Q 1 :

Suppose a,b denote the distinct real roots of the quadratic polynomial x2+20x-2020 and suppose c,d denote the distinct complex roots of the quadratic polynomial x2-20x+2020. Then the value of ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d) is                      [2020]

  • 0

     

  • 8000

     

  • 8080

     

  • 16000

     

(4)

Consider the quadratic polynomials in the form of equation

x2+20x-2020=0    ⋯(i)

x2-20x+2020=0    ⋯(ii)

Since, a and b are roots of the equation (i), then

a+b=-20,  ab=-2020

∵ c and d are the roots of the equation (ii), then

c+d=20,  cd=2020

Now,

ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)

=a2c-ac2+a2d-ad2+b2c-bc2+b2d-bd2

=a2(c+d)+b2(c+d)-c2(a+b)-d2(a+b)

=(c+d)(a2+b2)-(a+b)(c2+d2)

=(c+d)((a+b)2-2ab)-(a+b)((c+d)2-2cd)

=20[(20)2+4040]+20[(20)2-4040]

=20×800=16000



Q 2 :

Let -π6<θ<-π12. Suppose α1 and β1 are the roots of the equation x2-2xsecθ+1=0 and α2 and β2 are the roots of the equation x2+2xtanθ-1=0. If α1>β1 and α2>β2, then α1+β2 equals                         [2016]

  • 2(secθ-tanθ)

     

  • 2secθ

     

  • -2tanθ

     

  • 0

     

(3)

x2-2xsecθ+1=0⇒x=secθ±tanθ

and x2+2xtanθ-1=0⇒x=-tanθ±secθ

∵ -π6<θ<-π12⇒secπ6>secθ>secπ12

and -tanπ6<tanθ<-tanπ12

Also tanπ12<-tanθ<tanπ6

Since, α1,β1 are roots of x2-2xsecθ+1=0 and α1>β1

∴ α1=secθ-tanθ and β1=secθ+tanθ

Since, α2,β2 are roots of x2+2xtanθ-1=0 and α2>β2

∴ α2=-tanθ+secθ,  β2=-tanθ-secθ

∴ α1+β2=secθ-tanθ-tanθ-secθ=-2tanθ



Q 3 :

The quadratic equation p(x)=0 with real coefficients has purely imaginary roots. Then the equation p(p(x))=0 has                   [2014]

  • one purely imaginary root

     

  • all real roots

     

  • two real and two purely imaginary roots

     

  • neither real nor purely imaginary roots

     

(4)

Quadratic equation with real coefficients and purely imaginary roots can be considered as

p(x)=x2+a=0  where a>0 and a∈ℝ

Then p[p(x)]=0⇒(x2+a)2+a=0

⇒x4+2ax2+(a2+a)=0

⇒x2=-2a±4a2-4(a2+a)2=-2a±4a2-4a2-4a2

=-2a±-4a2⇒x2=-a±ai

⇒x=-a±ai=α±iβ, where α,β≠0

∴ p[p(x)]=0 has complex roots which are neither purely real nor purely imaginary.



Q 4 :

If a∈R and the equation -3(x-[x])2+2(x-[x])+a2=0 (where [x] denotes the greatest integer ≤x) has no integral solution, then all possible values of a lie in the interval:               [2014]

  • (-2,-1)

     

  • (-∞,-2)∪(2,∞)

     

  • (-1,0)∪(0,1)

     

  • (1,2)

     

(3)

Consider -3(x-[x])2+2[x-[x]]+a2=0

⇒3{x}2-2{x}-a2=0    (∵x-[x]={x})

⇒3({x}2-23{x})=a2, a≠0

⇒a2=3({x}-13)2-13    (∵0≤{x}<1)

-13≤{x}-13<23⇒0≤3({x}-13)<43

-13≤3({x}-13)-13<1

For non-integral solution 0<a2<1

⇒a∈(-1,0)∪(0,1)



Q 5 :

Let α and β be the roots of x2-6x-2=0, with α>β. If an=αn-βn  for n≥1, then the value of a10-2a82a9 is               [2011]

  • 1

     

  • 2

     

  • 3

     

  • 4

     

(3)

∵ α,β are the roots of x2-6x-2=0

∴ α2-6α-2=0

⇒α10-6α9-2α8=0

⇒α10-2α8=6α9    ⋯(i)

Similarly β10-2β8=6β9    ⋯(ii)

On subtracting (ii) from (i),

α10-β10-2(α8-β8)=6(α9-β9)

⇒a10-2a8=6a9

⇒a10-2a82a9=3



Q 6 :

Let (x0,y0) be the solution of the following equations (2x)ln2=(3y)ln3;  3lnx=2lny. Then x0 is                   [2011]

  • 16

     

  • 13

     

  • 12

     

  • 6

     

(3)

Given : (2x)ln2=(3y)ln3

⇒ln2·ln(2x)=ln3·ln(3y)

⇒ln2·ln(2x)=ln3·(ln3+lny)    ⋯(i)

Also given : 3lnx=2lny

⇒lnx·ln3=lny·ln2⇒lny=lnx·ln3ln2    ⋯(ii)

From equation (i) and (ii), we get

ln2·ln(2x)=ln3[ln3+lnx·ln3ln2]

⇒(ln2)2ln(2x)=(ln3)2ln2+(ln3)2lnx

⇒(ln2)2ln(2x)=(ln3)2(ln2+lnx)

⇒(ln2)2ln(2x)-(ln3)2ln(2x)=0

⇒[(ln2)2-(ln3)2]ln(2x)=0⇒ln(2x)=0

⇒2x=1⇒x=12



Q 7 :

Let p and q be real numbers such that p≠0, p3≠q and p3≠-q. If α and β are nonzero complex numbers satisfying α+β=-p and α3+β3=q, then a quadratic equation having αβ and βα as its roots is                      [2010]

  • (p3+q)x2-(p3+2q)x+(p3+q)=0

     

  • (p3+q)x2-(p3-2q)x+(p3+q)=0

     

  • (p3-q)x2-(5p3-2q)x+(p3-q)=0

     

  • (p3-q)x2-(5p3+2q)x+(p3-q)=0

     

(2)

Given : α+β=-p and α3+β3=q

⇒(α+β)3-3αβ(α+β)=q

⇒-p3-3αβ(-p)=q⇒αβ=p3+q3p

Now for required quadratic equation,

Sum of roots =αβ+βα=α2+β2αβ

=(α+β)2-2αβαβ=p2-2(p3+q3p)p3+q3p=p3-2qp3+q

and Product of roots =αβ·βα=1

∴ Required equation is x2-(p3-2qp3+q)x+1=0

⇒(p3+q)x2-(p3-2q)x+(p3+q)=0



Q 8 :

Let α,β be the roots of the equation x2-px+r=0 and α2,2β be the roots of the equation x2-qx+r=0. Then the value of r is             [2007]

  • 29(p-q)(2q-p)

     

  • 29(q-p)(2p-q)

     

  • 29(q-2p)(2q-p)

     

  • 29(2p-q)(2q-p)

     

(4)

Since α and β are the roots of x2-px+r=0

∴ α+β=p    ⋯(i)

      and  αβ=r    ⋯(ii)

Also α2 and 2β are the roots of x2-qx+r=0

∴ α2+2β=q⇒α+4β=2q           ⋯(iii)

Solving (i) and (iii) for α and β, we get

β=13(2q-p)  and  α=23(2p-q)

On substituting the values of α and β in equation (ii),

we get 29(2p-q)(2q-p)=r.



Q 9 :

Let a,b,c be the sides of a triangle where a≠b≠c and λ∈R. If the roots of the equation x2+2(a+b+c)x+3λ(ab+bc+ca)=0 are real, then        [2006]

  • λ<43

     

  • λ>53

     

  • λ∈(13,53)

     

  • λ∈(43,53)

     

(1)

∵ a,b,c are sides of a triangle and a≠b≠c

∴ |a-b|<|c|⇒a2+b2-2ab<c2       ⋯(i)

Similarly,

b2+c2-2bc<a2    ⋯(ii),  c2+a2-2ca<b2    ⋯(iii)

On adding (i), (ii) and (iii) we get

a2+b2+c2<2(ab+bc+ca)

⇒a2+b2+c2ab+bc+ca<2         ⋯(iv)

∵ Roots of the given equation are real

∴ (a+b+c)2-3λ(ab+bc+ca)≥0

⇒a2+b2+c2ab+bc+ca≥3λ-2             ⋯(v)

From (iv) and (v), 3λ-2<2⇒λ<43



Q 10 :

If one root is square of the other root of the equation x2+px+q=0, then the relation between p and q is                     [2004]

  • p3-q(3p-1)+q2=0

     

  • p3-q(3p+1)+q2=0

     

  • p3+q(3p-1)+q2=0

     

  • p3+q(3p+1)+q2=0

     

(1)

x2+px+q=0

Let roots be α and α2, then

α+α2=-p,  α·α2=q⇒α=q1/3

∴ (q)1/3+(q1/3)2=-p

On taking cube on both sides, we get

q+q2+3q(q1/3+q2/3)=-p3

⇒q+q2-3pq=-p3

⇒p3+q2-q(3p-1)=0



Q 11 :

For the equation 3x2+px+3=0, p>0, if one of the root is square of the other, then p is equal to                        [2000]

  • 1/3

     

  • 1

     

  • 3

     

  • 2/3

     

(3)

Let α,α2 be the roots of 3x2+px+3=0

∴ α+α2=-p3  and  α3=1

⇒(α-1)(α2+α+1)=0      ⇒α=1  or  α2+α=-1

If α=1, then p=-6, which is not possible as p>0

If α2+α=-1⇒-p3=-1⇒p=3



Q 12 :

If b>a, then the equation  (x-a)(x-b)-1=0 has                  [2000]

  • both roots in (a,b)

     

  • both roots in (-∞,a)

     

  • both roots in (b,+∞)

     

  • one root in (-∞,a) and the other in (b,+∞)

     

(4)

Given : (x-a)(x-b)-1=0, b>a

or x2-(a+b)x+(ab-1)=0

Let f(x)=x2-(a+b)x+(ab-1)

D=(a+b)2-4(ab-1)

   =(a-b)2+1>0

Since coefficient of x2 i.e. 1>0, ∴ f(x) represents upward parabola,

intersecting x-axis at two points corresponding to two real roots, D being positive.

Also f(a)=f(b)=-1

⇒curve is below x-axis at a and b

∴ a and b both lie between the roots.

Therefore, the graph of given equation is as shown.

It is clear from graph, that one root of the equation lies in (-∞,a) and other in (b,∞).



Q 13 :

If α and β(α<β) are the roots of the equation x2+bx+c=0, where c<0<b, then                       [2000]

  • 0<α<β

     

  • α<0<β<|α|

     

  • α<β<0

     

  • α<0<|α|<β

     

(2)

Given : c<0<b and α+β=-b           ⋯(i)

αβ=c    ⋯(ii)

From (ii), c<0⇒αβ<0⇒Either α is negative or β is negative and second quantity is positive.

From (i), b>0⇒-b<0⇒α+β<0

⇒the sum is negative

⇒(Modulus of negative quantity)>(Modulus of positive quantity)

But given α<β. Therefore, it is clear that α is negative and β is positive

and modulus of α is greater than modulus of β

⇒α<0<β<|α|



Q 14 :

The product of all positive real values of x satisfying the equation x(16(log5x)3-68log5x)=5-16 is ________.                 [2022]



(1)

Taking log with base 5 on both sides

(16(log5x)3-68(log5x))(log5x)=-16

Let log5x=t

⇒16t4-68t2+16=0

⇒4t4-17t2+4=0

⇒(4t2-1)(t2-4)=0

⇒t=±12, ±2

So log5x=±12, ±2⇒x=51/2, 5-1/2, 52, 5-2

Product =51/2·5-1/2·52·5-2=1



Q 15 :

For x∈ℝ, then the number of real roots of the equation 3x2-4|x2-1|+x-1=0 is ________.                     [2021]



(4)

3x2+x-1=4|x2-1|

Case 1: If x∈[-1,1],

             3x2+x-1=-4x2+4

⇒7x2+x-5=0  ∵  D=141>0  ∴Equation has two roots

Case 2: If x∈(-∞,-1]∪[1,∞)

3x2+x-1=4x2-4

⇒x2-x-3=0  ∵  D=13>0

∴ Equation has two roots

So, total 4 roots.



Q 16 :

The smallest value of k, for which both the roots of the equation x2-8kx+16(k2-k+1)=0 are real, distinct and have values at least 4, is              [2009]



(2)

The given equation is x2-8kx+16(k2-k+1)=0

∵ Both the roots are real and distinct.

∴ D>0 ⇒ (8k)2-4×16(k2-k+1)>0

⇒k>1                             ...(i)

∵ Both the roots are greater than or equal to 4

∴ α+β>8  and  f(4)≥0

⇒k>1                 ...(ii)

and 16-32k+16(k2-k+1)≥0

⇒k2-3k+2≥0 ⇒(k-1)(k-2)≥0

⇒k∈(-∞,1]∪[2,∞)             ...(iii)

Combining (i), (ii) and (iii), we get k≥2

∴ Smallest value of k=2



Q 17 :

Let R2 denote R×R. Let S={(a,b,c):a,b,c∈R and ax2+2bxy+cy2>0 for all (x,y)∈R2-{(0,0)}. Then which of the following statements is (are) TRUE?        [2024]

  • (2,72,6)∈S

     

  • If (3,b,112)∈S, then |2b|<1.

     

  • For any given (a,b,c)∈S, the system of linear equations ax+by=1, by+cy=-1 has a unique solution.

     

  • For any given (a,b,c)∈S, the system of linear equations (a+1)x+by=0, bx+(c+1)y=0 has a unique solution.

     

Select one or more options

(2, 3, 4)

Given that ax2+2bxy+cy2>0

and x,y∈ℝ-{0}

⇒c(yx)2+2b(yx)+a>0⇒c>0, D<0

4b2-4ac<0⇒b2<ac

(1)  (2,72,6)
         (72)2>2×6
        ∴ Option (1) is incorrect

(2)  If (3,b,112)∈S
         ⇒b2<3·112⇒b2<14⇒4b2<1
          ⇒|2b|<1  Option (2) is correct

(3)  ax+by=1

         bx+cy=-1

          D=|abbc|=ac-b2≠0

        ∴ unique solution, option (3) is correct

(4)  (a+1)x+by=0

          bx+(c+1)y=0

          D=|a+1bbc+1|

          =(a+1)(c+1)-b2=ac-b2+a+c+1

           Since ac-b2>0⇒b2<ac⇒ac is positive

           ⇒a and c are positive, then (ac-b2)+a+c+1>0

            ∴ unique solution⇒option (4) is correct



Q 18 :

If 3x=4x-1, then x=                                 [2013]

  • 2log322log32-1

     

  • 22-log23

     

  • 11-log43

     

  • 2log232log23-1

     

Select one or more options

(1, 2, 3)

3x=4x-1⇒xlog3=2(x-1)log2

⇒x=2log22log2-log3⇒x=2log322log32-1=22-log23

Also x=11-12log23=11-log43



Q 19 :

Let p, q be integers and let α,β be the roots of the equation x2-x-1=0, where α≠β. For n=0,1,2,…, let an=pαn+qβn.

FACT: If a and b are rational numbers and a+b5=0, then a=0=b.                    [2017]

 

Q.  a12= 

  • a11-a10

     

  • a11+a10

     

  • 2a11+a10

     

  • a11+2a10

     

(2)

α2=α+1

β2=β+1

an=pαn+qβn

     =p(αn-1+αn-2)+q(βn-1+βn-2)

      =an-1+an-2

∴ a12=a11+a10



Q 20 :

Let p,q be integers and let α,β be the roots of the equation x2-x-1=0, where α≠β. For n=0,1,2,…, let an=pαn+qβn.

FACT: If a and b are rational numbers and a+b5=0, then a=0=b.                  [2017]

 

Q.    If a4=28, then p+2q=

  • 21

     

  • 14

     

  • 7

     

  • 12

     

(4)

α=1+52,  β=1-52

a4=a3+a2

      =2a2+a1

       =3a1+2a0

28=p(3α+2)+q(3β+2)

28=(p+q)(32+2)+(p-q)(352)

∴ p-q=0  and  (p+q)·72=28

⇒p+q=8⇒p=q=4

∴ p+2q=12



Q 21 :

Let a,b,c,p,q be real numbers. Suppose α,β are the roots of the equation x2+2px+q=0 and α,1β are the roots of the equation ax2+2bx+c=0, where β2∉{-1,0,1}.

STATEMENT-1: (p2-q)(b2-ac)≥0

and

STATEMENT-2: b≠pa or c≠qa                                    [2008]

  • STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

     

  • STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

     

  • STATEMENT - 1 is True, STATEMENT - 2 is False

     

  • STATEMENT - 1 is False, STATEMENT - 2 is True

     

(2)

As a,b,c,p,q∈R and the two given equations have exactly one common root

⇒Either both equations have real roots or both equations have imaginary roots

⇒Either D1≥0 and D2≥0 or D1≤0 and D2≤0

⇒p2-q≥0 and b2-ac≥0 or p2-q≤0 and b2-ac≤0

⇒(p2-q)(b2-ac)≥0

∴ Statement 1 is true.

Also we have αβ=q and αβ=ca

∴ αβα/β=qc×a⇒β2=qac

As β≠1 or -1⇒β2≠1⇒qac≠1 or c≠qa

Again, as exactly one root α is common, and β≠1

∴ α+β≠α+1β⇒-2ba≠-2p⇒b≠ap

∴ Statement 2 is correct.

But Statement 2 is not a correct explanation of Statement 1.