1,Prove for positive a,b,c (b≠1) , alogbc = clogba
2,In a small sweet shop people usually buy either one cake or one box of chocolate ,One day the shop sold 57 cakes and 36 boxes of chocolate,How many customers were there that day if 12 people bought both a cake and a box of chocolate?
3,Show that [(n+1)/2] + [(n+2)/4] + [(n+4)/8] + [(n+8)/16] + ,,,,,,, = n
4,Solve |2x-1| = 3[x] + 2{x}
5,Solve (x[x] - x2 - 3[x] + 3x) is positive
6,Solve [x]3 - 2[x] + 1 = 0
7, Find range of |sin x| + |cos x| and sin2x - 5sinx -6
1057in question 2 if there isnt any twist answer is simply 57+36-12=81
in question 6....
[x]3 - 2[x] + 1 = 0
[x]3 - 2[x] = -1
[x]([x]2-2)=-1
observe that if [x] is greater than 1 or less than -1 then [x]2-2 will be greater than 1 which is not possible..
therefore [x]=1 is the only solution....
therefore 1≤x<2...
1057i hope u got till dis much....
[x]([x]2-2)=-1
now observe that both [x]and ([x]2-2) will be integers for all real values of x...
since their product is -1 they can only be 1,-1 or -1,1 respectively...
so x should be in the range -1 to 1
but if u put -1 in the equation [x] and ([x]2-2) both will be negative and their product will bcum positive....
so [x] shud be 1....
therefore x=[1,2)
1708Thanks Subho for Correction.
and Sorry johncenaiit for my typing Mistakes.
1@ketan,
i didn't understand your solution to [x]3-2[x]+1=0
can you pls explain a bit more?
1057question no 6...
its clear frm the question dat max exists at sinx=-1 and min at sinx=1...
1ANSWERS
2) 81
4) 0.25
5) (-∞,3)-I
6) [1,2)
7) for the first part, i forgot to mention the domain, it is [0,pi]
and answer is 1st part = [1,√2]
2nd part = [-10,0]
217. For the range of f=|sin x| + |cos x| (man111's answer was incomplete)
here it suffices to check for the cases when both sin x and cos x are positive
since otherwise, if both (sin , cos )are negative, the value f takes is the same under mapping
x --> x - \pi (but this makes both of them positive)
if exactly one of them is negative, then either x --> -x (when sin is negative) or x --> \pi - x,(when cos is negative) so we can restrict the domain of f to those x, which makes sin cos simultaneously positive, without having any change in range
so range of |sin x| + |cos x| is same of that of sin x + cos x, when both of them are positive i.e [1, √2]
EDIT: din see post 8 by johncenaiit :(
21The eqn. is equivalent to [n2+12]+[n4+12]+...=n
By Hermide's Identity we get: [x+12]=[2x]-[x]
So, the eqn. can be written as,
([n]-[n2])+([n2]-[n4])+....=[n]
111. a^{log_b{c}}=a^{\frac{log_a{c}}{log_a{b}}}=(a^{log_a{c}})^\frac{1}{log_a{b}}
\Rightarrow a^{log_b{c}}=c^\frac{1}{log_a{b}}
\Rightarrow a^{log_b{c}}=c^{log_b{a}}
1@man111,
|sin x| + |cos x| = √(1+|sin2x|)
→ 1≤|sin x| + |cos x|≤√2
In your answer to Q5,isn't the second step wrong ? you forgot to reverse the inequality
1708\hspace{-16}\mathbf{(5)\;\; x[x]-x^2-3[x]+3x>0}$\\\\ $\mathbf{[x].([x]-x)-3.(x-[x])>0}$\\\\ $\mathbf{([x]-3).([x]-x)>0}$\\\\ $\mathbf{(x-[x]).([x]-3)<0}$\\\\ $\mathbf{\left\{x\right\}.([x]-3)<0}$\\\\ Now Here $\mathbf{0<\left\{x\right\}<1}$\\\\ So $\mathbf{[x]-3<0\Leftrightarrow [x]<3}$\\\\ So $\mathbf{x<3}$
1708\hspace{-16}\mathbf{(6)\;\; \left[x\right]^3-2[x]+2=0}$\\\\ Let $\mathbf{[x]=t}$. Then \\\\ $\mathbf{t^3-2t+1=0}$\\\\ $\mathbf{\underbrace{\bold{t^3-t^2}}+\underbrace{\bold{t^2-2t+1}}=0}$\\\\ $\mathbf{t^2(t-1)+(t-1)^2=0}$\\\\ $\mathbf{(t-1).(t^2+t-1)=0}$\\\\ $\mathbf{t=1\;\;, t=\frac{-1\pm \sqrt{5}}{2}}$\\\\ Now $\mathbf{t=1\Leftrightarrow [x]=1}$\\\\ $\mathbf{1 \leq x<2}$\\\\ and $\mathbf{t=\frac{-1\pm \sqrt{5}}{2}\Leftrightarrow [x]=\frac{-1\pm \sqrt{5}}{2}}$\\\\ Not Possible bcz $\mathbf{[x]\in \mathbb{Z}}$
1708
\hspace{-16}$Here $\mathbf{f(x)=\sin^2-5\sin x+6}$\\\\ Put $\mathbf{\sin x=t}$. Then\\\\ $\mathbf{y=f(t)=t^2-5t+6}$ ,Where $\mathbf{-1\leq t\leq 1}$\\\\ So From Graph\\\\ $\mathbf{y_{Min}=f(1)=2}$\\\\ and $\mathbf{y_{Max.}=f(-1)=12}$
\hspace{-16}$For Calculation of Range of $\mathbf{f(x)=\mid \sin x\mid +\mid \cos x \mid }$\\\\ Now Using $\mathbf{C.S}$ Inequality, We Get\\\\ $\mathbf{\left(1^2+1^2\right).\left(\mid \sin x\mid^2+\mid \cos x\mid^2 \right)\geq \left(\mid \sin x\mid+\mid \cos x\mid\right)^2}$\\\\ So $\mathbf{\left(\mid \sin x\mid+\mid \cos x \mid \right)^2\right)\leq (\sqrt{2})^2}$\\\\ So $\mathbf{0 \leq \mid \sin x\mid+\mid \cos x \mid \leq \sqrt{2}}$
14) for x≥ 12,
2x-1 = 2x + [x]
=> [x] = -1 which is not possible in the given interval.
for x < 12,
1-2x = 2x + [x]
=> 5x-1 = {x}
=> 0≤ 5x-1 < 1
=> 15 ≤x<25
but in this interval [x] = 0
=> 1-2x=2x => x = 0.25 is the only solution