Privious year que paper of jee

JEE | NEET | Class 8 - 10 Download eSaral App



MATHEMATICS

SECTION - A –4=
Multiple Choice Questions: This section contains 20  2(2 – 1) + 5(5 + 2) +  + 4 – 15 = 0
multiple choice questions. Each question has 4 choices  4 + 25 +  – 2 + 10 + 4 – 15 = 0
(1), (2), (3) and (4), out of which ONLY ONE is correct. 30 – 3 = 0
1
 =
Choose the correct answer : 10
1. In an arithmetic progression if sum of 20 terms is
  +  +  = (2 – 1) + (5 + 2) + ( + 4)
790 and sum of 10 terms is 145, then S15 – S5 is
(when Sn denotes sum of n terms) 8
= 8 + 5 = + 5 = 5.8
(1) 400 (2) 395 10
(3) 385 (4) 405 n
n3
Answer (2) 3. lim
n→
 (n2 + k 2 )(n2 + 3k 2 )
k =1
20 
Sol. S20 = 2a + 19d  = 790    
2 (1) – (2) +
2a + 19d = 79 ...(1) 2 3 8 2 3 8
10     
S10 = 2a + 9d  = 145 (3) – (4) –
2 2 3 3 4
2a + 9d = 29 ...(2) Answer (1)
from (1) and (2) a = –8, d = 5 n
n3
15 5
S15 − S5 = 2a + 14d  − 2a + 4d 
Sol. lim 
n→ k =1 4  k 2  3k 2 
2 2 n 1 + 2 1 + 2 
 n  n 
15  5  
= −16 + 70 −  −16 + 20
2 2 n
1 1
= 405 – 10
= lim  
n→ n k =1 k 2
3k 2 
1 + 2 

1+ 2 
n 
= 395
 n 
2. If the foot of perpendicular from (1, 2, 3) to the line
x +1 y − 2 z − 4 1
dx
= = is (, , ) then find  +  +  =
2
(1) 6
5 1
(2) 5.8
( 1
)
0 3 1+ x2  + x2 
3 

( x2 + 1) –  x2 + 31 
(3) 4.8 (4) 5
Answer (2) 1
1 3
=  dx
( )  3 
Sol. 3 2 2  2 1
0 1 + x x +

 
1  
1  1 1 
=  –  dx
2
20
 x2 +  1  1+ x2 
   
( – 1) × 2 +( – 2) × 5 + ( – 3) × 1 = 0   3 
2 + 5 +  – 15 = 0
( ) ( )
1 1 1 1
= 3 tan–1 3x  – tan–1 x
Also, P lie on line 2 0 2 0
  + 1 = 2 3    1    
= – = –
 – 2 = 5 2  3  2  4  2 3 8


JEE Mains Shift-1 Answer Key www.esaral.com 1

, JEE | NEET | Class 8 - 10 Download eSaral App

4. The value of maximum area possible of a ABC 6. An ellipse whose length of minor axis is equal to
half of length between foci, then eccentricity is
such that A(0, 0) and B(x, y) and C(–x, y) such that 7
(1) (2) 17
y = –2x2 + 54x is (in sq. unit) 2
(1) 5800 (2) 5832 2 3
(3) (4)
(3) 5942 (4) 6008 5 7
Answer (2) Answer (3)
Sol. Sol.  ae = 2b

4b2
 = e2
a2
Or 4(1 – e2) = e2
2
 4 = 5e2  e =
5
Area of 
3  1
0 0 1 7. If g   = g   and
=
1
x y 1
 2  2
2
−x y 1
f (x) = g x + g ( 2 − x ) and f   3  = f   1  then
1 ( )
2  2  2
1
 ( xy + xy ) = xy (1) f (x) = 0 has exactly one root in (0, 1)
2
(2) f (x) = 0 has no root in (0, 1)
Area () = xy = x ( −2x 2 + 54x )
(3) f (x) = 0 has at least two roots in (0, 2)
d ( ) ( d
= −6x 2 + 108x )  = 0 at x = 0 and 18 (4) f (x) = 0 has 3 roots in (0, 2)
dx dx
Answer (3)
 at x = 0, minima
Sol.
and at x = 18 maxima
3  1
g   − g  
= 18 ( −2 (18) + 54 18) = 5832 g ( x ) − g (2 − x )  3  2 2 = 0
2
Area () f ( x ) = , f   =
2 2 2
5. The range of r for which circles (x + 1)2 + (y + 2)2 =
r2 and x2 + y2 – 4x – 4y + 4 = 0 coincide at two  1 3
g   − g  
distinct points  1 2  2  = 0, f  (1) = 0
Also f    =
2 2
(1) 3 < r < 7 (2) 5 < r < 9
3  1
1  f   = f   = 0
(3) r 4 (4) 0 < r < 3  2  2
2
1   3
Answer (1)  roots in  , 1 and 1, 
2   2
Sol. If two circles intersect at two distinct points
1 3
 |r1 – r2| < C1C2 < r1 + r2  f (x) is zero at least twice in  , 
2 2
| r – 2 | 9 + 16  r + 2
2− | x | −1
8. The domain of y = cos−1 + (log (3 − x )) is
|r – 2| < 5 and r + 2 > 5 4
[–, ) –{}, then value of  +  +  = ?
–5 < r – 2 < 5 r>3 …(2)
(1) 9 (2) 12
–3 < r < 7 …(1)
(3) 11 (4) 10
From (1) and (2)
Answer (3)
3<r<7


JEE Mains Shift-1 Answer Key www.esaral.com 2