Q 11 :    

Consider the equation x2+4xn=0, where n  [20, 100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to          [2025]

  • 5

     

  • 7

     

  • 6

     

  • 8

     

(3)

we have, x2+4xn=0

 x2+4x+4=n+4

 (x+2)2=n+4

 x=2±n+4

But 20n100

 24n+4104  24n+4104

 n+4{5,6,7,8,9,10}.

   6 integral values of 'n' are possible.



Q 12 :    

The product of all solutions of the equation e5(logex)2+3=x8, x>0, is:          [2025]

  • e2

     

  • e

     

  • e8/5

     

  • e6/5

     

(3)

We have, e5(logex)2+3=x8, x>0

Taking log on both sides, we get

5(ln x)2+3=8lnx  5(ln x)2+38lnx=0

Put ln x = t

 5t28t+3=0  t1+t2=85

  ln x1+ln x2=8/5  ln(x1x2)=8/5  x1x2=e8/5.



Q 13 :    

The product of all the rational roots of the equation (x29x+11)2(x4)(x5)=3, is equal to          [2025]

  • 28

     

  • 21

     

  • 7

     

  • 14

     

(4)

We have, (x29x+11)2(x4)(x5)=3

 (x29x+11)2(x9x+20)=3

Let x29x=t, then

(t+11)2(t+20)=3

 t2+121+22tt203=0

 t2+21t+98=0  (t+14)(t+7)=0

 t=7, 14

 x29x+7=0 and x29x+14=0

 x=9±532 and x=9±52

 Product of all rational roots = (9+52)(952)=7×2=14.



Q 14 :    

The number of real solution(s) of the equation x2+3x+2=min{|x3|,|x+2|} is :          [2025]

  • 2

     

  • 3

     

  • 1

     

  • 0

     

(1)

In(,12), |x + 2| is minimum and In(12,), |x – 3| is minimum.

From the graph, only two solutions.



Q 15 :    

The sum, of the squares of all the roots of the equation x2+|2x3|4=0, is          [2025]

  • 6(22)

     

  • 3(32)

     

  • 6(32)

     

  • 3(22)

     

(1)

Given, x2+|2x3|4=0

Case I: x2+(2x3)4=0 when x32

 x2+2x74=0  x=221          [  x32]

Case II : x2(2x3)4=0, x<32

 x22x1=0

 x=2±22=1±2  x=12          [  x<32]

   Required sum = (221)2+(12)2

                              =1262=6(22).



Q 16 :    

If α+iβ and γ+iδ are the roots x2(32i)x(2i2)=0, i=1, then αγ+βδ is equal to :          [2025]

  • –6

     

  • 6

     

  • –2

     

  • 2

     

(4)

We have, x2(32i)x(2i2)=0

x=(32i)±(32i)2+4(1)(2i2)2

   =(32i)±9412i+8i82=(32i)±34i2

   =(32i)±(1)2(2i)22(1)(2i)2

  =(32i)±(12i)22=(32i)±(12i)2

  x=32i+12i2 or (32i12i)2

  x=22i or 1+0i

So, αγ+βδ=2(1)+(2)(0)=2.

 



Q 17 :    

The number of solutions of the equation (9x9x+2)(7x7x+3)=0 is :          [2025]

  • 1

     

  • 3

     

  • 4

     

  • 2

     

(3)

We have, (9x9x+2)(7x7x+3)=0          ... (i)

Put in (i), we get

(9y29y+2)(2y27y+3)=0

 (3y2)(3y1)(2y1)(y3)=0

 y=23,13,12,3  x=94,9,4,19

   Number of solutions = 4.



Q 18 :    

If the set of all aR, for which the equation 2x2+(a5)x+15=3a has no real root, is the interval (α,β), and X={xZ : α<x<β}, then xXx2 is equal to :          [2025]

  • 2129

     

  • 2119

     

  • 2109

     

  • 2139

     

(4)

Since, the given equation has no real root

  b2ac<0  (a5)24(2)(153a)<0

 (a5)28(153a)<0

 a210a+25120+24a<0

 a214a95<0  a(5,19)

  xXx2=(12+22+32+42)+o2+(12+22+...+18)2

                       =4(5)(9)6+18(19)(37)6=128346=2139.



Q 19 :    

Let α, β be the roots of the equation x2axb=0 with Im(β)<Im(β). Let Pn=αnβn. If P3=57i, P4=37i, P5=117i and P6=457i, then |α4+β4| is equal to __________.          [2025]



31

We have, α+β=a and αβ=b

  P6=aP5+bP4

 457i=a×117i+b(37)i

 45=11a3b          ... (i)

and P5=aP4+bP3

 117i=a(37i)+b(57i)

 11=3a5b          ... (ii)

On solving equations (i) and (ii), we get a = 3, b = –4

  |α4+β4|=(α4β4)2+4α4β4

=63+4.44=63+1024=961=31.