ n } }

d) (1)n1(n1)!xn{ (-1) }^{ n-1 }\frac { (n-1)! }{ { x }^{ n } }

Answer:(d)

3) If f(0) = 0, f'(0) = 2, then the derivative of y = f(f(f(f(x)))) at x=0 is

a) 2

b)8

c)16

d)4

Ans: (c)

4) Suppose f(x) = eax+ebx{ e }^{ ax }+{ e }^{ bx }, where aba\neq b, and that **f"(x)-2f'(x)-15f(x)=0 for all x. Then the product ab is

(a) 25

(b)9

(c)-15

(d)-9

Answer (c)

5) If f(x) satisfies the relation

f(5x3y2)=5f(x)3f(y)2f(\frac { 5x-3y }{ 2 } )\quad =\quad \frac { 5f(x)-3f(y) }{ 2 } for all x,y\in R,

and f(0) = 3 and f'(0) = 2, then the period of sin(f(x)) is

a) 2π\pi

b)π\pi

c)3π\pi

d)4π\pi

Answer: (b)

Thanks and regards

Note by Manish Dash
4 years, 2 months ago

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1 vote

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2) d(logx)dx=1x\dfrac{d(\log_{}{x})}{dx}=\dfrac{1}{x}

d2(logx)dx2=1x2\dfrac{d^2(\log{}{x})}{dx^2} = -\dfrac{1}{x^2}

d3(logx)dx3=2x3\dfrac{d^3(\log{}{x})}{dx^3} = \dfrac{2}{x^3}

d4(logx)dx4=6x4\dfrac{d^4(\log{}{x})}{dx^4} = -\dfrac{6}{x^4}

d5(logx)dx5=24x5\dfrac{d^5(\log{}{x})}{dx^5} = \dfrac{24}{x^5}

\dots

dn(logx)dxn=(1)n1(n1)!xn\dfrac{d^n(\log{}{x})}{dx^n} = (-1)^{n-1}\dfrac{(n-1)!}{x^n}

Omkar Kulkarni - 4 years, 2 months ago

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  1. Apply the formula 2cosxcosy=cos(x+y)+cos(xy)2\cos x \cos y= \cos (x+y) +\cos(x-y) to simply it further and then observe the pattern by differentiating it repeatedly.

  2. Observe the pattern on repeated differentiation.

  3. The derivative of y=f'(f(f(f(x)))). f'(f(f(x))). f'(f(x)). f'(x) . Now substitute the given values.

  4. Simply differentiate and put the appropriate values in the given equation. You will get the values of a and b from there.

  5. The function which satisfies the function equation f(a+b2)=f(a)+f(b)2f\left( \dfrac{a+b}{2} \right)=\dfrac{f(a)+f(b)}{2} can be taken as f(x)=A.xf(x)=A.x and from the given values you can find the value of A. IN this case the value of A will come out to be 2. Hence the fundamental period of sin(f(x)) will be π\pi.

I've given the hints and the approaches of all your doubts. Now you can try and surely it will help you.

All the best! @Manish Dash

Sandeep Bhardwaj - 4 years, 2 months ago

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In question 5, you said f(x)=Axf(x) = A \cdot x
In these types of question,i am always been confused in identifying f(x)f(x).
How do you come to know that f(x) will be AxA \cdot x .

Akhil Bansal - 4 years, 1 month ago

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Thank you very much @Sandeep Bhardwaj Sir!

Manish Dash - 4 years, 1 month ago

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1) 2cosxcos3x=cos2x+cos4x2\cos x\cos 3x = \cos 2x+\cos 4x

ddx(cos2x+cos4x)=21(sin2x+21sin4x)\dfrac{d}{dx}(\cos 2x+\cos 4x) = -2^{1}(\sin 2x + 2^{1}\sin 4x)

d2dx2(cos2x+cos4x)=22(cos2x+22cos4x)\dfrac{d^{2}}{dx^{2}}(\cos 2x+\cos 4x) = -2^{2}(\cos 2x + 2^{2}\cos 4x)

d3dx3(cos2x+cos4x)=23(sin2x+23sin4x)\dfrac{d^{3}}{dx^{3}}(\cos 2x+\cos 4x) = 2^{3}(\sin 2x + 2^{3}\sin 4x)

d4dx4(cos2x+cos4x)=24(cos2x+24cos4x)\dfrac{d^{4}}{dx^{4}}(\cos 2x+\cos 4x) = 2^{4}(\cos 2x + 2^{4}\cos 4x)

d5dx5(cos2x+cos4x)=25(sin2x+25sin4x)\dfrac{d^{5}}{dx^{5}}(\cos 2x+\cos 4x) = -2^{5}(\sin 2x + 2^{5}\sin 4x)

\dots

d20dx20(cos2x+cos4x)=220(cos2x+220cos4x)\dfrac{d^{20}}{dx^{20}}(\cos 2x+\cos 4x) = 2^{20}(\cos 2x + 2^{20}\cos 4x)

Omkar Kulkarni - 4 years, 2 months ago

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I don't understand derivatives enough to do the other three, but this may help:

3) f(0)=0f'(0)=0

f(f(0))=0f'(f(0))=0

f(f(f(0)))=f(0)f'(f(f(0)))=f(0)

Both sides of all the above equations are equal.

4) ddx(eax+ebx)=aeax+bebx\frac{d}{dx}(e^{ax}+e^{bx})=ae^{ax}+be^{bx}

Omkar Kulkarni - 4 years, 2 months ago

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nope. The very last expression is wrong, because aeaxex(aea)ae^{ax} \neq e^{x} (ae^{a}) , and such is the same for bb.

Efren Medallo - 4 years, 2 months ago

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Oh right. Sorry. Thanks!

Omkar Kulkarni - 4 years, 2 months ago

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f (x)=\sqrt {1+cos^{2}\times(x^{2})}

Suvdeep Nayak - 1 year, 7 months ago

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How to find f dash x

Suvdeep Nayak - 1 year, 7 months ago

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