Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Elite's Grid Prep Forum ]]>

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

]]>a) cx^2+bx+a

b) bx^2+cx+A=0

c) cx^2+ax+b=0

d) bx^2+ax+c=0 ]]>

-- 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. ]]>

a) pq = r

b) qr = p

c) pr = q

d) pqr = 1 ]]>

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 ]]>

a) 9/4

b) -9/4

c) -1/4

d) 1/4 ]]>

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

]]>a) -3

b) -27/4

c) 27/4

d) not ]]>

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 ]]>

a) 0

b) 1

c) 2

d) infinite ]]>

so in that case, a = b = c = 0 ]]>

a) 0

b) 1

c) 2

d) 3 ]]>

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 ]]>

a) only one positive root

b) 2 +ve and 1 -ve

c) no +ve root

d) not ]]>

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

a) a = 11

b) a not equal to 1

c) b = 1

d) b not equal to 1 ]]>

(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 ]]>

a) 1

b) 0

c) 2

d) not ]]>

a) 2

b) 3

c) 4

d) none of the above ]]>

then D < 0

4(4a-1)^2-4(15a^2-2k-7) < 0

a^2-6a+8 < 0

2 < a < 4

so integral value is 3 ]]>

a) q = 1/p

b) q = b

c) q = c/b

d) q = b/c ]]>

p+q+r = 3a

pq + qr + rp = 3b

pqr = c

q = 2pr/p+r

1 = 2c/q / [3b-c/q]

3b = 3c/q

q=c/b

approach 2

roots are in HP

so 1/p,1/q,1/r are the roots of

(1/x)^3-3z(1/x)^2+3b(1/x)-c=0

so -cx^3+3bx^2-3ax+1=0 are in AP

so 1/p+1/q+1/r=3b./c

so 3/q=3b/c

so q=c/b