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Exemplar Solution Class 11th

The value of $s i n \frac{2 π}{18} + s i n \frac{5 π}{18} + s i n \frac{7 π}{18} + s i n \frac{4 π}{9}$ is given by:

(a) 1 

(b) -1 

(c) $c o s \frac{3 π}{9} + s i n \frac{π}{18}$  

(d) $c o s \frac{2 π}{9} + s i n \frac{π}{18}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s s i n \frac{π}{18} + s i n \frac{π}{9} + s i n \frac{2 π}{9} + s i n \frac{5 π}{18} \\ = (s i n \frac{5 π}{18} + s i n \frac{π}{18}) + (s i n \frac{2 π}{9} + s i n \frac{π}{9}) \\ = 2 s i n (\frac{\frac{5 π}{18} + \frac{π}{18}}{2}) . c o s (\frac{\frac{5 π}{18} - \frac{π}{18}}{2}) + 2 s i n (\frac{\frac{2 π}{9} + \frac{π}{9}}{2}) . c o s (\frac{\frac{2 π}{9} - \frac{π}{9}}{2}) \\ = 2 s i n \frac{π}{6} . c o s \frac{π}{9} + 2 s i n \frac{π}{6} . c o s \frac{π}{18} \\ = 2 \times \frac{1}{2} c o s \frac{π}{9} + 2 \times \frac{1}{2} c o s \frac{π}{18} = c o s \frac{π}{9} + c o s \frac{π}{18} \\ = s i n (\frac{π}{2} - \frac{π}{9}) + s i n (\frac{π}{2} - \frac{π}{18}) = s i n \frac{7 π}{18} + s i n \frac{8 π}{18} \\ = s i n \frac{7 π}{18} + s i n \frac{4 π}{9} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Class 11th

If $\overset{?}{? z} + 1 ? = z + 2 (1 + i)$ , then find $z$ .

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z + 1 | = z + 2 (1 + i) \\ L e t z = x + i y \\ S o, | x + i y + 1 | = (x + i y) + 2 (1 + i) \\ \Rightarrow | (x + 1) + i y | = x + i y + 2 + 2 i \\ \Rightarrow | (x + 1) + i y | = (x + 2) + (y + 2) i \\ \Rightarrow \sqrt[]{{(x + 1)}^{2} + y^{2}} = (x + 2) + (y + 2) i [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} . i^{2} + 2 (x + 2) (y + 2) i \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 y - 4 + 2 (x + 2) (y + 2) i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y a n d 2 (x + 2) (y + 2) = 0 \\ \Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 \dots (i) \\ a n d (x + 2) (y + 2) = 0 \dots (i i) \\ x + 2 = 0 o r y + 2 = 0 \\ ∴ x = - 2 o r y = - 2 \\ N o w p u t x = - 2 i n e q n . (i) \\ \Rightarrow 2 y^{2} - 2 \times (- 2) + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 y + 5 = 0 \\ b^{2} - 4 a c = {(4)}^{2} - 4 \times 2 \times 5 = 16 - 40 = - 24 < 0 n o r e a l r o o t s . \\ N o w p u t y = - 2 i n e q n . (i) \\ \Rightarrow 2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \\ \Rightarrow 8 - 2 x - 8 + 1 = \end{array}$

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Exemplar Solution Class 11th

Number of solutions of the equation $t a n x + s e c x = 2 c o s x$ lying in the interval $[0, 2 π]$ is:

(a) 0

(b) 1

(c) 2

(d) 3

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n x + s e c x = 2 c o s x \\ \Rightarrow \frac{s i n x}{c o s x} + \frac{1}{c o s x} = 2 c o s x \\ \Rightarrow 1 + s i n x = 2 c o s^{2} x \Rightarrow 2 c o s^{2} x - s i n x - 1 = 0 \\ \Rightarrow 2 (1 - si n^{2} x) - s i n x - 1 = 0 \Rightarrow 2 - 2 si n^{2} x - s i n x - 1 = 0 \\ \Rightarrow - 2 si n^{2} x - s i n x + 1 = 0 \Rightarrow 2 si n^{2} x + s i n x - 1 = 0 \\ Since t h e e q u a t i o n i s q u a d r a t i c e q u a t i o n i n s i n x . S o i t w i l l h a v e 2 s o l u t i o n s . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

If $s i n θ = - \frac{2}{5}$ and $θ$ lies in the third quadrant, then the value of $c o s θ$ is:

(a) $- \frac{1}{5}$  

(b) $- \frac{\sqrt[]{21}}{5}$  

(c) $\frac{1}{5}$  

(d) $\frac{\sqrt[]{21}}{5}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ = - \frac{4}{5}, θ l i e s i n t h i r d q u a d r a n t \\ c o s θ = \sqrt[]{1 - s i n^{2} θ} = \sqrt[]{1 - {(- \frac{4}{5})}^{2}} = \sqrt[]{1 - \frac{16}{25}} = \sqrt[]{\frac{9}{25}} = \frac{+ 3}{- 5} \\ ∴ c o s θ = - \frac{3}{5}, θ l i e s i n t h i r d q u a d r a n t \\ c o s θ = 2 c o s^{2} \frac{θ}{2} - 1 [? π < θ < \frac{3 π}{2}, ∴ \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}] \\ \Rightarrow \frac{- 3}{5} = 2 c o s^{2} \frac{θ}{2} - 1 \\ \Rightarrow 2 c o s^{2} \frac{θ}{2} = 1 - \frac{3}{5} = \frac{2}{5} \Rightarrow c o s^{2} \frac{θ}{2} = \frac{2}{5 \times 2} = \frac{1}{5} \\ \Rightarrow c o s \frac{θ}{2} = \pm \frac{1}{\sqrt[]{5}} \Rightarrow c o s \frac{θ}{2} = - \frac{1}{\sqrt[]{5}} [? \frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

If $α + β = \frac{π}{4}$ , then the value of $(1 + t a n α) (1 + t a n β)$  is:

(a) 1

(b) 2

(c) -2

(d) Not defined

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : α + β = \frac{π}{4} \\ W e k n o w t h a t \\ t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α . t a n β} \\ \Rightarrow t a n α + t a n β = 1 - t a n α . t a n β \\ \Rightarrow t a n α + t a n β + t a n α . t a n β = 1 \\ \Rightarrow 1 + t a n α + t a n β + t a n α . t a n β = 1 + 1 \\ \Rightarrow 1 (1 + t a n α) + t a n β (1 + t a n α) = 2 \\ \Rightarrow (1 + t a n α) (1 + t a n β) = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

If $s i n θ + c o s θ = 1$ , then the value of $s i n 2 θ$ is:

(a) 1

(b) $\frac{1}{2}$  

(c) 0

(d) -1

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ + c o s θ = 1 \\ \Rightarrow {(s i n θ + c o s θ)}^{2} = {(1)}^{2} \\ \Rightarrow s i n^{2} θ + c o s^{2} θ + 2 s i n θ c o s θ = 1 \\ \Rightarrow 1 + s i n 2 θ = 1 \\ \Rightarrow s i n 2 θ = 1 - 1 = 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

The value of $s i n 5 0^{\circ} - s i n 7 0^{\circ} + s i n 1 0^{\circ}$ is equal to:

(a) 1

(b) 0

(c) $- \frac{1}{4}$

(d) 1

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : sin 5 0^{0} - sin 7 0^{0} + sin 1 0^{0} \\ (sin 5 0^{0} - sin 7 0^{0}) + sin 1 0^{0} = 2 c o s \frac{5 0^{0} + 7 0^{0}}{2} . s i n \frac{5 0^{0} - 7 0^{0}}{2} + sin 1 0^{0} \\ = 2 c o s 6 0^{0} . (- sin 1 0^{0}) + sin 1 0^{0} \\ = - 2 \times \frac{1}{2} sin 1 0^{0} + sin 1 0^{0} \\ = - sin 1 0^{0} + sin 1 0^{0} = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

Solve the equation $z = \overset{?}{z} + 1 + 2 i$ .

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z | = z + 1 + 2 i \\ L e t z = x + y i \\ | z | = (z + 1) + 2 i \\ S q u a r i n g b o t h s i d e s \\ {| z |}^{2} = {(z + 1)}^{2} + 4 i^{2} + 4 (z + 1) i \\ \Rightarrow {| z |}^{2} = {| z |}^{2} + 1 + 2 z - 4 + 4 (z + 1) i \\ \Rightarrow 0 = - 3 + 2 z + 4 (z + 1) i \\ \Rightarrow 3 - 2 z - 4 (z + 1) i = 0 \\ \Rightarrow 3 - 2 (x + y i) - 4 (x + y i + 1) i = 0 \\ \Rightarrow 3 - 2 x - 2 y i - 4 x i - 4 y i^{2} - 4 i = 0 \\ \Rightarrow 3 - 2 x + 4 y - 2 y i - 4 x i - 4 i = 0 \\ \Rightarrow (3 - 2 x + 4 y) - i (2 y + 4 x + 4) = 0 \\ \Rightarrow 3 - 2 x + 4 y = 0 \Rightarrow 2 x - 4 y = 3 \dots (i) \\ \Rightarrow 2 y + 4 x + 4 = 0 \Rightarrow 2 x + y = - 2 \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i), w e g e t y = - 1 a n d x = - \frac{1}{2} \\ H e n c e, t h e v a l u e o f z = x + y i = (- \frac{1}{2} - i) . \end{array}$

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Exemplar Solution Class 11th

Show that the complex number $z$ , satisfying the condition $arg (\frac{1 - z}{1 + z}) = \frac{π}{4}$ , lies on a circle.

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z = x + i y \\ G i v e n t h a t : a r g (\frac{z - 1}{z + 1}) = \frac{π}{4} \\ \Rightarrow a r g (z - 1) - a r g (z + 1) = \frac{π}{4} [? a r g (z_{1}) - a r g (z_{2}) = a r g \frac{z_{1}}{z_{2}}] \\ \Rightarrow a r g (x + i y - 1) - a r g (x + i y + 1) = \frac{π}{4} \\ \Rightarrow a r g [(x - 1) + i y] - a r g [(x + 1) + i y] = \frac{π}{4} \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} - t a n^{- 1} \frac{y}{x + 1} = \frac{π}{4} [? a r g (x + y i) = t a n^{- 1} \frac{y}{x}] \\ \Rightarrow t a n^{- 1} [\frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + \frac{y}{x - 1} \times \frac{y}{x + 1}}] = \frac{π}{4} \\ \Rightarrow \frac{x y + y - x y + y}{x^{2} - 1 + y^{2}} = t a n \frac{π}{4} \\ \Rightarrow \frac{2 y}{x^{2} + y^{2} - 1} = 1 \\ \Rightarrow x^{2} + y^{2} - 1 = 2 y \\ \Rightarrow x^{2} + y^{2} - 2 y - 1 = 0 w h i c h i s a c i r c l e . \\ H e n c e, z l i e s o n a c i r c l e . \end{array}$

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Exemplar Solution Class 11th

If the real part of $\frac{z + 2}{z - 1}$  is 4, then show that the locus of the point representing  $z$ in the complex plane is a circle.

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z = x + i y \\ ∴ \bar{z} = x - i y \\ S o \frac{\bar{z} + 2}{\bar{z} - 1} = \frac{x - i y + 2}{x - i y - 1} \\ = \frac{(x + 2) - i y}{(x - 1) - i y} \times \frac{(x - 1) + i y}{(x - 1) + i y} \\ = \frac{(x + 2) (x - 1) + (x + 2) y i - (x - 1) y i - i^{2} y^{2}}{{(x - 1)}^{2} - i^{2} y^{2}} \\ = \frac{x^{2} + 2 x - x - 2 + (x + 2 - x + 1) y i + y^{2}}{{(x - 1)}^{2} + y^{2}} \\ = \frac{x^{2} + y^{2} + x - 2}{{(x - 1)}^{2} + y^{2}} + \frac{3 y}{{(x - 1)}^{2} + y^{2}} i \\ Real p a r t = 4 \\ ∴ \frac{x^{2} + y^{2} + x - 2}{{(x - 1)}^{2} + y^{2}} = 4 \\ \Rightarrow x^{2} + y^{2} + x - 2 = 4 [{(x - 1)}^{2} + y^{2}] \\ \Rightarrow x^{2} + y^{2} + x - 2 = 4 [x^{2} + 1 - 2 x + y^{2}] \\ \Rightarrow x^{2} + y^{2} + x - 2 = 4 x^{2} + 4 - 8 x + 4 y^{2} \\ \Rightarrow x^{2} - 4 x^{2} + y^{2} - 4 y^{2} + x + 8 x - 2 - 4 = 0 \\ \Rightarrow - 3 x^{2} - 3 y^{2} + 9 x - 6 = 0 \\ \Rightarrow x^{2} + y^{2} - 3 x + 2 = 0 \\ W h i c h r e p r e s e n t s a c i r c l e . H e n c e, z l i e s o n a c i r c l e . \end{array}$

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If $\overset{?}{? z} + 1 ? = z + 2 (1 + i)$ , then find $z$ .

The value of $s i n 5 0^{\circ} - s i n 7 0^{\circ} + s i n 1 0^{\circ}$ is equal to:

(a) 1

(b) 0

(c) $- \frac{1}{4}$

(d) 1

Solve the equation $z = \overset{?}{z} + 1 + 2 i$ .

Show that the complex number $z$ , satisfying the condition $arg (\frac{1 - z}{1 + z}) = \frac{π}{4}$ , lies on a circle.

If the real part of $\frac{z + 2}{z - 1}$  is 4, then show that the locus of the point representing  $z$ in the complex plane is a circle.