Some Interesting Problems in Calculus

\text{1)Evaluate:}

a)\lim_{n\rightarrow \infty}\frac{1}{2n}\log\binom{2n}{n}

b)\lim_{n\rightarrow \infty}\left[\frac{1}{a+n}+\frac{1}{2a+n}+\cdots +\frac{1}{na+n}\right]

\text{2)Let } a_{1}=1 \text{ and }a_{n}=n(a_{n-1}+1)\ \forall n\geq 2 .\text{Find }

\lim_{n\rightarrow \infty}\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right)\cdots \left(1+\frac{1}{a_{n}}\right)

\text{3)Let }a_0 =0<a_1<a_2<\cdots <a_n \text{ be real numbers.}
\text{Suppose }p(t)\text{ is a real valued polynomial of degree n such that}

\int\limits^{a_{j+1}}_{a_{j}}p(t)\, dt=0 \qquad \text{ for all } 0\leq j \leq n-1

\text{Show that for }0\leq j\leq n-1 \text{ the polynomian p(t)has }
\text{exactly one root in the interval }(a_j,a_{j+1})

\text{4)Let f(u) be a continuos function and,let \[u\]}\rightarrow GIF
\text{Show that for any }x>1
\int \limits ^{x}_{1}\[u\](\[u\]+1)f(u)\, du=2\sum\limits ^{\[x\]}_{i=1}i\int \limits^{x}_{i}f(u)\, du

\text{5)Let f(x) be a function differentiable n+1 times for some }n>0

f(1)=f(0)=f^{(0)}(0)=\cdots =f^{(n)}(0)=0

\text{Prove that }f^{(n+1)}(x)=0 \text{ for some }x\in (0,1)

\text{6)Prove that for }n>0:

\int \limits ^{\pi/2}_{0}\frac{\sin(2n+1)x}{\sin x}\, dx=\frac{\pi}{2}

\text{7)Show that}

\int \limits^{\pi}_{0}\left|\frac{\sin nx}{x}\right|\, dx \geq \frac{2}{\pi}\left(1+\frac{1}{2}+\cdots +\frac{1}{n}\right)

\text{8)Using the identity }\log x = \int^x_1\frac{dt}{t}\qquad x>0 \text{ ,show that for }n\geq 1

\frac{1}{n+1}\leq \log\left(1+\frac{1}{n}\right)\leq \frac{1}{n}

\text{9)Let }a_1>a_2>a_3\cdots >a_r>0

\text{Compute }\lim_{n \rightarrow \infty}(a_1^n+a_2^n+\cdots +a_r^n)^{1/n}

\text{10)Find the maximum value of }\cos\alpha_1\cdot\cos\alpha_2\cdots \cos \alpha_n
\text{under the restriction }0\leq \alpha_1,\alpha_2,\cdots ,\alpha_n\leq \frac{\pi}{2} \text{ and }}
\cot\alpha_1\cdot\cot\alpha_2\cdots \cot \alpha_n =1.

15 Answers

1708
man111 singh ·

\hspace{-16}(6)\;\; $Let $\bf{I_{2k+1} = \int_{0}^{\frac{\pi}{2}}\frac{\sin (2k+1)x}{\sin x}dx.........(1)}$\\\\\\ Replace $\bf{k\rightarrow (k+1)}$\\\\\\ $\bf{I_{2k+3} = \int_{0}^{\frac{\pi}{2}}\frac{\sin (2k+3)x}{\sin x}dx................(2)}$\\\\\\ So $\bf{I_{2k+3}-I_{2k+1} = \int_{0}^{\frac{\pi}{2}}\frac{\sin (2k+3)x-\sin (2k+1)x}{\sin x}dx}$\\\\\\ So $\bf{I_{2k+3}-I_{2k+1} = \int_{0}^{\frac{\pi}{2}}\frac{2\cos (2k+2)x\cdot \sin x}{\sin x}dx =0}$\\\\\\ So $\bf{I_{2k+3} = I_{2k+1}.}$\\\\\\ Now Replace $\bf{k\rightarrow (k-1)}$\\\\\\ $\bf{I_{2k+1} = I_{2k-1}}$\\\\\\ Using Recursively, We Get\\\\\\ $\bf{I_{2k+3} = I_{2k+1} = I_{2k-1}=I_{2k-3}=........=I_{3}}$\\\\\\ Now $\bf{I_{1} = \int_{0}^{\pi}\frac{\sin (2\cdot 1+1)}{\sin x}dx = \int_{0}^{\frac{\pi}{2}}\frac{3\sin x -4\sin ^3 x}{\sin x}=\frac{\pi}{2}}$\\\\\\

\hspace{-16}$So $\bf{I_{2k+3} = I_{2k+1} = I_{2k-1}=I_{2k-3}=........=I_{3} =\frac{\pi}{2}}$\\\\\\So $\bf{I_{2k+1}=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2k+1)x}{\sin x}dx = \frac{\pi}{2}}$

341
Hari Shankar ·

341
Hari Shankar ·

1.

341
Hari Shankar ·

9.

341
Hari Shankar ·

3.

341
Hari Shankar ·

5.

341
Hari Shankar ·

7.

341
Hari Shankar ·

8.

1357
Manish Shankar ·

Good sets of questions :)

Try these guys..

1133
Sourish Ghosh ·

I got some answers...

1) a - log2
1) b - log(1+a)a
2) 5/2
9) a1

1357
Manish Shankar ·

an+1(n+1)!=ann!+1n!

Which gives

an+1(n+1)!=1+1+1/2!+1/3!+...+1/n!

this might help

341
Hari Shankar ·

Ok. Will try post lunch

865
Soumyadeep Basu ·

For no. 4, expand both the LHS and RHS into integrals between successive integers. Then try to rearrange the terms of the RHS to get the LHS.

341
Hari Shankar ·

4.

341
Hari Shankar ·

Sorry Soumya I had been typing that out for a long time I didnt know you had posted this.

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