Quant Boosters  Hemant Malhotra  Set 11

If any quadratic equation ax^2 + bx + c = 0 has more than two roots then it becomes an identity.
so in that case, a = b = c = 0

Q7) Number of common roots between two equations x^3 + 3x^2 + 4x + 5 = 0 and x^3 + 2x^2 + 7x + 3 = 0
a) 0
b) 1
c) 2
d) 3

f(x)=x^3+3x^2+4x+5
g(x)=x^3+2x^2+7x+3
f(x)g(x)
x^23x+2=0
x = 1, 2
( now when u got this root there may be the case that ye value initial equatuion ko satisfy na kare so we should always check this_ these roots are know as false roots ...
put x=1 and x=2 in initial equation also

Q8) x^3  6x^2 + 15x + 3 = 0 has
a) only one positive root
b) 2 +ve and 1 ve
c) no +ve root
d) not

Roll's theroem
it states that if two value a and b are such that f(a)>0 and
f(b) < 0
then there must be a real root in between a and b
Example 
if f(x)=f(x)=2^xx^2+1
f(3)=0
f(3.2) < 0
f(4)=2^416+1>0
so one root will lie between 3 and 4
now f(1)=1/21+1>0
f(2)=1/44+1
so one root lie between 1 and 2
and f(3)=0 so one root 3
so 3 rootsevery equation of an odd degree has at least one real root
x^36x^2+15x+3=0 has atleast one real root and max real roots=3
now question
f(x)=x^36x^2+15x+3
f'(x) ((differentiation)
3x^212x+15=3((( x2)^2+1))
so f'(x)>0 means function is increasing so it will cut x axis only once so one real root only
so this equation has one real root either positive or negative
now f(0)=3
f(1) < 0
so one root will lie between o and 1 so negative real root

Q9) x^3  ax^2 + bx  a = 0 has three real roots then which of the following is true
a) a = 11
b) a not equal to 1
c) b = 1
d) b not equal to 1

x^3  ax^2 + bx  a = 0
(xp)(xq)(xr)=x^3(p+q+r)x^2+(pq+qr+rx)xpqr
now compare
b=pq+qr+rp
a=p+q+r=pqr
(p+q+r/pqr)=1
sk 1/pq+1/qr+1/rp=1
so pq+qr+rp > 3 so b does not equal to 1

Q10) root(x+1)root(x1)=root(4x1) find number of real values of x
a) 1
b) 0
c) 2
d) not

When you solve, you will get x=5/4 but that value will not satisfy initial equation so zero roots

Q11) Find integral value of a for which x^2  2(4a  1)x + 15a^2  2a  7 > 0 is valid for any x
a) 2
b) 3
c) 4
d) none of the above

if f(x) > 0 for all values of x
then D < 0
4(4a1)^24(15a^22k7) < 0
a^26a+8 < 0
2 < a < 4
so integral value is 3

Q12) If roots p,q,r in HP of equation x^3  3ax^2 + 3bx  c = 0 then
a) q = 1/p
b) q = b
c) q = c/b
d) q = b/c

approach 1
p+q+r = 3a
pq + qr + rp = 3b
pqr = c
q = 2pr/p+r
1 = 2c/q / [3bc/q]
3b = 3c/q
q=c/bapproach 2
roots are in HP
so 1/p,1/q,1/r are the roots of
(1/x)^33z(1/x)^2+3b(1/x)c=0
so cx^3+3bx^23ax+1=0 are in AP
so 1/p+1/q+1/r=3b./c
so 3/q=3b/c
so q=c/b

Q13) For all real values of find min and max value of (x^23x+4)/(x^2+3x+4)

x^23x+4/(x^2+3x+4) = y
now make a quadratic in x
x^2(y1)+3x(y+1)+4(y1) = 0
now real x so D > = 0
9(y+1)^216(y1)^2 > = 0
so (7y1)(y7) < = 0
so 1/7 < = y < = 7

Q14) Number of possible value of integer p for which x^2 + px + 16 = 0 has integral roots
a) 4
b) 6
c) 2
d) not

For roots to be integral Discriminant should be perfect square
D=k^2
a^264=k^2
a^2k^2=64
Now easy ? Answer is 6

Q15) In how many ways we arrange 6 boys and 4 girls such that
a) all girls are together
b) all boys are together
c) all boys are together , all girls together
d) no girls are together

a) all girls are together so make a group of them and take it as a one unit
so they could be arranged in 4! ways so now we have BBBBBB(G) so total 7 persons , number of ways to arrange those 7!
so total ways =7! * 4b) all boys are together so make it as one entity so they could be arranged in 6! ways
so total number of ways = 6! * 5!c) all boys are together and all girls are together
so BBBBBBGGGG
6!*4!
and GGGGBBBBBB
so 4 * 6!
so total ways = 2 * 4! * 6!d) no two girls are together
so _ B _ B _ B _ B _ B _ B _
first arrange 6 boys so ways 6!
now 6 boys will leave 7 spaces
now select 4 spaces and arrange 4 girls there
so 7c4*4!
so total ways = 6! * 7c4 * 4!

Q16) Five boys and five girls form a line with the boys and girls alternating . The number of ways of making the line