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Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2017 Exam Questions

Multiple Choice Questions

1. stack lim space with straight x space rightwards arrow straight pi over 2 below space fraction numerator cot space straight x space minus cos space straight x over denominator left parenthesis straight pi minus 2 straight x right parenthesis cubed end fraction space equals
  • 1/4

  • 1/24

  • 1/16

  • 1/8


C.

1/16

limit as straight x rightwards arrow straight pi over 2 of space fraction numerator cot space straight x space left parenthesis 1 minus sin space straight x right parenthesis over denominator negative 8 space open parentheses straight x space minus begin display style straight pi over 2 end style close parentheses cubed end fraction
space equals space limit as straight x rightwards arrow fraction numerator begin display style straight pi end style over denominator begin display style 2 end style end fraction of space fraction numerator tan begin display style space end style begin display style open parentheses straight pi over 2 minus straight x close parentheses end style over denominator 8 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction space fraction numerator open parentheses 1 minus space cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses close parentheses over denominator open parentheses begin display style straight pi over 2 minus straight x end style close parentheses end fraction
space equals space 1 over 8.1.1 half space equals space 1 over 16
396 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

nx squared space plus space straight x left parenthesis 1 space plus space 3 plus 5 plus..... space plus space left parenthesis 2 straight n minus 1 right parenthesis right parenthesis space plus
space left parenthesis 1.2 space plus space 2.3 space plus space..... space plus space left parenthesis straight n minus 1 right parenthesis. straight n right parenthesis minus 10 straight n space equals space 0
rightwards double arrow space nx squared space plus space straight x space left parenthesis straight n squared right parenthesis space plus space fraction numerator straight n left parenthesis straight n squared minus 1 right parenthesis over denominator 3 end fraction space minus space 10 straight n space equals space 0
rightwards double arrow space straight x squared space plus space straight x space left parenthesis straight n right parenthesis space plus space fraction numerator left parenthesis straight n squared minus 1 right parenthesis over denominator 3 end fraction minus 10 space equals space 0 space less than subscript straight beta superscript straight alpha
left parenthesis straight alpha space minus space straight beta right parenthesis squared space equals space 1
rightwards double arrow space left parenthesis straight alpha space plus space straight beta right parenthesis squared space minus space 4 αβ space equals space 1
rightwards double arrow space straight n squared space minus space 4 space open parentheses fraction numerator straight n squared minus 1 over denominator 3 end fraction minus 10 close parentheses space equals space 1
rightwards double arrow space straight n space equals space 11
2171 Views

3.

The function straight f colon space straight R space rightwards arrow with space on top space open square brackets negative 1 half comma 1 half close square brackets space defined space as space straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight x over denominator 1 plus space straight x squared end fraction comma space is

  • neither injective nor surjective.

  • invertible

  • injective but not surjective.

  • surjective but not injective


D.

surjective but not injective



straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight x over denominator 1 space plus space straight x squared end fraction semicolon space straight f colon space straight R space rightwards arrow open square brackets negative 1 half comma 1 half close square brackets
From above diagram of f(x), f(x) is surjective but not injective.
371 Views

4.

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 sum from straight n space equals 1 to 10 of space straight f space left parenthesis straight n right parenthesis 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 termsstraight t subscript straight n space equals space fraction numerator left parenthesis straight n squared space plus 5 straight n right parenthesis over denominator 2 end fraction
straight S subscript straight n space equals space straight capital sigma space straight t subscript straight n space
equals space space stack sum space fraction numerator left parenthesis straight n squared space plus space 5 straight n right parenthesis over denominator 2 end fraction with space below and space on top
straight S subscript straight n space equals space 1 half space open square brackets fraction numerator straight n space left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction plus fraction numerator 5 straight n space left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets
straight S subscript 10 space equals space 330

330 Views

5.

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

straight theta space equals space fraction numerator 20 space minus 2 straight r over denominator straight r end fraction
Area space space equals space 1 half straight r squared straight theta space equals 1 half straight r squared open parentheses fraction numerator 20 minus 2 straight r over denominator straight r end fraction close parentheses
straight A space equals space 10 straight r minus straight r squared
dA over dr space equals space 0
10 minus 2 straight r space equals space 0
straight r space equals 5
straight A space equals space 50 minus 25 space equals space 25

332 Views

6.

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

7.

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 space open square brackets fraction numerator 1 minus straight t over denominator straight t end fraction minus straight t close square brackets space equals space 2 left parenthesis 2 straight t minus 1 right parenthesis space plus space 9
left curly bracket space Let space cos squared space straight x space equals space straight t right curly bracket

⇒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
cos space 4 straight x space equals space 2 space open parentheses negative 1 third close parentheses squared minus 1 space equals space minus 7 over 9

542 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

  • left parenthesis negative square root of 2 comma space minus square root of 3 right parenthesis
  • left parenthesis 3 square root of 2 space comma space 2 square root of 3 right parenthesis
  • left parenthesis 2 square root of 2 space comma 3 space square root of 3 right parenthesis
  • left parenthesis square root of 3 comma square root of 2 right parenthesis

C.

left parenthesis 2 square root of 2 space comma 3 space square root of 3 right parenthesis

Equation of hyperbola is straight x squared over straight a squared minus straight y squared over straight b squared space equals space 1
foci is (±2, 0) hence ae = 2, ⇒ a2e2 = 4

b2 = a2(e2 – 1)
∴ a2 + b2 = 4
Hyperbola passes through √2,√3
therefore space 2 over straight a squared space minus 3 over straight b squared space equals space 1 space... space left parenthesis 2 right parenthesis
On space sol space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis
straight a squared space equals 8 space left parenthesis is space rejected right parenthesis space and space straight a squared space equals space 1 space and space straight b squared space equals 3
therefore space straight x squared over 1 minus straight y squared over 3 space equals 1
Equation space of space tangent space is space fraction numerator square root of 2 straight x over denominator 1 end fraction minus fraction numerator square root of 3 straight y over denominator 3 end fraction space equals 1
Hence space left parenthesis 2 square root of 2 comma 3 space square root of 3 right parenthesis space satisfy space it.

1078 Views

9.

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

10.

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

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