functions

in 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...

i got it now.....thanks ketan for an amazing solution....

question no 6...
its clear frm the question dat max exists at sinx=-1 and min at sinx=1...

ANSWERS
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]

7. 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 :(

The 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]

1. 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}}

@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

\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 $\mathbf{x<3}$

\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 $\mathbf{[x]\in \mathbb{Z}}$

\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 $\mathbf{0 \leq \mid \sin x\mid+\mid \cos x \mid \leq \sqrt{2}}$

thanks a lot ..pls try the rest also.....