\\1)\: solve\: \frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}\\ \:\:2)\: if\:f(n)=\sum_{r=1}^{n}\frac{1}{r}...then \:show \:that\:\frac{n}{2}<f(2^n-1)<n\\ 3)\:if\:(1+x)^n=\sum_{r=0}^{n}^{n}C_{r}x^r\:find \:teh\: sum\:of\:\sum_{s=0}^{r}\left ( (-1)^s.(r-s+1).^{n}C_{s} \right )\\ 4)\:if\:\alpha \:and\:\beta \:are\:distinct\:+ve\:roots\:of\:x^2-2ax+ab=0\:show\:taht\:\\ 1+2\left ( \frac{b}{a} \right )+3\left ( \frac{b}{a} \right )^2+.......+n\left ( \frac{b}{a} \right )^{n-1}<\left ( \frac{\alpha +\beta }{\alpha -\beta } \right )^2
help.
11 .
THE ANS IS
x = \frac{a^2 + b^2 }{a+ b}
posting the complete solution
on slight rearrangement
\frac{x - a}{b} - \frac{b}{x- a} = \frac{a}{x- b }-\frac{x-b}{a}
\frac{\left(x- a \right)^ 2 - (b^2) }{b \left(x- a \right)} = \frac{a^ 2 - \left(x-b \right)^ 2}{( x- b). a}
again rearranging we get
\frac{\left(x- a \right)^ 2 - (b^2) }{(a^ 2) - \left(x- b \right)^2 } = \frac{b (x-a)}{a( x- b). }
now use a^ 2 - b^2 pharmoola
\frac{\left(x- a+ b \right)\left(x- a- b \right)}{\left(a+ x- b \right)\left(a-x + b \right)} = \frac{\left(x-a + b \right)(x - a- b )}{\left(a + x - b \right)\left(-( -a + x- b) \right)}
\frac{ x- a + b }{-a - x+ b } = \frac{b\left(x-a \right)}{a\left(x-b \right)}
now on rearranging we get
( x- a + b )a ( x- b) = b ( x- a ) .( b- a - x )
simplifying both sides
( ax - a^2 + ab ) ( x- b) = ax^2 - axb - a^ 2x + a^2 b + abx - ab^2
---1
( b^2 - ba - bx ) ( x- a ) = b^ 2x - b^2a - bax + ba^2 - bx^2 + bax---2
equate 1 and 2
ax^ 2 - a^ 2 x + a^2 b - ab^2 = b^2 x - bx^2 + a^2b - ab^2
ax^ 2 - a^ 2 x = b^2 x - bx^2
ax - a^2 = b^2 - bx
ax + bx = a^2 + b^2
x ( a + b ) = a^2 + b^2
x =\frac{a^2 + b^2 }{ a+ b}
1no need to do so lengthy
on shifting terms from lhs to rhs and rhs to lhs we hav
\frac{x-a}{b}-\frac{a}{x-b}=\frac{b}{x-a}-\frac{x-b}{a}
now siplify seperately rhs and lhs
u get the ans
now that thing was quiet obvious
my q is since it comes out to be a function of 3 degree will der be any more roots bec in this we r cancelling numerator having x
1i have done same thing as u have done and were do we get a third degree equation
1\\\texttt{Q-4 } \\\texttt{we have dicriminant > 0 } \\ a > b \\ \\\texttt{let S(n) be the left hand site of inequality } \\ \\ S(n) < \lim_{n\rightarrow \infty}S(n)=\frac{(\alpha + \beta )^2}{(\alpha -\beta )^2}
1\frac{(x-a)(x-b)-ab}{b(x-b)}=\frac{ab-(x-a)(x-b)}{a(x-a)}\\ \frac{(x-a)(x-b)-ab}{b(x-b)}=-\frac{(-ab+(x-a)(x-b))}{a(x-a)}\\
numerator on both sides gets cancelled
hence we get x=a2+b2a+b
will der be any more roots?
1by simple observation we can see that x= o is a root since it satisfies
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