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# JEE Mathematics Solved Question Paper 2017

#### Multiple Choice Questions

1.

The function

• neither injective nor surjective.

• invertible

• injective but not surjective.

• surjective but not injective

D.

surjective but not injective

From above diagram of f(x), f(x) is surjective but not injective.
371 Views

2.

If, for a positive integer n, the quadratic equation,
x(x + 1) + (x + 1) (x + 2) + .....
+ (x + n -1 ) (x + n) = 10n
has two consecutive integral solutions, then n is equal to :

• 11

• 12

• 9

• 10

A.

11

2171 Views

3.

Let a, b, c ∈ R. If f(x) = ax2 + bx + c is such
that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, ∀x,y ∈ R,then  is equal to

• 225

• 330

• 165

• 190

B.

330

f(x) = ax2 + bx + c
f(1) = a + b + c = 3
Now f(x + y) = f(x) + f(y) + xy
put y = 1
f(x + 1) = f(x) + f(1) + x
f(x + 1) = f(x) + x + 3
Now
f(2) = 7
f(3) = 12
Now
Sn = 3 + 7 + 12 + ..... tn ...(1)
Sn = 3 + 7 + ......tn – 1 + tn ...(2)
On subtracting (2) from (1)
tn = 3 + 4 + 5 + ....... upto n terms

330 Views

4.

For any three positive real numbers a, b and c, (25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then

• b , c and a are in G.P

• b, c and a are in A.P

• a, b and c are in A.P

• a, b and c are in G.P

B.

b, c and a are in A.P

225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0
(15a)2 + (3b)2 + (5c)2 – (15a)(3b) – (3b)(5c) – (15a) (5c) = 0
1/2[(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)2] = 0
15a = 3b , 3b = 5c , 5c = 15a
5a = b , 3b = 5c , c = 3a
a/1 = b/5 = c/3
a = λ, b = 5λ, c = 3λ
a, c, b are in AP

814 Views

5.

The value of
(21C110C1) + (21C210C2) + (21C310C3) + (21C410C4) + .... +
(21C1010C10) is

• 220 – 210

• 221 – 211

• 221 – 210

• 220 – 29

A.

220 – 210

(21C1 + 21C2 .........+ 21C10)– (10C1 + 10C2 ......... 10C10)
=1/2[(21C1 + ....+ 21C10) + (21C11 +.....21C20)]
– (210 – 1)
=1/2[221 – 2] – (210 – 1)
= (220 – 1) – (210 – 1)
= 220 – 210

1178 Views

6.

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.Then the maximum area (in sq. m) of the flower-bed, is

• 30

• 12.5

• 10

• 25

D.

25

Total length = r + r + rθ = 20

332 Views

7.
• 1/4

• 1/24

• 1/16

• 1/8

C.

1/16

396 Views

8.

hyperbola passes through the point P(√2, √3) and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point

C.

Equation of hyperbola is
foci is (±2, 0) hence ae = 2, ⇒ a2e2 = 4

b2 = a2(e2 – 1)
∴ a2 + b2 = 4
Hyperbola passes through √2,√3

1078 Views

9.

The following statement
(p → q ) → [(~p → q) → q] is

• a fallacy

• a tautology

• equivalent to ~ p → q

• equivalent to p → ~q

B.

a tautology

(p → q) → [(~p → q) →q]
(p → q) → ((p → q) → q)
(p → q) → ((~p → ~q) → q)
(p → q) → ((~p→ q) → (~q→ q))
(p→ q) v (p → q) which is tautology

397 Views

10.

If 5(tan2x – cos2x) = 2cos 2x + 9, then the value of cos4x is

• -7/9

• -3/5

• 1/3

• 2/9

A.

-7/9

⇒5(1 – t – t2) = t(4t + 7)
⇒ 9t2 + 12t – 5 = 0
⇒ 9t2 + 15t – 3t – 5 = 0
⇒ (3t – 1) (3t + 5) = 0
⇒ t = t/3 as t≠-5/3.
cos2x = 2(1/3)-1 = -1/3

542 Views