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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2013 Exam Questions

Multiple Choice Questions

1.

The number of values of k, for which the system of equations
(k+1) x + 8y = 4k
kx + (k+3)y = 3k -1
has no solution, is 

  • infinite 

  • 1

  • 2

  • 3


B.

1

Condition for the system of equations has no solution,

straight a subscript 1 over straight a subscript 2 space equals straight b subscript 1 over straight b subscript 2 space not equal to straight c subscript 1 over straight c subscript 2
therefore space fraction numerator straight k plus 1 over denominator straight k end fraction space equals space fraction numerator 8 over denominator straight k plus 3 end fraction space not equal to fraction numerator 4 straight k over denominator 3 straight k minus 1 end fraction
Take space fraction numerator straight k plus 1 over denominator straight k end fraction space equals space fraction numerator 8 over denominator straight k plus 3 end fraction
rightwards double arrow space straight k squared space plus space 4 straight k space plus 3 space equals space 8 straight k
rightwards double arrow straight k squared minus 4 straight k space plus 3 space equals space 0
rightwards double arrow left parenthesis straight k minus 1 right parenthesis left parenthesis straight k minus 3 right parenthesis space space equals 0
straight k space equals space 1 comma 3
If space straight k space equals 1 comma space then space fraction numerator 8 over denominator 1 plus 3 end fraction space equals space fraction numerator 4.1 over denominator 2 end fraction comma space false
Therefore, k = 3
Hence, only one value of k exists.

269 Views

2.

The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point

  • (-5,2)

  • (2,-5)

  • (5,-2)

  • (-2,5)


C.

(5,-2)

(x − 3)2+ y2+ λy = 0
The circle passes through (1, − 2)
⇒ 4 + 4 − 2λ = 0 ⇒ λ = 4
(x − 3)2+ y2+ 4y = 0
⇒ Clearly (5, − 2) satisfies.

181 Views

3.

If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a, b, c ∈ R, have a common root, then a : b : c is

  • 1:2:3

  • 3:2:1

  • 1:3:2

  • 3:1:2


A.

1:2:3

Given equations are 
x2 +2x+3 =0  ... (i)
ax2 +bx +c =0 .. (ii)
 since, Eq(i) has imaginary roots, So eq (ii) will also have both roots same as eq (i)
thus,

straight a over 1 space equals space straight b over 2 space equals space straight c over 3
Hence, a:b:c  is 1:2:3

177 Views

4.

A ray of light along straight x space plus square root of 3 straight y end root space equals space square root of 3 get reflected upon reaching X -axis, the equation of the reflected ray is 

  • straight y equals space straight x plus square root of 3
  • square root of 3 straight y end root space equals space straight x minus square root of 3
  • straight y space equals space square root of 3 straight x end root minus square root of 3
  • square root of 3 straight y end root space equals space straight x minus 1

B.

square root of 3 straight y end root space equals space straight x minus square root of 3

Given equation of line

straight x plus square root of 3 straight y space equals square root of 3 space end root space space space space space.... space left parenthesis straight i right parenthesis
straight y equals space 1 minus fraction numerator straight x over denominator square root of 3 end fraction
Slope of incident ray is space minus fraction numerator 1 over denominator square root of 3 end fraction 
So, slope of reflected ray must be fraction numerator 1 over denominator square root of 3 end fraction and the point of incident left parenthesis square root of 3 comma 0 right parenthesis
So equation of reflected ray
straight y minus 0 space equals space fraction numerator 1 over denominator square root of 3 end fraction space left parenthesis straight x minus square root of 3 right parenthesis
rightwards double arrow space square root of 3 straight y end root space equals space straight x minus square root of 3

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5.

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is

  • 256

  • 220

  • 219

  • 211


C.

219

Given, n(A) =2, n(B) = B
The number of subsets of AXB having 3 or more elements,
=straight C presuperscript 8 subscript 3 space plus to the power of 8 straight C subscript 4 space plus space..... plus to the power of 8 straight C subscript 8
space equals space 2 to the power of 8 space minus to the power of 8 straight C subscript 0 space minus to the power of 8 straight C subscript 1 space minus space to the power of 8 straight C subscript 2
space equals space 256 minus 1 minus 8 minus 28 space equals space 219

286 Views

6.

If x, y, z are in A.P. and tan−1 x, tan−1 y and tan−1 z are also in A.P., then

  • x= y= z

  • 2x =3y = 6z

  • 6x = 3y= 2z

  • 6x = 4y = 3z


A.

x= y= z

If x, y, z are in A.P.
2y = x + z and
tan−1 x, tan−1 y, tan−1 z are in A.P.
2 tan−1 y = tan−1 x + tan−1
z ⇒ x = y = z.

207 Views

7.

The real number k for which the equation, 2x3 +3x +k = 0 has two distinct real roots in [0,1]

  • lies between 1 and 2

  • lies between 2 and 3

  • lies between -1 and 0

  • does not exist


D.

does not exist

Let f(x) = 2x3+3x+k
On differentiating w.r.t x, we get
f'(x) = 6x2 + 3> 0, ∀ x ε R
⇒ f(x) is strictly increasing function
⇒ f(x) = 0 has only one real root, so two roots are not possible.

183 Views

8. limit as straight x space rightwards arrow 0 of fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis left parenthesis 3 plus cos space straight x right parenthesis over denominator straight x space tan space 4 straight x end fraction space is space equal space to space
  • -1/4

  • 1/2

  • 1

  • 2


D.

2

Let space l space equals space space limit as straight x rightwards arrow 0 of space fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis over denominator straight x squared end fraction fraction numerator left parenthesis 3 plus cos space straight x right parenthesis over denominator 1 end fraction. fraction numerator straight x over denominator tan space 4 straight x end fraction
equals space limit as straight x rightwards arrow 0 of space fraction numerator 2 space sin squared straight x over denominator straight x squared end fraction. fraction numerator 3 plus cosx over denominator 1 end fraction. fraction numerator straight x over denominator tan space 4 straight x end fraction
equals space 2 space limit as straight x rightwards arrow 0 of open parentheses fraction numerator sin space straight x over denominator straight x end fraction close parentheses squared. space limit as straight x rightwards arrow 0 of left parenthesis 3 plus space cos space straight x right parenthesis. space limit as straight x rightwards arrow 0 of fraction numerator 4 straight x over denominator space 4 space tan space 4 straight x end fraction
2.4.1 fourth space equals space 2
188 Views

9.

Consider :
Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.

  • Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I 

  • Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I

  • Statement -I is True; Statement -II is False.

  • Statement -I is False; Statement -II is True


B.

Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I

p q ~p  ~q p^~q ~p^q (p^~q)^(~p^q)
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
F
T
F
F
F
F
T
F
F
F
F
F
Fallacy

p q ~p  ~q p ⇒ q ~ q ⇒ ~ p (p ⇒ q) ⇔ (~ q ⇒ ~ p)
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
T
F
T
T
T
F
T
T
T
T
T
T
Tautology
S2 is not an explanation of S1
289 Views

10.

The sum of first 20  terms of the sequence 0.7,0.77,0.777...... is

  • 7 over 81 space left parenthesis 179 minus 10 to the power of negative 20 end exponent right parenthesis
  • 7 over 9 left parenthesis 99 minus 10 to the power of negative 20 end exponent right parenthesis
  • 7 over 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis
  • 7 over 9 left parenthesis 99 plus 10 to the power of negative 20 end exponent right parenthesis

C.

7 over 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis

Let S = 0.7 + 0.77 +0.777 + .... upto 20 terms

7 over 10 plus 77 over 10 squared plus 777 over 10 cubed plus..... space upto space 20 space terms
space equals space open square brackets 1 over 10 space plus 11 over 10 squared plus 111 over 10 cubed plus........ plus space upto space 20 space terms close square brackets
equals space 7 over 9 open square brackets 9 over 10 plus 99 over 100 plus 999 over 1000 plus..... plus space upto space 20 space terms close square brackets
space equals 7 over 9 open square brackets open parentheses 1 minus 1 over 10 close parentheses plus open parentheses 1 minus 1 over 10 squared close parentheses plus open parentheses 1 minus 1 over 10 cubed close parentheses plus space..... upto space 20 space terms close square brackets

7 over 9 open square brackets left parenthesis 1 plus 1 plus..... upto space 20 space terms right parenthesis minus open parentheses 1 over 10 plus fraction numerator begin display style 1 end style over denominator begin display style 10 squared end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style 10 cubed end style end fraction plus.... upto space 20 space terms close parentheses close square brackets
space equals space 7 over 9 space open square brackets 20 minus fraction numerator begin display style 1 over 10 open curly brackets 1 minus open parentheses 1 over 10 close parentheses to the power of 20 close curly brackets end style over denominator 1 minus begin display style 1 over 10 end style end fraction close square brackets
space equals space 7 over 9 open square brackets 20 minus 1 over 9 open curly brackets 1 minus open parentheses 1 over 10 close parentheses to the power of 20 close curly brackets close square brackets space equals space 7 over 9 open square brackets 179 over 9 plus 1 over 9 open parentheses 1 over 10 close parentheses to the power of 20 close square brackets
equals space 7 over 181 space open square brackets 179 plus left parenthesis 10 right parenthesis to the power of negative 20 end exponent close square brackets

958 Views

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