1) If one of the roots of the cubic expression x^3-ax^2+11x-6=3;What are the other roots. 2) If x^3-ax^2+bx+10 is perfectly divided by (x-3) then,(a-3b)=? 3) How many numbers in the set {-4,-3,0,2] satisfy the condition mod(y-4) is smaller or equal to 6 and mod(y+4) smaller than 5? 4) If the equation 2x^2-7x+1=0 and ax^2+bx+2=0 have a commomn root,then 5)The nummber of real solution of e^x=x is 6)The number of real solution of x-(1/x^2-4) = -2(1/x^2-4) 7)The number of integral roots of the equation x^3-x^2+2x-17=0 8)If (x+2) is a factor of x^4-4x^2+2ax+3=0,then what is the value of a? 9)Given that as^4+bs^3+cs^2+ds+e=0 is a biquadratic equation in s and a is not equal to 0. Value of (1-s1)(1-s2)(1-s3)(1-s4) is? 10) How many real roots will be there of the equation -x^2+5 modx+6=0 __________________
n/a |
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Soln.As f(3)=0
=>27-9a+3b+10=0
=>3(b-3a)=-37
=>(b-3a)=-37/3
not getin a-3b
Regards,
Dipanjan
n/a
Soln.(d)not
regards,
Dipanjan
n/a
10) How many real roots will be there of the equation -x^2+5 modx+6=0
a)0 b) 2 c) 4 d) NOT
ans is 4
-x^2+5 x+6=0 and -x^2-5 x+6=0
n/a
Given x=3 is root for the eqn =>f(3)=0.
From that `a` will be obtained as 6;
the eqn now becomes
x^3-6x^2+11x-6=0
The solutions are 3,2and1.
n/a
The answer of Q:10 is not correct.
n/a
For x<0;
-x2-5x+6=0
solving this x=-6...(x not equal to 1 as x must be <0)
for x>0;
-x2+5x+6=0
solving this x=6...(x not equal to -1 as x>0)
So the answer is two real roots.
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