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 Multiple Choice QuestionsMultiple Choice Questions

11.

If the roots of the quadratic equation x2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q − p is

  • 2

  • 3

  • 0

  • 1


B.

3

x2 + px + q = 0
tan 30° + tan 15° = − p
tan 30° ⋅ tan 15° = q

tan space 45 to the power of straight o space equals space fraction numerator tan space 30 to the power of straight o space plus space tan space 15 to the power of straight o over denominator 1 minus tan space 30 to the power of straight o space tan space 15 to the power of straight o end fraction space
equals space fraction numerator negative straight p over denominator 1 minus straight q end fraction space equals 1

⇒ − p = 1 − q
⇒ q − p = 1
∴ 2 + q − p = 3

145 Views

12.

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

13.

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

  • ab =1


D.

ab =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

14.

If both the roots of the quadratic equation x2 – 2kx + k2 + k – 5 = 0 are less than 5, then k lies in the interval

  • (5, 6]

  • (6, ∞)

  • (-∞, 4)

  • [4, 5]


C.

(-∞, 4)

fraction numerator negative straight b over denominator 2 straight a end fraction space less than space 5
straight f left parenthesis 5 right parenthesis space greater than space 0
rightwards double arrow space straight k space element of space left parenthesis negative infinity comma space 4 right parenthesis
418 Views

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

  • -π/2


C.

0

z1 + z2| = |z1| + |z2| ⇒ z1 and z2 are collinear and are to the same side of origin; hence arg z1 – arg z2 = 0

128 Views

16.

If the equation anxn +an-1xn-1 +....... +a1x =0, a1 ≠ 0, n≥2, has a positive root x =  α, then the equation nanxn-1 + (n-1)an-1xn-2 +......+a1 = 0 has a positive root, which is

  • greater than α

  • smaller than α

  • greater than or equal to α

  • equal to α


B.

smaller than α

f(0) = 0, f(α) = 0
⇒ f′(k) = 0 for some k∈(0, α).

152 Views

17.

A body weighing 13 kg is suspended by two strings 5 m and 12 m long, their other ends being fastened to the extremities of a rod 13 m long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are 

  • 12 kg and 13 kg

  • 5 kg and 5 kg

  • 5 kg and 12 kg

  • 5 kg and 13 kg


C.

5 kg and 12 kg



straight T subscript 2 space cos space open parentheses straight pi over 2 minus straight theta close parentheses space equals space straight T subscript 1 space cosθ
rightwards double arrow space straight T subscript 1 space cos space straight theta space equals space straight T subscript 2 space sin space straight theta
straight T subscript 1 space sin space straight theta space plus space straight T subscript 2 space cos space straight theta space equals space 13
because space OC space equals space CA space space equals CB
rightwards double arrow space angle space AOC space equals space angle OAC space and space angle COB space equals space angle OBC
therefore space sin space straight theta space equals space sin space straight A space equals space 5 over 13 space and space cos space straight theta space equals 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 space and space cos space straight theta space equals space 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 straight T subscript 2
straight T subscript 2 space open parentheses 5 over 12.5 over 13.12 over 13 close parentheses space equals space 13
straight T subscript 2 space open parentheses fraction numerator 169 over denominator 12.13 end fraction close parentheses space equals space 13
straight T subscript 2 space equals space 12 space kgs
rightwards double arrow space straight T subscript 1 space equals space 5 space kgs
106 Views

18.

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
  • 1 half space left parenthesis straight alpha minus straight beta right parenthesis squared


A.

straight a squared over 2 left parenthesis straight alpha minus straight beta right parenthesis squared
limit as straight x rightwards arrow straight alpha of space fraction numerator 1 minus cos space straight a space left parenthesis straight x minus straight alpha right parenthesis left parenthesis straight x minus straight beta right parenthesis over denominator left parenthesis straight x minus straight alpha right parenthesis squared end fraction space
space equals space limit as straight x rightwards arrow straight alpha of fraction numerator 2 space sin squared space open parentheses straight a begin display style fraction numerator left parenthesis straight x minus straight alpha right parenthesis left parenthesis straight x minus straight beta right parenthesis over denominator 2 end fraction end style close parentheses over denominator left parenthesis straight x minus straight alpha right parenthesis squared end fraction
space equals space limit as straight x rightwards arrow straight alpha of space fraction numerator 2 over denominator left parenthesis straight x minus straight alpha right parenthesis squared end fraction space straight x fraction numerator begin display style sin squared end style begin display style space end style begin display style open parentheses straight a fraction numerator left parenthesis straight x minus straight alpha right parenthesis left parenthesis straight x minus straight beta right parenthesis over denominator 2 end fraction close parentheses end style over denominator begin display style fraction numerator straight a squared space left parenthesis straight x minus straight alpha right parenthesis squared left parenthesis straight x minus straight beta right parenthesis squared over denominator 4 end fraction end style end fraction space straight x space fraction numerator straight a squared space left parenthesis straight x minus straight alpha right parenthesis squared left parenthesis straight x minus straight beta right parenthesis squared over denominator 4 end fraction
space equals space fraction numerator straight a squared left parenthesis straight alpha minus straight beta right parenthesis squared over denominator 2 end fraction
100 Views

19.

All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval

  • −2 < m < 0

  • m > 3

  • −1 < m < 3 

  • 1 < m < 4 


C.

−1 < m < 3 

Equation x2 − 2mx + m2 − 1 = 0
(x − m)2 − 1 = 0
(x − m + 1) (x − m − 1) = 0
x = m − 1, m + 1 − 2 < m − 1 and m + 1 < 4

m > − 1 and m < 3 − 1 < m < 3.

152 Views

20.

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
  • straight x over 2 minus 3 over 8 straight x squared

C.

negative 3 over 8 straight x squared
left parenthesis 1 minus straight x right parenthesis to the power of 1 divided by 2 end exponent space open square brackets 1 plus 3 over 2 straight x space plus 3 over 2 open parentheses 3 over 2 minus 1 close parentheses straight x squared minus 1 minus 3 open parentheses 1 half straight x close parentheses minus 3 left parenthesis 2 right parenthesis open parentheses 1 half straight x close parentheses squared close square brackets
equals space left parenthesis 1 minus straight x right parenthesis to the power of 1 divided by 2 end exponent space open square brackets negative 3 over 8 straight x squared close square brackets space equals space minus space 3 over 8 straight x squared
227 Views