JEE (Main + Advanced) 2024

JEE (Main + Advanced) 2024

ENTHUSIAST COURSE
ENTHUSIAST COURSE
RACE # 01-SOLUTION SPECIAL RACE ON FOM, QE, LOG, S & S MATH EMATI CS
1. Ans. (A) 5. Ans. (B)
2 Multiplying 1st equation by 'x'
Tn 
3  n  1 n  2  x4 + ax2 + x = 0 .......(1)
x4 + ax2 + 2 = 0 .......(2)
2 1 1  ———————————
 
3  n  1 n  2 
 x – 2 = 0  x = 2 (common root)
  8 + 2a + 1 = 0
 S   Tn 9
n 1  a
2
1
 6. Ans. (B,D)

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3 4 3 2
2. Ans. (A) x + ax + bx – 32x + 16 = 0
  16
12 2 2 32 4 2
s    ..... ()1/ 4  2
70 71 72 73
1 1 1 1
s 12 2 2 32    2
    ...    
7 7 72 73
4
——————————— 2  AM = HM
1 / 
6s 3 5 9
 1   2  3  .....     2
7 7 7 7 4 2 2
x + ax + bx – 32x + 16
6s 1 3 5 9 = x4 – 8x3 + 24x2 – 32x + 16
     ..... 7. Ans. (C,D)
7 2 7 7 2 73 7 4
36s 2 2 2 log 2x 2  2x  3 (x 2  2x) = 1  x2 – 2x = 2x2 +
 2  1   2  3  .... 2x + 3
7 7 7 7
 x = –1, –3
36s 4 49
  s but for x = – 3
49 3 27  x2 + 6x + 8 < 0
3. Ans. (A)  only solution is x = –1
Two parabolas opening upwards do not 8. Ans. (A,C)
intersect, if they have same axis Put m = 1
b q So Sn+1 – Sn – S1 = n(1)
   or aq = bp  Tn+1 = n + S1 = n + 1
2a 2p
 Tr = r
 maximum value is given by (c – r)2  T100 = 100
4. Ans. (A) 100
100  101
x3 – 2x2 + 3x – 2 = (x – 1) (x2 – x + 2) and S100 =  r   5050
r 1 2
 = 1,  +  = 1,  = 2
Aliter :
     Sm+n – Sm – Sn = mn
    2
         1  1   S2 – S 1 – S1 = 1
 S2 = 1 + 2 = 3
5 13  T2 = 2
 2
4 4 and S3 – S2 – S1 = 2(1)
 S3 = 2 + 3 + 1 = 6

MATHEMATICS / R # 01-SOLUTION 1/3

 JEE (Main + Advanced) 2024

ENTHUSIAST COURSE
T 3 = S 3 – S2 = 3 14. Ans. (A)(P); (B)(R); (C)(P); (D)(Q)
and S4 – S2 – S2 = (2)(2) abc 3
 S4 = 10 (A) 
3 1 1 1
 T 4 = S4 – S3 = 4  
a b c
 S4 = 4 + 6 = 10
1 1 1
So the given sequence is 1, 2, 3, 4 .......  a  b  c      9
a b c
(100)(101)
T100 = 100 and S100   5050 ab
2 (B)  ab
2
9. Ans. (A,B,D)
bc
Let ƒ(x) = (x – log23)(x – log34)  bc
2
+ (x – log34)(x – log42)
ca
+ (x – log42)(x – log23)  ac
2
ƒ(log23) = (log23 – log34)(log23 – log42) > 0

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Multiply
ƒ(log34) = (log34 – log42)(log34 – log23) < 0  a  b  b  c  c  a   8
ƒ(log42) = (log42 – log23)(log42 – log34) > 0 abc
Paragraph for Question 10 to 12
a b c
10. Ans. (A)  
b c d  3

r (C) 3 b c d
     = 1 + ( + ) + ( + )2 + ........  
r0 a b c
1 1 15  a b c  b c d 
           9
1       1  25 14  b c d  a b c 
375 bc ca ab
11. Ans. (D)  2 2 1/ 3
(D) a 2
b c  bc ca ab 
 2 . 2 . 2 
   3 a b c 
A1 + A2 +......+A30 = 30A = 30   1
 2  bc ca ab
  3
12. Ans. (A) a 2 b2 c2
10 10 10 10 15. Ans. 4
 
y   t 2  2rt  r 2   t 2  2t  r   r 2
5b  2 5(b  1)  7
r 1 r 1 r 1 r 1
a 
10 11 t  b 1 b 1
 10t 2  2 r 2
 10t 2  110t   r 2
2 7
a  5  b–1=±1
y is minimum at b 1
b 110 11 b – 1 = ±7
t  
2a 20 2  4 values.
13. Ans. (A) 16. Ans. 5
2 For log3(ax2 + 4x + a) to be valid
an(n + 1) – n an+1 = n (n + 1)
a n a n 1 if ax2 + 4x + a > 0  x
 n  a > 0 and 4 – a2 < 0  a > 2
n n 1
a  n  n  1   ax 2  4x  a 
a1  n 1    Given inequality  log 3   1
 n 1  2  x 1 
2

 n  1  a n 1  ax 2  4x  a  3(x 2  1)
 50  n    (a – 3)x2 + 4x + (a – 3) > 0  x
 2  n 1  a – 3 > 0 and 16 – 4(a – 3)2 < 0
Now check  a > 3 and a – 3 > 2  a > 5
Finally a > 5
2/3 MATHEMATICS / R # 01-SOLUTION
 JEE (Main + Advanced) 2024

ENTHUSIAST COURSE
17. Ans. 256 18. Ans. 5
82 Let equation is a(x – )(x – ) = 0, where 
3x–1 + 3–x–1 =  are prime numbers.
27
Sum of coefficients = a( – 1)( – 1) which a
3x 3 x 82 prime number.
 
3 3 27 Possible only when a = 1;
82  – 1 = 1 &  – 1 Prime number.
3x + 3–x = 3x = A  =2
9
 – 1 both are prime numbers, then  = 3
1 82
A 
A 9
2
9A – 82A + 9 = 0
9A2 – 81A – A + 9 = 0
1
A = 9,

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9
1
3x = 9,
9
x = 2, –2
 x1 = 2
x2 = –2
log 71 101.log37 47.log101 625. log 47 71
x3 
log 27 5.log37 27
log 71 625.log 37 71 log37 625
  =4
log37 5 log37 5
x1  x 2
  x3  = (4)2+2 (4)4 = 256

MATHEMATICS / R # 01-SOLUTION 3/3