Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2012 Exam Questions

Multiple Choice Questions

21.

If g(x) = integral subscript 0 superscript straight x cos space 4 straight t space dt comma then g(x +π) equals

  • g(x)/g(π)

  • g(x) +g(π)

  • g (x) - g(π)

  • g(x). g(π)


B.

g(x) +g(π)

C.

g (x) - g(π)

Integral straight g space left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x space cos space 4 straight t. dt
To find g(x+π) in terms of g(x) of g(π)
straight g left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x space cos space 4 straight t space dt
rightwards double arrow space straight g left parenthesis straight x plus space straight pi right parenthesis space equals space integral subscript straight t space equals 0 end subscript superscript straight t space equals space straight x plus straight pi end superscript space cos space 4 straight t space dt
equals space integral subscript 0 superscript straight x space cos space 4 straight t space dt space plus space integral subscript straight x superscript straight x plus straight pi end superscript space cos space 4 straight t space dt
equals space straight g left parenthesis straight x right parenthesis space plus straight I subscript 1
straight l subscript 1 space equals space integral subscript straight x superscript straight x plus straight pi space end superscript space cos space 4 straight t space dt
equals integral subscript 0 superscript straight pi space cos space 4 straight t space dt
space equals space straight g left parenthesis straight pi right parenthesis
straight g left parenthesis straight x plus straight pi right parenthesis space equals space straight g left parenthesis straight x right parenthesis space plus straight g left parenthesis straight pi right parenthesis
But space the space value space of space straight I subscript 1 space is space zero
rightwards double arrow space straight g left parenthesis straight x plus straight pi right parenthesis space equals space straight g left parenthesis straight x right parenthesis minus straight g left parenthesis straight pi right parenthesis

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22.

Let ABCD be a parallelogram such that AB with rightwards arrow on top space equals space straight q with rightwards arrow on top comma space AD with rightwards arrow on top space equals space straight p with rightwards arrow on top and ∠BAD be an acute angle. If straight r with rightwards arrow on top  is the vector that coincides with the altitude directed from the vertex B the side AD, then straight r with rightwards arrow on top is given byLet ABCD be a parallelogram such that AB = q,AD = p and ∠BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by (1)

  • straight r with rightwards arrow on top space equals space 3 straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals negative space straight q with rightwards arrow on top space plus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals space straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals space 3 straight q with rightwards arrow on top space minus fraction numerator 3 open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top

D.

straight r with rightwards arrow on top space equals space 3 straight q with rightwards arrow on top space minus fraction numerator 3 open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top


AE with rightwards arrow on top space equals space vector space component space of space straight q with rightwards arrow on top space on space straight p with rightwards arrow on top

AE with rightwards arrow on top space equals fraction numerator left parenthesis straight p with rightwards arrow on top. straight q with rightwards arrow on top right parenthesis over denominator left parenthesis straight p with rightwards arrow on top. straight q with rightwards arrow on top right parenthesis end fraction straight p with rightwards arrow on top
therefore space From space increment ABE comma AB with rightwards arrow on top space plus BE with rightwards arrow on top space equals space AE with rightwards arrow on top
rightwards double arrow space straight q with rightwards arrow on top space plus straight r with rightwards arrow on top space equals space fraction numerator left parenthesis straight p with rightwards arrow on top. straight q with rightwards arrow on top right parenthesis over denominator left parenthesis straight p with rightwards arrow on top. straight q with rightwards arrow on top right parenthesis end fraction straight p with rightwards arrow on top
rightwards double arrow straight r with rightwards arrow on top space equals space minus straight q with rightwards arrow on top space fraction numerator left parenthesis straight p with rightwards arrow on top. straight q with rightwards arrow on top right parenthesis over denominator left parenthesis straight p with rightwards arrow on top. straight q with rightwards arrow on top right parenthesis end fraction straight p with rightwards arrow on top
645 Views

23.

Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.
Statement 1: f′(4) = 0
Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5).

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is false


C.

Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

f(x) = 7 – 2x; x < 2
= 3; 2 ≤ x ≤ 5
= 2x – 7; x > 5
f(x) is constant function in [2, 5]
f is continuous in [2, 5] and differentiable in (2, 5) and f(2) = f(5)
by Rolle’s theorem f′(4) = 0
∴ Statement 2 and statement 1 both are true and statement 2 is correct explanation for statement 1.

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24.

Let straight a with hat on top space and space straight b with hat on top be  two unit vectors. If the vectors straight c equals space straight a with hat on top space plus 2 straight b with hat on top and straight d space equals space 5 straight a with hat on top space minus 4 straight b with hat on top are perpendicular to each other, then the angle between straight a with hat on top space and space straight b with hat on top is 

  • π/6

  • π/2

  • π/3

  • π/4


C.

π/3

straight a with hat on top space and space straight b with hat on top space are space unit space vecotr comma space straight i. straight e comma space vertical line straight a with hat on top vertical line space equals space vertical line straight b with hat on top vertical line space equals 11
ii right parenthesis space straight C space equals straight a with hat on top space plus 2 straight b with hat on top space and space straight d space equals space 5 straight a with hat on top space minus 4 straight b with hat on top
(iii) c and d are perpendicular to each other, i.e., c.d =0
To find Angle between a and b
Now, c.d = 0
left parenthesis straight a with hat on top space plus 2 straight b with hat on top right parenthesis. left parenthesis 5 straight a with hat on top space minus 4 straight b with hat on top right parenthesis space equals space 0
rightwards double arrow space 5 straight a with hat on top. straight a with hat on top space minus space 4 space straight a with hat on top. straight b with hat on top space plus space 10 straight b with hat on top. straight a with hat on top space minus 8 space straight b with hat on top. straight b with hat on top space equals space 0
rightwards double arrow space 6 straight a with hat on top. straight b with hat on top space equals 3
rightwards double arrow space straight a with hat on top. straight b with hat on top space equals space 1 half
rightwards double arrow space vertical line stack straight a vertical line with hat on top vertical line straight b with hat on top vertical line space cos space straight theta space equals space cos space straight pi over 3
rightwards double arrow space Angle space between space straight a with hat on top space and space straight a with hat on top space is straight pi over 3
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25.

If: R →R is a function defined by straight f space left parenthesis straight x right parenthesis space equals space left square bracket straight x right square bracket space cos space open parentheses fraction numerator 2 straight x minus 1 over denominator 2 end fraction close parentheses straight pi where [x] denotes the greatest integer function, then f is

  • continuous for every real x

  • discontinous only at x = 0

  • discontinuous only at non-zero integral values of x

  • continuous only at x =0


A.

continuous for every real x

straight f left parenthesis straight x right parenthesis space equals space left square bracket straight x right square bracket cos space open parentheses fraction numerator 2 straight x minus 1 over denominator 2 end fraction close parentheses straight pi
space equals space left square bracket straight x right square bracket space cos space open parentheses πx minus straight pi over 2 close parentheses space equals space left square bracket straight x right square bracket space cos space open parentheses straight pi over 2 minus πx close parentheses
straight f left parenthesis straight x right parenthesis space equals space left square bracket straight x right square bracket space sin space πx
therefore, [x] sin π x is continuous for every real x.
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26.

If the integral integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x space equals space straight x
space plus space straight a space log space vertical line space sin space straight x minus space 2 space cos space straight x vertical line space plus space straight k comma  then an equal to 

  • -1

  • -2

  • 1

  • 2


D.

2

integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x space equals space straight x space plus space straight a space log space vertical line space sin space straight x minus space 2 space cos space straight x vertical line space plus space straight k comma space... space left parenthesis straight i right parenthesis
Now, let us assume that I,

straight I space equals integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x
Multiplying by cos x in numerator and denominator, we get
straight I space equals space integral fraction numerator 5 space sin space straight x over denominator sin space straight x minus 2 space cos space straight x end fraction straight d space straight x
This special integration requires special substitution of type
N' = A (D') +straight B space open parentheses fraction numerator dD apostrophe over denominator dx end fraction close parentheses
⇒ Let 5 sin x = (A + 2B) sin x + (B- 2A) cos x
Comapring the coefficients of sin x and cosx, 
we get
A +2B =5 and B- 2A = 0
Solving the above two equations in A and B
we get
A = 1 an B= 2
⇒ 5 sin x = ( sin x - 2 cos x)+ 2 (cos x + 2 sin x)
rightwards double arrow space straight I space equals space integral fraction numerator 5 space sin space straight x over denominator sin space straight x minus space 2 space cos space straight x end fraction dx
space equals space integral fraction numerator sin space straight x minus space 2 space cos space straight x right parenthesis space plus space 2 space left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis space dx over denominator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis end fraction
rightwards double arrow space straight I space equals space integral fraction numerator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis plus space 2 space left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis over denominator sin space straight x minus space 2 cos space straight x end fraction space dx
rightwards double arrow space straight I space equals space integral fraction numerator sin space straight x minus space 2 space cos space straight x over denominator sin space straight x minus space 2 space cos space straight x end fraction dx space plus space 2 space integral fraction numerator left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis over denominator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis end fraction dx
rightwards double arrow space straight I space equals integral space 1 space dx space plus space 2 space integral fraction numerator straight d left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis over denominator left parenthesis sin space straight x minus space 2 space cos space straight x right parenthesis space end fraction
rightwards double arrow space straight I space space equals space straight x space plus 2 space log space vertical line sin space straight x minus space 2 space cos space straight x right parenthesis vertical line space plus straight k space... space left parenthesis ii right parenthesis

where k is the constant of integration. Now, by comparing the value of l in eq. (i) and (ii) we get a = 2
121 Views

27.

Let P and Q be 3 × 3 matrices with P ≠ Q. If P3= Qand P2Q = Q2P, then determinant of(P2+ Q2) is equal to

  • -2

  • 1

  • 0

  • -1


C.

0

P3= Q3
P3– P2Q = Q3– Q2P
P2(P – Q) = Q2(Q – P)
P2(P – Q) + Q2(P – Q) = O
(P2+ Q2)(P – Q) = O
⇒ |P2+ Q2| = 0

309 Views

28.

The population p(t) at time t of a certain mouse species satisfies the differential equation fraction numerator dp space left parenthesis straight t right parenthesis over denominator dt end fraction space equals space 0.5 space left parenthesis straight t right parenthesis space minus 450. if p (0) = 850, then the  time at which the population becomes zero is

  • 2 log 18

  • log 9

  • 1 half space log space 18
  • log 18


A.

2 log 18

fraction numerator straight d left parenthesis straight p left parenthesis straight t right parenthesis right parenthesis over denominator dt end fraction space equals space 1 half straight p left parenthesis straight t right parenthesis space minus 450
fraction numerator begin display style straight d left parenthesis straight p left parenthesis straight t right parenthesis right parenthesis end style over denominator begin display style dt end style end fraction space equals space fraction numerator straight p space left parenthesis straight t right parenthesis minus 900 over denominator 2 end fraction
2 integral fraction numerator straight d left parenthesis straight p left parenthesis straight t right parenthesis right parenthesis over denominator straight p left parenthesis straight t right parenthesis minus 900 end fraction equals integral dt
2 space ln space vertical line straight p left parenthesis straight t right parenthesis minus 900 vertical line space equals space straight t plus straight c
straight t equals 0
rightwards double arrow space 2 space ln space 50 space equals space 0 plus straight c
therefore space 2 space ln space vertical line straight p left parenthesis straight t right parenthesis minus 900 vertical line space equals space straight t space plus space 2 space ln space 50
straight p left parenthesis straight t right parenthesis space equals space 0 space
rightwards double arrow 2 space ln space 900 space equals space straight t plus 2 space ln space 50
straight t space equals space 2 space left parenthesis In space 900 minus In space 50 right parenthesis space equals space 2 space ln space open parentheses 900 over 50 close parentheses space
equals space 2 space ln space 18
493 Views

29.

Let a, b ∈ R be such that the function f given by f(x) = ln |x| + bx
2+ ax, x ≠ 0 has extreme values at x = –1 and x = 2.
Statement 1: f has local maximum at x = –1 and at x = 2.
Statement 2: straight a space equals space 1 half space and space straight b space equals space fraction numerator negative 1 over denominator 4 end fraction

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is false


B.

Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

(i) A function f, such that f(x)= log |x| +bx2 +ax, x≠0
(ii) The function 'f' has extrema at x = -1 and x =2 i.e, f'(1) = f'(2) = 0 and f''(-1) ≠ 0≠f''(2)
Now, given function f is given by 
f(x) = log |x| +bx2 +ax
rightwards double arrow space straight f apostrophe left parenthesis straight x right parenthesis space equals space 1 over straight x space plus 2 bx space plus straight a
rightwards double arrow space straight f apostrophe apostrophe space left parenthesis straight x right parenthesis space equals space fraction numerator negative 1 over denominator straight x squared end fraction space plus 2 straight b
Since 'f' has extrema at x = - 1 and x =2
Hence, f'(-1) = 0 =f'(2)
f'(-1) = 0 
⇒ a-2b =1 ..... (i)
and f'(2) = 0 
⇒ a+ 4b = -1/2
solving eq. (i) and (ii), we get
a =1/2 and b = -1/4
straight f apostrophe apostrophe space left parenthesis straight x right parenthesis space equals space fraction numerator negative 1 over denominator straight x squared end fraction space plus fraction numerator negative 1 over denominator 2 end fraction space equals space minus space open parentheses fraction numerator straight x squared space plus 2 over denominator 2 straight x squared end fraction close parentheses
⇒ f'' has local maxima at both x = - 1 and x =2
Thus, a statement I is correct. Also, while solving for the statement I, we found values of a and b, which justify that statement 2 is also correct.

143 Views

30.

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is

  • 880

  • 629

  • 630

  • 879


D.

879

181 Views

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