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Limits and Derivatives · Question 2

JEE Main 2026 (Online) 6th April Evening ShiftMCQMedium📈 Topic in 13% papers+4−1
1

Let x 2 ( (x-2) ) (rx2+(p-2)x-2p )(x-2)2=5 for some r, p R. If the set of all possible values of q, such that the roots of the equation rx2-px+q=0 lie in (0, 2), be the interval ( , ], then 4( + ) equals :

  1. 11
  2. 13
  3. 17
  4. 21

Answer

(3) 17

Solution

[figure]

x 2 (x-2)x-2=1, so the given limit is x 2 rx2+(p-2)x-2px-2=5

For a finite limit the numerator must vanish at x=2:

4r+2(p-2)-2p=0 4r-4=0 r=1

Then x2+(p-2)x-2p=(x-2)(x+p)

x 2(x+p)=2+p=5 p=3

Equation: x2-3x+q=0; let h(x)=x2-3x+q. Both roots in (0, 2) requires:

(i) D 0: 9-4q 0 q 94

(ii) h(0)>0: q>0

(iii) h(2)>0: 4-6+q>0 q>2

(iv) vertex x= 32 lies in (0, 2), true

q (2, 94 ] =2, = 94

4( + )=8+9=17

Answer: (3)

JEE Main 2026 (Online) 6th April Evening Shift

Keep practising — Limits and Derivatives

2

The product of all possible values of , for which x 0 ( 1- ( x) (( +1)x) (( +2)x) 2(( +1)x) )=2, is :

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  2. 1
  3. -1
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3

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  3. 3
  4. 1
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  2. 6
  3. 8
  4. 10
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5

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  1. (e10 - 1)2e2(e2 - 1)
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  1. 3
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  3. 9
  4. 7
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8

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  1. 9
  2. 18
  3. 1
  4. 2
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9

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  1. 6575
  2. 1175
  3. 25
  4. 625
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10

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  1. 53
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  3. 13
  4. 115
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