Quant Boosters - Hemant Malhotra - Set 11

[hemant_malhotra]

Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source - Elite's Grid Prep Forum

[hemant_malhotra]

Q1) Find maximum number of real roots (+ve , -ve) for the equations x^3 + 6x^2 + 11x - 6 = 0

[hemant_malhotra]

The max number of +ve real roots of a polynomial f(x)=0 is d number of changes of signs .
the max number of -ve real roots of f(x)=0 is number of changes of sign in f(-x)=0
so f(x)=x^3+6x^2+11x-6
positive real roots = + + + - so 1 sign change so max number of +ve real roots=1
for negative real roots f(-x)=-x^3+6x^2-11x-6
sign changes are as - + - - : 2 changes so 2 negative real roots max

this rule is known as Descarte's rule of sign. This will give u only max number of real roots .. there is possibility that roots are less than that

[hemant_malhotra]

Q2) Find equation whose roots are reciprocal of the roots of ax^2 + bx + c = 0
a) cx^2+bx+a
b) bx^2+cx+A=0
c) cx^2+ax+b=0
d) bx^2+ax+c=0

[hemant_malhotra]

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 x-2 .. so we have to follow just reverse process.

[hemant_malhotra]

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

[hemant_malhotra]

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^3-p^3+qp-r=0
so pq=r

[hemant_malhotra]

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

[hemant_malhotra]

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(a-1)=0

case1- a-1=0 so a=1
2a+b=0 so b=-2

case2-
b=0

but this is given that b is not zero
so a=1 and b=-2
so x^2+ax+b ( min)
= -D/4a=4b-a^2/4=-9/4

min value of ax^2+bx+c is -D/4a

[hemant_malhotra]

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

[hemant_malhotra]

let p is common root
2p^2+kp-5=0
p^2-3p-4=0
now solve these equations
p^2/(-4k-15)=p/(-5+8)=1/(-6-k)
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

[hemant_malhotra]

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

[hemant_malhotra]

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

[hemant_malhotra]

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

[hemant_malhotra]

f(x)=x^3+3x^2+4x+5
g(x)=x^3+2x^2+7x+3
f(x)-g(x)
x^2-3x+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

[hemant_malhotra]

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

[hemant_malhotra]

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^x-x^2+1
f(3)=0
f(3.2) < 0
f(4)=2^4-16+1>0
so one root will lie between 3 and 4
now f(-1)=1/2-1+1>0
f(-2)=1/4-4+1
so one root lie between -1 and -2
and f(3)=0 so one root 3
so 3 roots

every equation of an odd degree has at least one real root
x^3-6x^2+15x+3=0 has atleast one real root and max real roots=3
now question
f(x)=x^3-6x^2+15x+3
f'(x) ((differentiation)
3x^2-12x+15=3((( x-2)^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

[hemant_malhotra]

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

[hemant_malhotra]

x^3 - ax^2 + bx - a = 0
(x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rx)x-pqr
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

[hemant_malhotra]

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