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

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

b) ${ 2 }^{ 20 }(\cos { 2x } +{ 2 }^{ 20 }\cos { 4x } )$

c) ${ 2 }^{ 20 }(\sin { 2x } +{ 2 }^{ 20 }\sin { 4x } )$

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

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

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

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

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

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

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) = ${ e }^{ ax }+{ e }^{ bx }$, where $a\neq b$, 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

5) If f(x) satisfies the relation

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

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

a) 2$\pi$

b)$\pi$

c)3$\pi$

d)4$\pi$

Thanks and regards

Note by Manish Dash
5 years, 2 months ago

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

- 2 years, 7 months ago

How to find f dash x

- 2 years, 7 months ago

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

3) $f'(0)=0$

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

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

Both sides of all the above equations are equal.

4) $\frac{d}{dx}(e^{ax}+e^{bx})=ae^{ax}+be^{bx}$

- 5 years, 2 months ago

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

- 5 years, 2 months ago

Oh right. Sorry. Thanks!

- 5 years, 2 months ago

1. Apply the formula $2\cos x \cos y= \cos (x+y) +\cos(x-y)$ 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 $f\left( \dfrac{a+b}{2} \right)=\dfrac{f(a)+f(b)}{2}$ can be taken as $f(x)=A.x$ 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 $\pi$.

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

- 5 years, 2 months ago

Thank you very much @Sandeep Bhardwaj Sir!

- 5 years, 1 month ago

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

- 5 years, 2 months ago

1) $2\cos x\cos 3x = \cos 2x+\cos 4x$

$\dfrac{d}{dx}(\cos 2x+\cos 4x) = -2^{1}(\sin 2x + 2^{1}\sin 4x)$

$\dfrac{d^{2}}{dx^{2}}(\cos 2x+\cos 4x) = -2^{2}(\cos 2x + 2^{2}\cos 4x)$

$\dfrac{d^{3}}{dx^{3}}(\cos 2x+\cos 4x) = 2^{3}(\sin 2x + 2^{3}\sin 4x)$

$\dfrac{d^{4}}{dx^{4}}(\cos 2x+\cos 4x) = 2^{4}(\cos 2x + 2^{4}\cos 4x)$

$\dfrac{d^{5}}{dx^{5}}(\cos 2x+\cos 4x) = -2^{5}(\sin 2x + 2^{5}\sin 4x)$

$\dots$

$\dfrac{d^{20}}{dx^{20}}(\cos 2x+\cos 4x) = 2^{20}(\cos 2x + 2^{20}\cos 4x)$

- 5 years, 2 months ago

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

$\dfrac{d^2(\log{}{x})}{dx^2} = -\dfrac{1}{x^2}$

$\dfrac{d^3(\log{}{x})}{dx^3} = \dfrac{2}{x^3}$

$\dfrac{d^4(\log{}{x})}{dx^4} = -\dfrac{6}{x^4}$

$\dfrac{d^5(\log{}{x})}{dx^5} = \dfrac{24}{x^5}$

$\dots$

$\dfrac{d^n(\log{}{x})}{dx^n} = (-1)^{n-1}\dfrac{(n-1)!}{x^n}$

- 5 years, 2 months ago

- 5 years, 2 months ago

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