If P, Q and R are angles of △PQR, then the value of- 1

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

Advertisement

31.

If P, Q and R are angles of PQR, then the value of

- 1cosRcosQcosR- 1cosPcosQcosP- 1 is equal to

  • - 1

  • 0

  • 12

  • 1


B.

0

Let  =- 1cosRcosQcosR- 1cosPcosQcosP- 1Multiplying C1 by p and then doingC1  C1 + qC2 + rC3, we get = 1p- 1cosRcosQcosR- 1cosPcosQcosP- 1     = 1p- p + qcosR +rcosQcosRcosQpcosR - q + rcosP- 1cosPpcosQ + qcosP - rcosP- 1     = 1p0cosRcosQ0- 1cosP0cosP- 1     = 0 Using the projection formula,     p = qcosR + rcosQ 


Advertisement
32.

The number of real values of a for which the system of equations

x + 3y + 5z = αx

5x + y + 3z = αy

3x + 5y + z = αz

has infinite number of solutions is

  • 1

  • 2

  • 4

  • 6


33.

If z = 11 + 2i- 5i1 - 2i- 35 +3i5i5 - 3i7, then i = - 1

  • z is purely real

  • z is purely imaginary

  • z + z¯ = 0

  • z - z¯


34.

If one of the cube roots of 1 be w then

11 + w2w21 - i- 1w2 - 1- i- 1 + w- 1 is equal to

  • w

  • i

  • 1

  • 0


Advertisement
35.

a - bb - cc - ab - cc - aa - bc - aa - bb - c is equal to

  • 0

  • - 1

  • 1

  • 2


36.

w is an imaginary cube root of unity and

x +w2w1ww21 +x1x +  ww2 = 0, then one of the value of x is

  • 1

  • 0

  • - 1

  • 2


37.

If A = 12- 4- 1, then A- 1 is

  • 17- 1- 241

  • 1712- 4- 1

  • 17- 1- 24- 1

  • does not exist


38.

Let w be the complex number cos2π3 + isin2π3 Then, the number of distinct complex number z satisfying

z + 1ww2wz + w21w21z + w = o is equal to

  • 1

  • 0

  • 2

  • 3


Advertisement
39.

Let a, b and c be positive real numbers. The following system of equations in x, y and z,

x2a2 + y2b2 - z2c2 = 1, x2a2 - y2b2 + z2c2 = 1and - x2a2 + y2b2 + z2c2 = 1 has

  • finitely many solutions

  • no solution

  • unique solution

  • infinitely many solutions


40.

If r = rr31nn + 1, then r = 1nr is equal to

  • r = 1nr2

  • r = 1nr3

  • r = 1nr

  • r = 1nr4


Advertisement