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Transcript of exercise math
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therefore f(x): I torx e R.2
Thus, 1a - -rof a- -2.
2. Letu : x2 * 4x, then u -
(x * 2), -
4 >
168 Solufions fo Testing Quesfions
- -Z satisfies the r
-4 and
y: (u*5)(u+1)+3ut5- r,+9u* 10:x
- -2), therefore ./min : eqz + 9e4) + 10 :3. y- --:', + T =+ U - L)*' * yx+ (y - n)- o.' xzax*l
When y -
1, then x -
n -
1, so 1 is in the range of y.When y * 1, then above quadratic equation in x has real roors I ry # 1 in its range. So its discriminant is non-negative. Therefore
A-y2-4(y-lXy -n) >0=+ 3yr-4(n+t)y*4n S,_
3 ) --rL"rL
3)
therefore (a"l6(n2
- n *
l6(n + 1)2 -a!,m
Lecture \c*'e: ',f,.
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(ii) (x* y +r)2 > x2*y'*22,3(x*y *z)) y*2y*32 imp,_uur,that
13 .*' + y2 + z2 + z *2y *32 3(1 + {z)@bc)*,therefore
6--- t62(1 + "[Z)r'
where the equalities hold when and only when a -
b -
c. Thus,
5:T/l'maxt62(1 + O),
: The given conditions gives
f(o)_ -11+ al+220
f(x) -
x2-2x-lr( x2 +3
: { x2-2xtl x2-4x-
AC: r/(r+b)2*c,
-
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t72 Solutions to Testing Questions
Thus, f (x) > 0 for all x IR, So c6in - -2.
Since a > 0, f (x) takes its minimum value when x
l,f (xo) I -
(3a)b-(a+Dz a2(3a)
Lecture Notes On Mathemi
Since'S x?oos :S x?oog3 t- x?oog A *,0i
anyl0.
l/(1)l -a-b -
f(1)> -b- l/(o)l; and
l,f(xo)l . l/(1)l # a2 + b2 -
ab < 3a(a -
b) +> (a + bt: : r