1sorry, I made a mistake
the answar is 3√2
\hspace{-16}$\textbf{(1)\;\; If $\mathbf{x,y\in\mathbb{R}}$ and $\mathbf{x^2+y^2=1}$. Then Max. value of $\mathbf{\mid x-y \mid +\mid x^3-y^3 \mid=}$ }
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21let x=sinA y=cosA
Without loss of generality, let sinA≥cosA
so,the given expression in equivalent to:
sinA-cosA+sin3A-cos3A=(sinA-cosA)(1+sin2A+cos2A+sinAcosA)
=(sinA-cosA)(2+sinAcosA)
=2(sinA-cosA) + sinAcosA(sinA-cosA)
=2(sinA-cosA) + sin2A(sinA-cosA)/2
it will attain max. value when sinA-cosA is max.
so it is ≤2(1-0) +0(1-0)/2
=2
Ans. 2
21ok
how can u say this
=2(sinA-cosA) + sin2A(sinA-cosA)/2
it will attain max. value when sinA-cosA is max.
?
21W.L.O.G
x≥y
so |x-y| + |x3- y3| = x-y + x3- y3
so maxima will occour when x>0 and y<0, let x= a>0, -y= b>0
so we have to maximize a+ a3+ b+ b3, when a2+ b2 = 1 (*)
W.L.O.G let a≥b
so (1+ a2) ≥ (1+ b2)
By chebycheffs inequality
a(1+a2)+ b(1+ b2)2 ≤ (a+b)2(1+a2+ 1+ b2)2
and also , by AM-GM , (a+b)24≤ a2+ b22
Giving (a+b) ≤√2 , combine this with (*) , to get the answer as 3/√2, equality when x=y= 1/√2
341Somehow I am getting a different answer. I may have goofed up but:
\because \ (x^3-y^3) = (x-y)(x^2+xy+y^2) and (x^2+xy+y^2)\ge 0
x-y and x3-y3 have the same sign.
So WLOG x≥y and the expression is therefore(x-y)+(x^3-y^3) = (x-y)[1+(x^2+xy+y^2)] = (x-y)(2+xy)
Also xy = \frac{1-(x-y)^2}{2}
So the expression is \frac{t(5-t^2)}{2} where t=x-y≥0
Now, this is increasing in \left[0, \sqrt{\frac{5}{3}} \right] and decreasing thereafter.
So the max occurs when t= \sqrt{\frac{5}{3}} < \sqrt 2
and equals
\frac{5 \sqrt{5}}{3 \sqrt{3}}
1708Thanks bhatt Sir. (Right answer)
also Thanks Subhodip, Arnab for valuable efford.