doubts

1.pls explain these
a)[-2x/pi]+1/2 = -([2x/pi]+1/2)
b)-3≤|[x]|≤2 implies -2≤x<3

2. What is the formula to find number surjections from A to B?

3.If f"(x) = -f(x) and g(x) = f'(x) and F(x) = (f(x/2))2+(g(x/2))2 and given that F(5) = 5, then F(10) = ?

16 Answers

1
johncenaiit ·

4. If f(x) is symmetrical about x=1, find the real values of x satisfying the equation f(x)=f(x+1x+2)

5.If f(x) is an even function and satisfies the relation
x2f(x)-2f(1/x) = g(x),x≠0,where g is odd. then f(2) = ?

262
Aditya Bhutra ·

3. hint find derivative of F(x)

4. either x=x+1x+2 or x+x+1x+2=2

5. replace x with (-x). then conclude about g(x) and then finally f(x)

1
johncenaiit ·

Pls elaborate ans 4 and 5.....i didn't get that....

1
johncenaiit ·

6.Find the period of tanpi2[x]

7.Number of real roots of 3x+4x+5x-6x = 0 are
(1,2,>2,0)

8.minimum value of |x-1|+|x-2|+|x-3| =
(0,1,2,3)

9. Let S_{n} = \sum_{r=1}^{n}{r!} (n>6), then S_{n}-7\left[\frac{S_{n}}{7} \right] =
([n/7],n!-7[n!/7],5,3)

10.If f(x) is a real valued function discontinuous at all integral points lying in [0,n] and if (f(x))2 = 1 for all 0≤x≤n then number of functions f(x) are
(2n+1,6x3n,2x3n-1,3n+1)

1708
man111 singh ·

(5)

(7)

\hspace{-16}$Here $\mathbf{3^x+4^x+5^x=6^x}$\\\\ Now Divide both side by $\mathbf{(5.5)^x}\;,$ We Get\\\\ $\mathbf{\underbrace{\left(\frac{3}{5.5}\right)^x+\left(\frac{4}{5.5}\right)^x+\left(\frac{5}{5.5}\right)^x}_{L.H.S}=\underbrace{\left(\frac{6}{5.5}\right)^x}_{R.H.S}}$\\\\ Now $\mathbf{L.H.S}$ is a fun. of decreasing fun. and $\mathbf{R.H.S}$\\\\ is an Increasing fun.\\\\ So these two function meet at only one point\\\\ So only one value of $\mathbf{x}$ Which is $\mathbf{x=3}$\\\\

(8)

\hspace{-16}$Let $\mathbf{y=f(x)=\mid x-1 \mid +\mid x-2 \mid+\mid x-3 \mid}$\\\\ $\mathbf{y=\underbrace{\bold{\mid x-1 \mid+\mid 3-x \mid}}+\mid x-2 \mid}$\\\\ Using $\mathbf{\triangle}$ Inequality in underbrace portion::\\\\ Which is $\mathbf{\mid x-a \mid+\mid b-x \mid\geq \mid b-a \mid}$\\\\ Inequality Hold When $\mathbf{(x-a).(b-x)\geq 0\Leftrightarrow a\leq x\leq b}$\\\\ So $\mathbf{y=\underbrace{\bold{\mid x-1 \mid+\mid 3-x \mid}}+\mid x-2 \mid\geq \mid 2\mid+\mid x-2 \mid\geq 0}$\\\\ So $\mathbf{y\geq 2\Leftrightarrow y\in [2,\infty)}$\\\\ Minimum occur at $\mathbf{x=2}$

262
Aditya Bhutra ·

4. f(a)=f(b) and f(x) is symmetrical about x=c ,
then a=b (obvious) or AM of (a,b) =c (logical)

6. i dont think it is periodic since if [x]=2n+1 (n→N) then tan is not defined

10.is it 2*3n-1

1708
man111 singh ·

ans (9) = 5

1
johncenaiit ·

@man111, correct ans to 9....

1057
Ketan Chandak ·

@johncenaiit plz post question 9 clearly....

1
johncenaiit ·

@ketan..... latex is not working in my pc...when i type the equation in the editor and then press copy to document /copy to clipboard, nothing is happening....!!!!

why is this problem???

1057
Ketan Chandak ·

@johncenaiit....try this....
right click on the equation and copy image URL....
close the latex window and come out in the text box and paste the URL using image link....

1
johncenaiit ·

Thanks a lot ketan....

\texttt{Let } S_{n} = \sum_{r=1}^{n}{r!} (n>6),\texttt{ then }S_{n}-7\left[\frac{S_{n}}{7} \right] =

(A) \left[\frac{n}{7} \right]
(B) n!-7\left[\frac{n!}{7} \right]

(C) 5

(D) 3

262
Aditya Bhutra ·

reqd. to cal - 7* {Sn/7}

for r≥7 , r! is divisible by 7
hence it will not affect the result .

summing up 6!+5!+...+1! and then finding 7*{Sn/7}, we get 5

1
johncenaiit ·

some one pls explain the answer to Q10

1708
man111 singh ·

\hspace{-16}$Here I am Calculating for $\mathbf{n=2}$\\\\ So Question is like that..............\\\\ If $\mathbf{f(x)}$ is a real valued function discontinuous at all integral points lying\\\\ in $\mathbf{\left[0,2\right]}$ and if $\mathbf{(f(x))^2 = 1\forall x\in \left[0,2\right]}$ then number of functions $\mathbf{f(x)}$ are\\\\ $\underline{\underline{\bold{Ans::}}}\Rightarrow$ So Here $\mathbf{x=1}$ is only Integer points in $\mathbf{0}$ to $\mathbf{2}$.\\\\ So $\mathbf{(f(x))^2=1\Leftrightarrow f(x)=\pm 1\forall x\in \left[0,2\right]}$\\\\ So \begin{Bmatrix} \bold{f(x)=\pm 1} ,& \bold{0\leq x<1} \\\\ \bold{f(x)=\pm 1} ,& \bold{x=1}\\\\ \bold{f(x)=\pm 1} ,& \bold{1<x\leq 2} \end{Bmatrix}$\\\\\\ $So Out of Total $\mathbf{8}$ function only $\mathbf{2}$ function is Continuous\\\\ Which is $\mathbf{f(x)=+1\forall x\in \left[0,2\right]}$\\\\ OR $\mathbf{f(x)=-1\forall x\in \left[0,2\right]}$\\\\ So Total Discont. function in $\mathbf{x\in \left[0,2\right]}$ is $\mathbf{8-2=6}$\\\\

It is not a Complete Proof. Here i just take it for one case.

I also wait for Generalistaion.

I thinks that we use somethings like the partition of sets of function

(I am aslo trying for that things)

Thanks

1
fahadnasir nasir ·

very very nisei

Your Answer

Close [X]