Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2006 Exam Questions

Multiple Choice Questions

11.

Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of 2π/3  at its centre is

  • x2+y2 = 3/2

  • x2 + y2 = 1

  • x2+y2 = 27/4

  • x2+y2 = 9/4


D.

x2+y2 = 9/4

cos space straight pi over 3 space equals space fraction numerator square root of straight h squared plus straight k squared end root over denominator 3 end fraction
rightwards double arrow space straight h squared space plus straight k squared space equals 9 over 4
458 Views

12.

If the expansion in powers of x of the function fraction numerator 1 over denominator left parenthesis 1 minus ax right parenthesis left parenthesis 1 minus bx right parenthesis end fraction is a0 + a1x + a2x2 + a3x3 + … , then an is

  • fraction numerator straight b to the power of straight n minus straight a to the power of straight n over denominator straight b minus straight a end fraction
  • fraction numerator straight a to the power of straight n minus straight b to the power of straight n over denominator straight b minus straight a end fraction
  • fraction numerator straight a to the power of straight n plus 1 end exponent minus straight b to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction
  • fraction numerator straight b to the power of straight n plus 1 end exponent minus straight a to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction

D.

fraction numerator straight b to the power of straight n plus 1 end exponent minus straight a to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction

(1-ax)-1(1-bx)-1 = (1+ax+a2x2+.....)(1+bx+b2x2+....)
therefore coefficient of xn = bn +abn-1 +a2bn-2 +.....+an-1b +an =fraction numerator straight b to the power of straight n plus 1 end exponent minus straight a to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction

182 Views

13.

If (a, a2 ) falls inside the angle made by the lines y =x/2, x >0 and y = 3x, x > 0, then a belongs to

  • (0,1/2)

  • (3, ∞)

  • (1/2, 3)

  • (-3, -1/2)


C.

(1/2, 3)

411 Views

14.

Let a1, a2, a3, … be terms of an A.P. If fraction numerator straight a subscript 1 space plus straight a subscript 2 space plus......... straight a subscript straight p over denominator straight a subscript 1 space plus straight a subscript 2 space plus.......... straight a subscript straight q end fraction space equals space straight p squared over straight q squared comma space straight p not equal to straight q comma space then space straight a subscript 6 over straight a subscript 21 equals

  • 41/11

  • 7/2

  • 2/7

  • 11/47


D.

11/47

fraction numerator begin display style straight p over 2 open square brackets 2 straight a subscript 1 space plus left parenthesis straight p minus 1 right parenthesis straight d close square brackets end style over denominator begin display style straight q over 2 open square brackets 2 straight a subscript 1 space plus left parenthesis straight q minus 1 right parenthesis straight d close square brackets end style end fraction space equals space straight p squared over straight q squared
rightwards double arrow space fraction numerator 2 straight a subscript 1 space plus left parenthesis straight p minus 1 right parenthesis straight d over denominator 2 straight a subscript 1 space plus left parenthesis straight q minus 1 right parenthesis straight d end fraction space equals straight p over straight q
fraction numerator straight a subscript 1 space plus space open parentheses begin display style fraction numerator straight p minus 1 over denominator 2 end fraction end style close parentheses over denominator straight a subscript 1 space plus open parentheses begin display style fraction numerator straight q minus 1 over denominator 2 end fraction end style close parentheses straight d end fraction space equals straight p over straight q
For space straight a subscript 6 over straight a subscript 21 comma space straight p equals 11 comma space 1 space equals space 41
rightwards arrow space straight a subscript 6 over straight a subscript 21 space equals 11 over 41
174 Views

15.

If the lines 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 are two diameters of a circle of area 49π square units, the equation of the circle is

  • x2 + y2 + 2x − 2y − 47 = 0

  • x2 + y2 + 2x − 2y − 62 = 0

  • x2 + y2 − 2x + 2y − 62 = 0

  • x2 + y2 − 2x + 2y − 47 = 0


D.

x2 + y2 − 2x + 2y − 47 = 0

Point of intersection of 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 is (1 , − 1), which is the centre of the circle and radius = 7.
∴ Equation is (x − 1)2 + (y + 1)2 = 49
⇒ x2 + y2 − 2x + 2y − 47 = 0.

184 Views

16.

The value of integral subscript 1 superscript straight a left square bracket straight x right square bracket straight f apostrophe left parenthesis straight x right parenthesis space dx comma space straight a space greater than 1,where [x] denotes the greatest integer not exceeding x is

  • af(a) − {f(1) + f(2) + … + f([a])}

  • [a] f(a) − {f(1) + f(2) + … + f([a])}

  • [a] f([a]) − {f(1) + f(2) + … + f(a)}

  • af([a]) − {f(1) + f(2) + … + f(a)}


B.

[a] f(a) − {f(1) + f(2) + … + f([a])}

Let a = k + h, where [a] = k and 0 ≤ h < 1

integral subscript 1 superscript straight a left square bracket straight x right square bracket straight f apostrophe left parenthesis straight x right parenthesis space dx space integral subscript 1 superscript 2 1 straight f apostrophe left parenthesis straight x right parenthesis space dx space plus integral subscript 2 superscript 3 2 straight f apostrophe space left parenthesis straight x right parenthesis dx space plus
space....... integral subscript straight k minus 1 end subscript superscript straight k left parenthesis straight k minus 1 right parenthesis space dx space plus integral subscript straight k superscript straight k plus straight h end superscript kf apostrophe left parenthesis straight x right parenthesis space dx

{f(2) − f(1)} + 2{f(3) − f(2)} + 3{f(4) − f(3)}+…….+ (k−1) – {f(k) − f(k − 1)} + k{f(k + h) − f(k)}

= − f(1) − f(2) − f(3)……. − f(k) + k f(k + h)
= [a] f(a) − {f(1) + f(2) + f(3) + …. + f([a])}

117 Views

17.

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is

  • 3x2/2

  • x3/8

  • x2/2

  • πx2


C.

x2/2


area space equals space 1 half space straight x squared space sin space straight theta
straight A subscript max space equals space 1 half space straight x squared space open parentheses at space sin space straight theta space equals 1 comma space straight theta space equals straight pi over 2 close parentheses
137 Views

18.

For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is

  • (20, 45)

  • (35, 20)

  • (45, 35)

  • (35, 45)


D.

(35, 45)

left parenthesis 1 minus straight y right parenthesis to the power of straight m space left parenthesis 1 plus straight y right parenthesis to the power of straight n space equals space left square bracket 1 minus to the power of straight m straight C subscript 1 straight y space plus to the power of straight m straight C subscript 2 straight y squared space minus........ right square bracket left square bracket 1 plus to the power of straight n straight C subscript 1 straight y space plus to the power of straight n straight C subscript 2 straight y squared plus... right square bracket
space equals space 1 space plus space left parenthesis straight n minus straight m right parenthesis space plus open curly brackets fraction numerator straight m left parenthesis straight m minus 1 right parenthesis over denominator 2 end fraction plus fraction numerator straight n left parenthesis straight n minus 1 right parenthesis 2 over denominator 2 end fraction minus mn close curly brackets straight y squared space plus....
therefore space straight a subscript 1 space equals straight n minus straight m space equals space 10 space and space straight a subscript 2 space equals space fraction numerator straight m squared plus straight n squared space minus straight m minus straight n minus 2 mn over denominator 2 end fraction space equals space 10
So comma space straight n minus straight m equals 10 space and space left parenthesis straight m minus straight n right parenthesis squared space minus left parenthesis straight m plus straight n right parenthesis space equals space 20
rightwards double arrow straight m plus straight n space equals space 80
therefore comma space straight m space equals space 35 comma space straight n space equals space 45
180 Views

19.

If z2 + z + 1 = 0, where z is a complex number, then the value ofopen parentheses straight z plus 1 over straight z close parentheses squared space plus open parentheses straight z squared space plus 1 over straight z squared close parentheses squared space plus open parentheses straight z cubed space plus 1 over straight z cubed close parentheses squared space plus..... open parentheses straight z to the power of 6 plus 1 over straight z to the power of 6 close parentheses squared space is

  • 18

  • 54

  • 6


E.

12

z2 + z + 1 = 0 ⇒ z = ω or ω2

straight z space plus 1 over straight z to the power of 4 space equals space minus space 1 comma space straight z to the power of 5 space plus 1 over straight z to the power of 5 space equals space minus space 1 space and space straight z to the power of 6 space plus 1 over straight z to the power of 6 space equals space 2

∴ The given sum = 1 + 1 + 4 + 1 + 1 + 4 = 12
109 Views

20.

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is

  • 5040

  • 6210

  • 385

  • 1124


C.

385

10C1 + 10C2 + 10C3 + 10C4
= 10 + 45 + 120 + 210 = 385
124 Views

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