1) If the equation x^2+2(p+1)x+9p-5=0 has only positive roots ,then which of the following is true? 2) The equation x^2-ax-21=0 and x^2-3ax+35=0 ,where a is positive have one common root .What is the value of a? 3) Which of the following equation not have real roots? 4) For what value of k ,the equations 2x+3y=5 and 4x+ky=A will not have any solution? 5) If x is smaller than 0,what is the maximum value of (9/x+x/9)? 6) If the coefficient of x^2 and the constant trem of a quadratic equation are interchanged,then which of the following will not get changed? 7) 5 mangoes ,6 guavas and 7 oranges costs Rs. 178.While 6 mangoes,4 guavas and 2 oranges costs Rs. 124 .What will be the cost of 3 mangoes +3 guavas and 3 oranges? __________________
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1) If the equation x^2+2(p+1)x+9p-5=0 has only positive roots ,then which of the following is true?
a) p greater and equal to 6.
b) p smaller than or equal to 0.
c) p smaller than and equal to -6.
d) NOT
ans:
The first condn is
[2(p+1)]^2 > 4(9p - 5)
=> (p^2 + 1 + 2p) > 9p - 5
=> p^2 -7p + 6 > 0
=> p^2 -6p -p + 6 > 0
=> p(p - 6) -1(p - 6) > 0
=> (p - 6)(p - 1) > 0
p >= 1 and p >= 6. . . .so the cond is p >= 6
or, p =< -1 and p =< -6. . . .so the cond is p =< -6
for p = 6, the eqn is
x^2 + 14x + 45 = 0 and both the roots are positive
for p = -6, the eqn is
x^2 + 10x - 50 = 0 but here both the roots are not positive
ans a) p greater and equal to 6.
Soln.: Let the common root is m
Frm 1st eqn. we get m-21/m=a
Frm 2nd eqn we get m+35/m=3a
Solving them we get a=4.
So option (c) is correct.....
Giv ur best to d world.
Nd d best will cm back to u.....
Regards,
Dipanjan.......
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Soln.: Easy one....As for 3rd eqn 3^2-4*5*3<0
It has no real roots....
Giv ur best to d world.
Nd d best will cmback to u.....
Regards,
Dipanjan......
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Soln.2/4=3/k
=>k=6
So option (c) is correct....
Giv ur best to d world.
Nd d best will cm back to u.....
Regards,
Dipanjan......
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Soln. As 9/x+x/9 =(x+9)^2/9x - 2 & x not equals to 0
So option (d) is correct....
Giv ur best to d world.
Nd d best will cm back to u......
Regards,
Dipanjan.......
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Soln.: Option (b) is correct...
Giv ur best to d world.
Nd d best will cm back to u....
Regards,
Dipanjan......
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Soln.: 5m+6g+7o=178 =>10m+12g+14o=356...(i)
6m+4g+2o=124...(ii)
(i)+(ii)
16(m+g+o)=480
=>3(m+g+o)=90
So option (a) is correct....
Giv ur best to d world.
Nd d best will cm back to u....
Regards,
Dipanjan......
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How did you solved the 4th one??
"To get the BEST, be the BEST"
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In question no. 6 -
Let the equation be ax^2+bx+c=0
And the two roots be m and n.
Product of the roots,mxn=c/a.
Now the coefficient of x^2 and the constant term changes.The equation becomes,cx^2+bx+a=0.
Let the roots be s and t.
Product of the roots ,sxt=a/c.
So how did you opted for b)??
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Also look at Question no.5.
The answer is b) -2.
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