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5. (2)
Now,2 2x y
116 9
A = 4, b = 3
23 7C 1
44
So, foci = ae,0 7,0
Radius of new wide = 2 2
0 7 3 0
= 4.Centre = (0, 3)
Equation of wide = 2 2 2x 0 y 3 4
= 2 2x y 6y 9 16
=2 2
x y 6y 7 0
6. (3)2
Adj A A
2
1 3
1 3 3 4 1
2 4 4
7. (3)
p q ~ p ~ q p q ~ q ~ p p^ ~ q ^ ~ p ^ q p q ~ q ~ p
T T F F T T F T
T F F T F F F T
F T T F T T F T
F F T T T T F T
8. (3)
Mid-points are 0,1 , 1,1 and 1,0
Vertices are 0,0 , 2,0 , 0,2
radius of incentre : 2 2 x co-ordinate 2 2
9. (1)
3f x 2x 3x K
2f x 6x 3 0 for all x Hence, only one root is positive at max
10. (4) f x dx x
5 3 3x x dx Let t x
1
2
3
-1
-2
-3
1 2 3-4 -1-2-3 4
(0, 0)
(0, 2)
(2, 0)
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2dt 3x dx
So,
3 3
5 3 2x f x
x x dx 3x dx3
1
t f t dt3
After Inequation by parts,
1 t t dt ft t dt dt3
1
t t dt as f x dx a3
Putting t as 3x
3 3 3 21 x x x 3x dx3
3 3 2 31 x a x x dx c3
11. (3)
k 1 x 8y 4k
kx k 3 y 3k 1
For no solutionk 1 8
0k k 3
And atleast one of4k 8 k 1 4k
and 03k 1 k 3 k k 3
The only value of k which satisfies is k = 3.
No. of values = 1.
12. (3)
Tn is the number of triangle formed by joinly vertices of n sided polygon
So, nn 3T C n 1
n 1 3T C
n 1 n n
n 1 n 3 3 2T T C C C 10
x 5
13. (3)
Circle 2 21S x y 5 2
Equation of tangent 25
y mx tan2
.(1)
Parabola 22S y 4 5x
5y mx
m .(2)
(1)&(2) one some equation
25 5
1 m2 m
4 2m m 2
2 2m 1 m 2 0
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So, m 1 So, y x 5 y is a tangent
M 1 , satisfies the equation 4 2m 3m 2 0 Hence, S-I is True
S-II is True
14. (2)
Point of intersection (9, 3)
Area =3
Right left
0
X X .dy
3
2
0
2y 3 y .dy
= 9
15. (3)tan A cot A
1 cot A 1 tan A
2 2sin A cos A
cosA sin A cos A sin A cos A sin A
3 3sin A cos A
sin A cos A sin A cos A
2 2sin A cos A sin A cos A
sinAcosA
secA.cosecA 1
16. (1)3
6
cosxI dx
cos x sin x
3
6
2I dx3 6 6
I12
17. (2)
1 1 2x 1
1sec tan x tan tan x
1 x
1 12
1 1 2
x
2y x 3 0
7/30/2019 IIT JEE maths solutions
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18. (4)
z
z
otg 2
19. (1)Self Explanatory
20. (3)
o30
x 3y 3
3, 0
1
y 0 x 33
x-axis
Hence, option (3) is correct
21. (4)S 0.7 0.77 0.777 ........(20terms)
7S 0.9 0.99 0.999 ....(20terms)9
7 2 37S 20 10 10 10 ... 20 terms9
20
1 1
1
10 1 107
S 209 1 10
20
7 1 1S 20 1
9 9 10
207 179 1081
Hence option (a) is correct
22. (4)4 5
5 54 5
1 2 1C C
3 3 3
5 5
10 1 11
3 3
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23. (4)
1 10
3 x 1x 1x
101 1
3 2x x
10 r r 1 1
10 3 2rC x x
10 r r 10 3 2
rC x
r 4 10
410.9.8.7
C 2104!
24. (4)
n A 2 n B 4
n A B 8 8 8 8 8
3 4 5 8C C C ..... C 219
25. (4)
AB BC CA 0
BC AB CA
AB CABM
2
Now, AM BM AB
AB CAAM 33
2
26. (2)
dyx
dx
2
2
xx 0
xy
xx 0
2
Slope
x x 0
x x 0
x 0 x 0 x 2 x 2 y 2 y 2
1P 2, 2 2P 2, 2
T : y 2 2 x 2 T : y 2 2 x 2 x intercept 1 x intercept 1
C
M
BA
5i 2 j 4k
3i 4k
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30. (2)
A B
CD
p
2 2p q
N
p
q
p
tanq
p
tanAn
p
tanAN
tan tan R
1 tan tan AN
ptan
q tan p pq
p q p tan AN1 tanq
p p tan qq AN q
q tan p
2 2p tan pq q tan pq
q tan p
2 2p q sinp sin qsin