Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

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51.

ABC is a triangle. Forces straight P with rightwards arrow on top comma space straight Q with rightwards arrow on top comma space straight R with rightwards arrow on top space acting along IA, IB and IC respectively are in equilibrium, where I is incentre of ∆ABC. Then P : Q : R is

  • sin A : sin B : sin C

  • sin space straight A over 2 space colon space Sin space straight B over 2 space colon space sin space straight C over 2
  • cos space straight A over 2 colon space cos space straight B over 2 space colon space Cos space straight C over 2
  • cos space straight A over 2 colon space cos space straight B over 2 space colon space Cos space straight C over 2


C.

cos space straight A over 2 colon space cos space straight B over 2 space colon space Cos space straight C over 2

Using Lami’s Theorem

therefore space straight P space colon space straight Q space colon straight R space equals space cos space straight A over 2 space colon space cos space straight B over 2 colon space cos straight C over 2

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52.

If C is the mid -point of AB and P is any point outside AB, then

  • PA with rightwards arrow on top space plus space PB with rightwards arrow on top space equals space 2 PC with rightwards arrow on top
  • PA with rightwards arrow on top space plus space PB with rightwards arrow on top space equals space PC with rightwards arrow on top
  • PA with rightwards arrow on top space plus space PB with rightwards arrow on top space plus space 2 PC with rightwards arrow on top space equals space 0
  • PA with rightwards arrow on top space plus space PB with rightwards arrow on top space plus space 2 PC with rightwards arrow on top space equals space 0
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53.

The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that

  • it passes through the origin

  • it makes angle π/2 + θ with the x-axis

  • it passes through (aπ/2 ,-a)

  • it passes through (aπ/2 ,-a)

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54.

The area enclosed between the curve y = loge (x + e) and the coordinate axes is

  • 1

  • 2

  • 3

  • 3

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55.

The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1, S2, S3 are respectively the areas of these parts numbered from top to bottom; then S1 : S2: S3 is

  • 1 : 2 : 1

  • 1 : 2 : 3

  • 2 : 1 : 2

  • 2 : 1 : 2

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56.

The line parallel to the x−axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx − 2ay − 3a = 0, where (a, b) ≠ (0, 0) is

  • below the x−axis at a distance of 3/2 from it

  • below the x−axis at a distance of 2 /3 from it

  • above the x−axis at a distance of 3/ 2 from it

  • above the x−axis at a distance of 3/ 2 from it

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57. integral space open curly brackets fraction numerator left parenthesis log space straight x minus 1 right parenthesis over denominator 1 plus left parenthesis log space straight x right parenthesis squared end fraction close curly brackets squared space dx space is space equal space to
  • fraction numerator log space straight x over denominator left parenthesis log space straight x right parenthesis squared plus 1 end fraction plus space straight C
  • fraction numerator straight x over denominator straight x plus 1 end fraction space plus straight C
  • fraction numerator xe to the power of straight x over denominator 1 plus straight x squared end fraction plus straight C
  • fraction numerator xe to the power of straight x over denominator 1 plus straight x squared end fraction plus straight C
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58.

Let f : R → R be a differentiable function having f (2) = 6, f′ (2) =(1/48) . Then limit as straight x space rightwards arrow 2 of space integral subscript 6 superscript straight f left parenthesis straight x right parenthesis end superscript space fraction numerator 4 straight t cubed over denominator straight x minus 2 end fraction dt space equals

  • 24

  • 36

  • 12

  • 12

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59.

Let f (x) be a non−negative continuous function such that the area bounded by the curve y = f (x), x−axis and the ordinates x = π/4  and x = β >  π/4 open parentheses straight beta space sin space straight beta space plus space straight pi over 4 space cos space straight beta space plus space square root of 2 straight beta close parentheses. Then f (π/2) is

  • open parentheses straight pi over 4 plus square root of 2 minus 1 close parentheses
  • open parentheses straight pi over 4 minus square root of 2 plus 1 close parentheses
  • open parentheses 1 minus straight pi over 4 minus square root of 2 plus 1 close parentheses
  • open parentheses 1 minus straight pi over 4 minus square root of 2 plus 1 close parentheses
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60.

If the angle θ between the line fraction numerator straight x plus 1 over denominator 1 end fraction space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 2 end fractionand the plane space 2 straight x minus space straight y plus space square root of straight lambda space straight z space plus 4 space equals 0 is such of sin θ = 1/3 the value of λ is

  • 5/3

  • -3/5

  • 3/4

  • 3/4

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