If P, Q and R are angles of an isosceles triangle and ∠P

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 Multiple Choice QuestionsMultiple Choice Questions

651.

cos2π7 + cos4π7 + cos6π7

  • is equal to zero

  • lies between 0 and 3

  • is a negative number

  • lies between 3 and 6


652.

The minimum value of 2sinx + 2cosx is

  • 21 - 1/2

  • 21 + 1/2

  • 22

  • 2


653.

If p = cosπ4- sinπ4sinπ4cosπ4 and X = 1212. Then, p3X is equal to

  • 01

  • - 1212

  • - 10

  • - 12- 12


654.

For 0  P, Q  π2, if sinP + cosQ = 2, then the value of tanP + Q2 is equal to

  • 1

  • 12

  • 12

  • 32


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655.

The value of

cos275° + cos245° + cos215° - cos230° - cos260° is

  • 0

  • 1

  • 12

  • 14


656.

The maximum and minimum values of cos6θ + sin6θ  are respectively

  • 1 and 14

  • 1 and 0

  • 2 and 0

  • 1 and 12


657.

Let fθ = 1 + sin2θ2 - sin2θ. Then, for all values of θ

  • fθ > 94

  • f(θ) < 2

  • fθ > 114

  • 2  f(θ)  94


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658.

If P, Q and R are angles of an isosceles triangle and P = π2,  then the value of

cosP3 - isinP33 + cosQ + isinQcosR - isinR        + cosP - isinPcosQ - isinQcosR - isinR

  • i

  • - i

  • 1

  • - 1


B.

- i

Given that, P, Q and R are angles of an isosceles triangle and P = π2

Q = R = π4                  P + Q + R = 180°

cosP3 - isinP33 + cosQ + isinQcosR - isinR        + cosP - isinPcosQ - isinQcosR - isinR= e- iP33 + eiQ . e- iR + e- iPe- iQ . e- iR               cosθ + isinθ = e= e- iP + eiQ - R + e- iP +Q +R= e- /2 + ei0 + e-               Q = R = π4, P = π2= - i + 1 + - 1= - i


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659.

If fx = sinx + 2cos2x, π4  x  3π4. Then, f attains its

  • minimum at x = π4

  • maximum at x = π2

  • minimum x = π2

  • mamum at x = sin-114


660.

If sin2θ + 3cosθ = 2 then cos3θ + sec3θ is equal to

  • 1

  • 4

  • 9

  • 18


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