Q 1 :

A value of b for which the equations

x2+bx-1=0

x2+x+b=0

have one root in common is                          [2011]

  • -2

     

  • -i3

     

  • i5

     

  • 2

     

(2)

Let α be the common root of given equations, then

α2+bα-1=0               ...(i)

and  α2+α+b=0             ...(ii)

On subtracting (ii) from (i), we get

(b-1)α-(b+1)=0

⇒α=b+1b-1

Substituting this value of α in equation (i), we get

(b+1b-1)2+b(b+1b-1)-1=0

⇒b3+3b=0

⇒b=0, i3, -i3



Q 2 :

For all 'x', x2+2ax+10-3a>0, then the interval in which 'a' lies is                   [2004]

  • a<-5

     

  • -5<a<2

     

  • a>5

     

  • 2<a<5

     

(2)

f(x)=ax2+bx+c has same sign as that of a if D<0.

Since x2+2ax+10-3a>0 ∀x

∴ D<0⇒4a2-4(10-3a)<0

⇒a2+3a-10<0

⇒(a+5)(a-2)<0⇒a∈(-5,2)



Q 3 :

Let f(x)=x4+ax3+bx2+c be a polynomial with real coefficients such that f(1)=-9. Suppose that i3 is a root of the equation 4x3+3ax2+2bx=0 where i=-1. If α1,α2,α3 and α4 are all the roots of the equation f(x)=0, then |α1|2+|α2|2+|α3|2+|α4|2 is equal to _______.            [2024]



(20)

Given that f(1)=-9⇒1+a+b+c=-9    ...(i)

and 4x3+3ax2+2bx=0

⇒x=0, or 4x2+3ax+2b=0    ...(ii)

⇒3i and -3i are roots of (ii)

⇒3i-3i=-3a4,  3i(-3i)=2b4

⇒a=0, b=6, c=-16  from (i)

⇒f(x)=0⇒x4+6x2-16=0

⇒x2=-6±36+642=-3±5=2,-8

⇒x=-2, 2, -22i, 22i

⇒|α1|2+|α2|2+|α3|2+|α4|2=20



Q 4 :

Let S be the set of all non-zero real numbers α such that the quadratic equation αx2-x+α=0 has two distinct real roots x1 and x2 satisfying the inequality |x1-x2|<1. Which of the following intervals is(are) a subset(s) of S?                [2015]

  • (-12,-15)

     

  • (-15,0)

     

  • (0,15)

     

  • (15,12)

     

Select one or more options

(1, 4)

Given, x1 and x2 are roots of αx2-x+α=0

∴ x1+x2=1α  and  x1x2=1

Also, |x1-x2|<1

⇒|x1-x2|2<1⇒(x1-x2)2<1

or  (x1+x2)2-4x1x2<1

⇒1α2-4<1   or  1α2<5

or   5α2-1>0    or    (5α-1)(5α+1)>0

∴  α∈(-∞,-15)∪(15,∞)         ...(i)

Also, D>0

⇒1-4α2>0    or    α∈(-12,12)            ...(ii)

From (i) and (ii), we get

α∈(-12,-15)∪(15,12)