Given that, equation |x| = ax + 1 has exactly one negative solution and no positive solutions then range of a:
(a) a>1 (b) a=1 (c) a≥1 (d) None of the proceeding
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Answers
Answer:
let P(x,0) be any point one the x-axis . Let the given points be A(2,3) and B(3/2,-1),then
AP=\begin{gathered}\sqrt{ {(x - 2)}^{2} + {(0 - 3)}^{2} } \\ BP = \sqrt{ {(x - \frac{3}{2} )}^{2} + {(0 - ( - 1))}^{2} } \\ according \: to \: the \: given \frac{ap}{bp} = \frac{2}{1} \\ that \: implies \: ap = 2bp \\ \sqrt{ {(x - 2)}^{2} + 9} = 2 \times \sqrt{ {(x - \frac{3}{2} }^{2}) + 1 } \\ {(x - 2)}^{2} + 9 = 4( {(x - \frac{3}{2}) }^{2} + 1) \\ {x - 4x}^{2} + 4 + 9 = 4( {x}^{2} - 3x + \frac{9}{4} + 1 \\ {x }^{2} - 4x + 13 = {4x}^{2} - 12x + 13 \\ { - 3x}^{2} + 8x = 0 \\ x( - 3x + 8) = 0 \\ x = 0 \: or \: - 3x + 8 = 0 \\ x = 0 \: or \: \frac{8}{3} \\ x = 0 \: or \: 2 \frac{2}{3} \\ hope \: it \: helps \: you \: please \: mark \: it \: as \: brainliest\end{gathered}
(x−2)
2
+(0−3)
2
BP=
(x−
2
3
)
2
+(0−(−1))
2
accordingtothegiven
bp
ap
=
1
2
thatimpliesap=2bp
(x−2)
2
+9
=2×
(x−
2
3
2
)+1
(x−2)
2
+9=4((x−
2
3
)
2
+1)
x−4x
2
+4+9=4(x
2
−3x+
4
9
+1
x
2
−4x+13=4x
2
−12x+13
−3x
2
+8x=0
x(−3x+8)=0
x=0or−3x+8=0
x=0or
3
8
x=0or2
3
2
hopeithelpsyoupleasemarkitasbrainliest
For the equation x
2
−(a+1)x+2a=0 to have exactly one root lying between the interval (0,3), the following conditions should be satisfied,
(i)f(0)×f(3)<0
⇒(2a)(9−3(a+1)+2a)<0
⇒a(6−a)<0
⇒a(a−6)>0
⇒a<0 and a>6
(ii) Discriminant, D≥0
⇒(a+1)
2
−8a≥0
⇒a
2
−6a+1≥0
⇒(a−3)
2
−8≥0
⇒(a−(3+2
2
))(a−(3−2
2
))≥0
⇒a≥(3+2
2
) and a≤(3−2
2
)
The part of intersection for both of these conditions gives the set of values of a
i.e. a<0 and a>6
Also we have to check for the end points.
For a=0 we get the roots to be 0 and 1. Hence 0 is also possible. But for a=6 we get the roots to be 3 and 4, which does not lies in (0,3).
Hence, option 'B' is correct.