Question

i) The argument of i^17 is ?

ii) The solution set of the inequality 2/x-3 < 0 is ?

iii) If SinΦ + CosΦ = 1 then the value of Sin2Φ is ?​

Answer 1

Step-by-step explanation:

i)

= > i^17 = > ( i^16 )i = > ( 1 )i = > i

= > 0 + 1i

Since the given point lies on y axis, thus argument of this number is π / 2.

ii)

= > 2 / ( x - 3 ) < 0

= > ( x - 3 )^2 × ( 2 ) / ( x - 3 ) < 0

= > 2x - 6 < 0

= > x < 3

Solution set is { ...- 1 , 0 , 1 , 2 } { x € Z }

iii)

= > sinA + cosA = 1

= > sin^2 A + cos^2 A + 2sinAcosA = 1 { sq. on both sides }

= > 1 + sin2A = 1 { sin^2 B + cos^2 B = 1 , 2sinAcosA = sin2A }

= > sin2A = 0

Answer 2

Step-by-step explanation:

(i) Argument of ${\iota}^{17}$

${ \iota}^{17} \\ \\ = { \iota}^{16 + 1} \\ \\ = { \iota}^{16} \times { \iota}^{1} \\ \\ = 1 \times \iota \: \: \: \: \: \: \: \: \: \: \: ( { \iota}^{4} = 1)\\ \\ = 0 + 1\iota$

Clearly, it lies on the Y axis.

Therefore, it's argument is $\dfrac{\pi}{2}$

(ii) Solution set of inequality 2/(x-3) < 0

$\frac{2}{x - 3} < 0 \\ \\ = > \frac{ {2(x - 3)}^{2} }{x - 3} < 0 \times {(x - 3)}^{2} \\ \\ = > 2x - 6 < 0 \\ \\ = > 2x - 6 + 6< 0 + 6 \\ \\ = > 2x < 6 \\ \\ = > \frac{2x}{2} < \frac{6}{2} \\ \\ = > x < 3$

Hence, solution set is {.....-5, -4....0,1,2}

where, x € Z

(iii) Value of $\sin (2\phi)$

$\sin( \phi) + \cos( \phi) = 1$

Squarring both the sides, we get

$= > { \sin }^{2} \phi + { \cos }^{2} \phi + 2 \sin( \phi) \cos( \phi) = 1 \\ \\ = > 1 + \sin(2 \phi) = 1 \\ \\ = > \sin(2 \phi) = 1 - 1 \\ \\ = > \sin(2 \phi) = 0$

Hence, $\sin(2\phi)= 0$

Concept Map :-

• ${ \sin}^{2} \alpha + { \cos }^{2} \alpha = 1$

• $2 \sin( \alpha ) \cos( \alpha ) = \sin(2 \alpha )$