revision problem for jee 2011 aspirants

\boxed {4}
\texttt{if} \ \alpha =e^{i\frac{2\pi}{7}} \ \texttt{and} \ f(x)=A_0+\sum_{i=1}^{20}{A_kx^k} \ \texttt{and the value of } \\ f(x)+f(\alpha x)+f(\alpha^2 x)+\cdots f(\alpha^6 x)= k(A_0+A_7x^7+A_{14}x^{14}) \\ \texttt{, then find the value of k}~~easy ~~

\boxed {5}
z_!,z_2,\cdots ,z_n \texttt{be non-zero complex numbers of equal modulus and satisfying equation } \\ z^n-|z|z^{n-1}+1=0,\texttt{then} \\ (a)Re\left(\sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_j}{z_l}}} \right)=0 \\ (b)Re\left(\sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_l}{z_j}}} \right)=0 \\ (c)Im\left(\sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_j}{z_l}}} \right)=0 \\ (d)Im\left(\sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_l}{z_j}}} \right)=0 \\

\boxed {6}

f:[0,1]\rightarrow \mathbb{R} \texttt{is a differentiable function such that f(0)=0 and |f'(x)|}\leq \texttt{k|f(x)| for all x}\\ \in[0,1],(k>0),\texttt{then which of the following is/are always true}

(a) f(x)=0,\forall x\in \mathbb{R} \\ (b) f(x)=0,\forall x\in [0,1] \\ (c) f(x)\neq 0,\forall x\in [0,1]\\ (d) f(1)=k

last two are pretty good

Q1

I=\int e^{secx}sec^{3}x(sin^{2}x+sinx+cosx+sinxcosx)dx

=\int e^{secx}(tan^{2}xsecx + sec^{2}xtanx +sec^{2}x+secxtanx)dx
=\int e^{secx}secxtanx(tanx + secx +\frac{secx}{tanx}+1)dx
=\int e^{secx}([tanx + secx] +[\frac{secx}{tanx}+1])d(secx)

\frac{d}{d(secx)}(tanx+secx) = \frac{secx}{tanx}+ 1

so

I = e^{secx}(secx+tanx) + c

3 is easy ,

substitute r = - (p+q)/3 in the eqn

so px² +qx - (p+q)/3 = 0

f(0) = - (p+q)/3

f(1) = 2(p+q)/3

so atleast 1 root between 0 , 1

so obviously that root is also in -1, 1

Q-5 solution :

\sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_j}{z_l}}}=\left( z_1+z_2+z_3+\cdots+z_n\right)\left(\frac{1}{z_1}+\frac{1}{z_2}+\cdots+\frac{1}{z_n} \right) \\ let \ |z_i|=a \\ z_i.\bar{z_i}=a^2 \\ \frac{1}{z_i}=\frac{\bar{z_i}}{a^2} \\ \sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_j}{z_l}}}=\left( z_1+z_2+z_3+\cdots+z_n\right)\frac{\left(\bar{z_1}+\bar{z_2}+\cdots +\bar{z_n} \right)}{a^2} \\ \sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_j}{z_l}}}=\frac{|\left({z_1}+{z_2}+\cdots +{z_n} \right)|^2}{a^2} \\ \texttt{but} \ \sum{z_i=|z|=a} \\ \sum_{j=1}^{n}{\sum_{l=1}^{n}{\frac{z_j}{z_l}}}=1

thats not right