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

Answer: (b)

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

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) = \({ e }^{ ax }+{ e }^{ bx }\), 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

Answer (c)

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\(\pi\)

Answer: (b)

Thanks and regards

## Comments

Sort by:

TopNewestObserve the pattern on repeated differentiation.

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

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

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 – Sandeep Bhardwaj · 1 year, 5 months ago

Log in to reply

@Sandeep Bhardwaj Sir! – Manish Dash · 1 year, 5 months ago

Thank you very muchLog in to reply

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 \(A \cdot x \). – Akhil Bansal · 1 year, 5 months ago

Log in to reply

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

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

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

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

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

\(\dots\)

\(\dfrac{d^n(\log{}{x})}{dx^n} = (-1)^{n-1}\dfrac{(n-1)!}{x^n}\) – Omkar Kulkarni · 1 year, 5 months ago

Log in to reply

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) \(\frac{d}{dx}(e^{ax}+e^{bx})=ae^{ax}+be^{bx}\) – Omkar Kulkarni · 1 year, 5 months ago

Log in to reply

– Efren Medallo · 1 year, 5 months ago

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

– Omkar Kulkarni · 1 year, 5 months ago

Oh right. Sorry. Thanks!Log in to reply

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

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

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

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

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

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

\(\dots\)

\(\dfrac{d^{20}}{dx^{20}}(\cos 2x+\cos 4x) = 2^{20}(\cos 2x + 2^{20}\cos 4x)\) – Omkar Kulkarni · 1 year, 5 months ago

Log in to reply

@Sandeep Bhardwaj , @Calvin Lin , @Brian Charlesworth – Manish Dash · 1 year, 5 months ago

Log in to reply