Equations

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Equations

by harsh.champ » Tue Feb 09, 2010 4:03 am
The number of non-negative real roots of 2^x - x - 1 = 0 equals

(A)0
(B)1
(C)2
(D)3
(E)4

The OA is C.

For reference,here is the soln.[spoiler]

2^x - x - 1 = 0
2^x - 1 = x
If we put x = 0, then this is satisfied and we put x = 1, then this is also satisfied.
Now we put x = 2, then this is not valid.[/spoiler]
Last edited by harsh.champ on Tue Feb 09, 2010 4:28 am, edited 1 time in total.
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by shashank.ism » Tue Feb 09, 2010 4:21 am
harsh.champ wrote:The number of non-negative real roots of 2x - x - 1 = 0 equals

(A)0
(B)1
(C)2
(D)3
(E)4

The OA is C.

For reference,here is the soln.[spoiler]

2x - x - 1 = 0
2x - 1 = x
If we put x = 0, then this is satisfied and we put x = 1, then this is also satisfied.
Now we put x = 2, then this is not valid.[/spoiler]

if x=0 or 1 eq is satisfied
for x<1 after that since 2^x is an exponential function.so it rises exponentially to infinity . hence only two solution
while x rises linearly to infinity

Ans C
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by ajith » Tue Feb 09, 2010 5:51 am
harsh.champ wrote:The number of non-negative real roots of 2^x - x - 1 = 0 equals

(A)0
(B)1
(C)2
(D)3
(E)4
2^x - x - 1 = 0
2^x = x+1

For x =0
1 = 0+1 so, x=0 is a root

for x =1
2 = 1+1

so x =1 is a root

C
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