Q 11 :

The set of all real numbers x for which x2-|x+2|+x>0, is

  • (-,-2)(2,)

     

  • (-,-2)(2,)

     

  • (-,-1)(1,)

     

  • (2,)

     

(2)

x2-|x+2|+x>0

When x+2>0,

Then  x2-(x+2)+x>0,  x2-x-2+x>0x2-2>0

      (x-2)(x+2)>0

      x(-2,-2)(+2,)                        ...(i)

When x+2<0, |x+2|=-(x+2),

  x2+x+2+x>0  xR                  ...(ii)

From (i) and (ii), we get x(-,-2)(2,)



Q 12 :

Let α and β be the roots of the equation x2-6x-2=0. If an=αn-βn, for n1, then the value of a10-2a82a9 is equal to

  • 3

     

  • - 3

     

  • 6

     

  • - 6

     

(1)

α is a root of x2-6x-2=0

Then  α2-6α-2=0

Multiplying by αn, it becomes,

       αn+2-6αn+1-2αn=0                          ...(i)

Similarly,  βn+2-6βn+1-2βn=0                 ...(ii)

Subtracting (ii) from (i), we get

       (αn+2-βn+2)-6(αn+1-βn+1)-2(αn-βn)=0

i.e.,  an+2-6an+1-2an=0

Thus,  an+2-2an2an+1=3

Set n=8 to obtain the desired value a10-2a82a9=3



Q 13 :

If one root of a quadratic equation is 11+3, then the quadratic equation is

  • 2x2+x-1=0

     

  • 2x2-2x-1=0

     

  • 2x2+2x+1=0

     

  • 2x2+2x-1=0

(4)

One root=11+3=1-3-2=-12+32

  Other root=-12-32

Sum=-1,    product=14-34=-12

  Required equation is x2+x-12=0

or    2x2+2x-1=0



Q 14 :

Let α and β be the roots of the equation, 5x2+6x-2=0. If Sn=αn+βn, n=1,2,3,, then

  • 6S6+5S5=2S4

     

  • 6S6+5S5+2S4=0

     

  • 5S6+6S5=2S4

     

  • 5S6+6S5+2S4=0

     

(3)

Given α,β are the roots of the equation 5x2+6x-2=0.

  α+β=-65  and  αβ=-25

Now,  Sn=αn+βnSn+1=αn+1+βn+1

Sn+1=(αn+βn)(α+β)-αβ(αn-1+βn-1)

           =Sn(-65)-(-25)Sn-1                [ α+β=-65,  αβ=-25  and  αn+βn=Sn]

  Sn+1=-65Sn+25Sn-1  Sn+1=25(Sn-1-3Sn)

  5Sn+1=2(Sn-1-3Sn), which satisfies option (3).



Q 15 :

Let aR and let α, β be the roots of the equation x2+6014x+a=0. If  α4+β4=-30, then the product of all possible values of a is _______.



(45)

x2+6014x+a=0                       ...(i)

As we have, α4+β4=(α2+β2)2-2α2β2

=((α+β)2-2αβ)2-2α2β2                ...(ii)

Now, α+β=-6014  and  αβ=a      [From (i)]

(α+β)2=(60)12=60  and  α2·β2=a2               ...(iii)

Also, α4+β4=-30                                                     ...(iv)

So, substituting values from (iii) and (iv) in (ii), we get

a2-260a+45=0  a2-415a+45=0

  (a-315)(a-15)=0

Product of values of a=315×15=3×15=45



Q 16 :

Let α, β be the roots of the equation x2+22x-1=0. The quadratic equation, whose roots are α4+β4 and 110(α6+β6), is:

  • x2-195x+9466=0

     

  • x2-180x+9506=0

     

  • x2-195x+9506=0

     

  • x2-190x+9466=0

     

(3)

We have,  x2+22x-1=0

  α+β=-22  and  αβ=-1

Now, α2+β2=(α+β)2-2αβ=(-22)2-2(-1)=8+2=10

Also, α4+β4=(α2+β2)2-2(αβ)2=(10)2-2(1)=100-2=98

and  α6+β6=(α2+β2)3-3α2β2(α2+β2)=(10)3-3(1)(10)=1000-30=970

So,  110(α6+β6)=97

Hence, equation whose roots are α4+β4 and 110(α6+β6) is

x2-(98+97)x+98×97=0  i.e., x2-195x+9506=0



Q 17 :

The solution set of the inequation x2+6x-7|x+4|<0 is

  • (-7,-4)

     

  • (-7,-4)(4,1)

     

  • (-7,1)

     

  • (-7,-4)(-4,1)

     

(4)

Given inequation is

x2+6x-7|x+4|<0                                    ...(i)

Since |x+4|>0  xR except for x=-4

  for x=-4, (i) does not exist.

  x2+6x-7<0  for given inequation

or   x2+7x-x-7<0

or   (x+7)(x-1)<0  or  x(-7,1)

  Solution set of (i) is (-7,-4)(-4,1)



Q 18 :

Sum of the roots of the equation |x-3|2+|x-3|-2=0 is equal to

  • 2

     

  • 4

     

  • 6

     

  • 16

     

(3)

We have,  |x-3|2+|x-3|-2=0

  |x-3|=1x-3=±1x=2, 4



Q 19 :

If α, βC are the distinct roots of the equation x2-x+1=0, then α101+β107 is equal to

  • 1

     

  • 2

     

  • - 1

     

  • 0

     

(1)

We have,  x2-x+1=0

  x=-1(-1)±1-42=1±3i2

  Let α=1+3i2  and  β=1-3i2

  α=-ω2  and  β=-ω,  where ω=-1+3i2

and  ω2=-1-3i2

Now,  α101+β107=(-ω2)101+(-ω)107

             =-ω202+(-ω)107

             =-ω-ω2=-(ω+ω2)=-(-1)    ( 1+ω+ω2=0)

             =1



Q 20 :

The sum of the roots of the equation  x+1-2log2(3+2x)+2log4(10-2-x)=0, is

  • log214

     

  • log212

     

  • log213

     

  • log211

     

(4)

We have,  x+1-2log2(3+2x)+2log2(10-2-x)=0

  (x+1)+log2(10·2x-12x)-log2(3+2x)2=0

  (x+1)+log2[10·2x-1(3+2x)2]-log22x=0

  (x+1)+log2[10·2x-19+(2x)2+6·2x]-x=0

  log2(10·2x-19+(2x)2+6·2x)=-1

  (10·2x-19+(2x)2+6·2x)=12

  20·2x-2=9+(2x)2+6·2x(2x)2-14·2x+11=0

  2x=14±(14)2-4(11)(1)2=14±1522

  2x=7±38

  x=log2(7+38)  or  x=log2(7-38)

  Sum of roots=log2(7+38)+log2(7-38)

         =log2[49-38]=log211



Q 21 :

The least positive value of 'a' for which the equation 2x2+(a-10)x+332=2a has real roots is _______.



(8)

Since, the given equation   2x2+(a-10)x+332=2a has real roots.

  D0(a-10)2-4(2)(332-2a)0

  a2+100-20a-132+16a0

  a2-4a-320(a-8)(a+4)0

  a(-,-4][8,)



Q 22 :

The product of all positive real values of x satisfying the equation x(16(log5x)3-68log5x)=5-16 is ________.



(1)

We have,  x(16(log5x)3-68log5x)=5-16

Taking log both sides and take log5x=t, we get

t(16t3-68t)=-16

  16t4-68t2+16=04t4-17t2+4=0

  t1+t2+t3+t4=0

  log5x1+log5x2+log5x3+log5x4=0x1x2x3x4=1



Q 23 :

p(x)=x4+ax3+bx2+cx+d, where p(1)=10, p(2)=20, p(3)=30, then the value of p(12)+p(-8)10-1980 is ________.



(4)

Let f(x)=p(x)-10x

  f(1)=f(2)=f(3)=0

  p(x)=(x-1)(x-2)(x-3)(x-λ)+10x

         p(12)=11×10×9(12-λ)+120

         p(-8)=(-9)×(-10)×(-11)(-8-λ)-80

  p(12)+p(-8)=11×10×9(12-λ+8+λ)+40

        =11×10×9×20+40=40×496

or  p(12)+p(-8)10=1984

  p(12)+p(-8)10-1980=4