Q 11 :

The number of real solutions of the equation x|x+3|+|x-1|-2=0 is:              [2026]

  • 3

     

  • 2

     

  • 5

     

  • 4

     

(1)

(I)  

x2+3x+x-1-2=0

x2+4x-3=0

x=-2+7 (rejected),  -2-7 (rejected)

(II)  

x2+3x+1-x-2=0

x2+2x-1=0

x=-1+2,  -1-2

(III)  

-x2-3x+1-x-2=0

x2+4x+1=0

x=-2-3,  -2+3 (rejected)



Q 12 :

The number of distinct real solutions of the equation x|x+4|+3|x+2|+10=0 is                [2026]

  • 1

     

  • 2

     

  • 3

     

  • 0

     

(1)

Case I x<-4

x(-x+4)+3(-x+2)+10=0

x2+7x-4=0

x=-7+652  or  -7-652

                    Reject                  Accept

Case II -4x<-2

x(x+4)+3(-(x+2))+10=0

x2+x+4=0

D<0    No solution

Case III x-1

x(x+4)+3(x+2)+10=0

x2+7x+16=0

D<0    No solution

No. of solutions=1



Q 13 :

The smallest positive integral value of a, for which all the roots of x4-ax2+9=0 are real and distinct, is equal to        [2026]

  • 7

     

  • 3

     

  • 4

     

  • 9

     

(1)

x4-ax2+9=0  ...(1)

let x2=t

t2-at+9=0  ...(2)

For roots of equation (1) to be real & distinct, roots of equation (2) must be positive & distinct.

(i) D>0a2-36>0a(-,-6)(6,)

(ii) -b2a>0a2>0a>0

(iii) f(0)>09>0a

By (i)  (ii)  (iii)

 a(6,)

 least integral value of a=7



Q 14 :

The sum of all the real solutions of the equation

log(x+3)(6x2+28x+30)=5-2log(6x+10)(x2+6x+9) is equal to.   [2026]

  • 0

     

  • 2

     

  • 4

     

  • 1

     

(1)

logx+3[(x+3)(6x+10)]=5-2log6x+10(x+3)2

1+logx+3(6x+10)=5-4log6x+10(x+3)

Let logx+3(6x+10)=A

A+4A=4 or A=2

logx+3(6x+10)=2

6x+10=(x+3)2

6x+10=x2+9+6x

x2=1, x=±1

So sum of roots =0



Q 15 :

Let α and β be the roots of the equation x2+2ax+(3a+10)=0 such that α<1<β. Then the set of all possible values of a is :           [2026]

  • (-,-115)

     

  • (-,-3)

     

  • (-,-2)(5,)

     

  • (-,-115)(5,)

     

(1)

  α<1<β

f(1)<0

1+2a+(3a+10)<0

5a+11<0

a<-115

  a(-,-115)



Q 16 :

The positive integer n, for which the solutions of the equation

x(x+2)+(x+2)(x+4)++(x+2n-2)(x+2n)=8n3

are two consecutive even integers, is :         [2026]

  • 3

     

  • 12

     

  • 9

     

  • 6

     

(1)

x(x+2)+(x+2)(x+4)++(x+2n-2)(x+2n)=8n3

r=1n(x+2r-2)(x+2r)=8n3

nx2+2xr=1n(2r-1)+4r=1nr(r-1)=8n3

nx2+2xn2+4n(n2-1)3-8n3=0

x2+2nx+4(n2-1)3-83=0

 |α-β|=2D|a|=2D=4

4n2-4(4(n2-1)3-83)=4

n2-4n23=-3

n2=9

n=3



Q 17 :

Let α, β be the roots of the quadratic equation 12x2-20x+3λ=0,  λZ. If 12|β-α|32, then the sum of all possible values of λ is.    [2026]

  • 1

     

  • 3

     

  • 6

     

  • 4

     

(2)

12|α-β|32

14|α-β|294

14(α+β)2-4αβ94

14259-4×λ494

-9136-λ-1936

1936λ9136

λ=1,2

Sum=3



Q 18 :

If α,β where α<β are the roots of the equation λx2-(λ+3)x+3=0 such that 1α-1β=13, then the sum of all possible values of λ is   [2026]

  • 8

     

  • 2

     

  • 6

     

  • 4

     

(3)

β-ααβ=13,  α+β=λ+3λ,  αβ=3λ

β-α=αβ3=1λ

on squaring

α2+β2-2αβ=1λ2  ...(1)

α2+β2+2αβ=(λ+3)2λ2  ...(2)

(2) - (1)                     4αβ=(λ+3)2-1λ2

                                 12λ=λ2+6λ+8λ2

λ2-6λ+8λ=0

λ=0,2,4

Sum of possible values of λ is 6