Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

If the letters of word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number

  • 601

  • 600

  • 602

  • 602

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12.

The value  of straight C presuperscript 50 subscript 4 space plus sum from straight r space equals 1 to 6 of straight C presuperscript 56 minus straight r end presuperscript subscript 3 space is

  • straight C presuperscript 55 subscript 4
  • straight C presuperscript 55 subscript 3
  • straight C presuperscript 56 subscript 3
  • straight C presuperscript 56 subscript 3
181 Views

13.

If A = open square brackets table row 1 0 row 1 1 end table close square brackets and I = open square brackets table row 1 0 row 0 1 end table close square brackets , then which one of the following holds for all n ≥ 1, by the principle of mathematical induction

  • An = nA – (n – 1)I

  • An = 2n-1A – (n – 1)I

  • An = nA + (n – 1)I

  • An = nA + (n – 1)I

131 Views

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14.

If the coefficient of x7  open square brackets ax squared space plus open parentheses 1 over bx close parentheses close square brackets to the power of 11in equals the coefficient of x-7 inopen square brackets ax squared space minus open parentheses 1 over bx close parentheses close square brackets to the power of 11then a and b satisfy the relation

  • a – b = 1

  • a + b = 1

  • a/b =1

  • a/b =1


D.

a/b =1

straight T subscript straight r plus 1 end subscript space in space the space expansion space space open square brackets ax squared space plus space 1 over bx close square brackets to the power of 11 space equals space to the power of 11 straight C subscript straight r space left parenthesis ax squared right parenthesis to the power of 11 minus straight r end exponent space open parentheses 1 over bx close parentheses to the power of straight r
space equals space to the power of 11 straight C subscript straight r space left parenthesis straight a right parenthesis to the power of 11 minus straight r end exponent space left parenthesis straight b right parenthesis to the power of negative straight r end exponent space left parenthesis straight x right parenthesis to the power of 22 minus 2 straight r minus straight r end exponent
rightwards double arrow space 22 minus 3 straight r space equals space 7
straight r space equals 5
therefore comma space coefficient space of space straight x to the power of 7 space equals space to the power of 11 straight C subscript 5 space left parenthesis straight a right parenthesis to the power of 6 space left parenthesis straight b right parenthesis to the power of negative 5 end exponent...... space left parenthesis 1 right parenthesis
Again space straight T subscript straight r plus 1 end subscript space in space the space exapansion space open square brackets ax space space minus space 1 over bx squared close square brackets to the power of 11 space equals space to the power of 11 straight C subscript straight r space left parenthesis ax right parenthesis to the power of 11 minus straight r end exponent space open parentheses negative 1 over bx squared close parentheses to the power of straight r
space equals space to the power of 11 straight C subscript straight r straight a to the power of 11 minus straight r end exponent space left parenthesis negative 1 right parenthesis to the power of straight r space straight x space left parenthesis straight b right parenthesis to the power of negative straight r end exponent space left parenthesis straight x right parenthesis to the power of negative 2 straight r end exponent space left parenthesis straight x right parenthesis to the power of 11 minus straight r end exponent
Now comma space 11 space minus 3 straight r space equals space 7
rightwards double arrow space 3 straight r space equals 18
rightwards double arrow straight r space equals 6
therefore comma space coefficient space of space straight x to the power of negative 7 end exponent space equals space to the power of 11 straight C subscript 6 straight a to the power of 5 space straight x space 1 space straight x space left parenthesis straight b right parenthesis to the power of negative 6 end exponent
rightwards double arrow to the power of 11 straight C subscript 5 space left parenthesis straight a right parenthesis to the power of 6 left parenthesis straight b right parenthesis to the power of negative 5 end exponent space equals space to the power of 11 straight C subscript 6 straight a to the power of 5 space straight x space left parenthesis straight b right parenthesis to the power of negative 6 end exponent
rightwards double arrow ab space equals space 1
113 Views

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15.

If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2| then argz1 – argz2 is equal to

  • π/2

  • 0

  • 0

128 Views

16. If space straight omega space equals space fraction numerator straight z over denominator straight z minus begin display style 1 third end style straight i space end fraction space and space vertical line straight omega vertical line space equals 1 comma space then space straight z space lies space on
  • an ellipse

  • a circle

  • a straight line

  • a straight line

208 Views

17.

Let α and β be the distinct roots of ax2 + bx + c = 0, then limit as straight x rightwards arrow straight alpha of space fraction numerator 1 minus cos space left parenthesis ax squared plus bx plus space straight c right parenthesis over denominator left parenthesis straight x minus straight alpha right parenthesis squared end fraction space is equal to

  • straight a squared over 2 left parenthesis straight alpha minus straight beta right parenthesis squared
  • 0

  • negative straight a squared over 2 left parenthesis straight alpha minus straight beta right parenthesis squared
  • negative straight a squared over 2 left parenthesis straight alpha minus straight beta right parenthesis squared
100 Views

18.

If x is so small that x3 and higher powers of x may be neglected, then fraction numerator left parenthesis 1 plus straight x right parenthesis to the power of 3 divided by 2 end exponent space minus open parentheses 1 plus begin display style 1 half end style straight x close parentheses cubed over denominator left parenthesis 1 minus straight x right parenthesis to the power of 1 divided by 2 end exponent end fraction spacemay be approximated as

  • 1 minus 3 over 8 straight x squared
  • 3 x 6 plus 3 over 8 straight x squared
  • negative 3 over 8 straight x squared
  • negative 3 over 8 straight x squared
227 Views

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19.

If straight x space equals space sum from straight n equals space 0 to infinity of straight a to the power of straight n comma space straight y space equals space sum from straight n equals 0 to infinity of straight b to the power of straight n comma space sum from straight n equals 0 to infinity of straight c to the power of straight n where a, b, c are in A.P. and |a| < 1, |b|<1, |c|< 1, then x, y, z are in

  • G.P.

  • A.P.

  • Arithmetic − Geometric Progression

  • Arithmetic − Geometric Progression

117 Views

20.

In a triangle, ABC, let ∠C = π/2 . If r is the in radius and R is the circumradius of the triangle ABC, then 2 (r + R) equals

  • b + c

  • a+b

  • a + b + c

  • a + b + c

141 Views

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