Quant Boosters  Hemant Malhotra  Set 11

Let we have an equation ax^2+bx+c = 0 and roots of this is p and q now find equation whose roots are reciprocal to roots of this given equation
 reciprocal so just replace x by 1/x
a(1/x)^2+b(1/x)+c=0 so cx^2+bx+a=0
let in same question we want equation whose roots are cube of give equation so
so just replace x by x^1/3
NOTE : in square we will replace x by x^1/2 , cube by x^1/3 , in cube root x by x^3 and so on :) ... let we have to find equation whose roots are 2 more than the roots of given equation then replace x by x2 .. so we have to follow just reverse process.

Q3) If sum of two roots of equation x^3  px^2 + qx  r is zero. then
a) pq = r
b) qr = p
c) pr = q
d) pqr = 1

sum of roots=a+b+c=p
given a+b=0
so c=p
we know that c is a root of equation so it will satisfy
so p^3p^3+qpr=0
so pq=r

Q4) If a and b (does not equal to zero) are roots of x^2 + ax + b = 0 find least value of x^2 + ax + b (x is real)
a) 9/4
b) 9/4
c) 1/4
d) 1/4

a and b are roots of x^2+ax+b
so a+b = a
so 2a+b=0
now product of roots=a*b=b
b(a1)=0case1 a1=0 so a=1
2a+b=0 so b=2case2
b=0but this is given that b is not zero
so a=1 and b=2
so x^2+ax+b ( min)
= D/4a=4ba^2/4=9/4min value of ax^2+bx+c is D/4a

Q5) If the equation 2x^2 + kx  5 = 0 and x^2  3x  4 = 0 have one root in common then k = ?
a) 3
b) 27/4
c) 27/4
d) not

let p is common root
2p^2+kp5=0
p^23p4=0
now solve these equations
p^2/(4k15)=p/(5+8)=1/(6k)
so p^2=(4k+15)/(k+6)
and p =3/(k+6)
so (3/k+6)^2=4k+15/(k+6)
so k=3 or 27/4

Q6) The number of values of pair (a,b) for which a(x+1)^2 + b( x^2  3x  2) + x + 1 = 0 is an identity in x
a) 0
b) 1
c) 2
d) infinite

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