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Class 11th

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New answer posted

a year ago

Class 11th Exemplar Solution

Solve the system of equations $Re (z^{2}) = 0, z = 2$ .

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : R e (z^{2}) = 0 a n d | z | = 2 \\ L e t z = x + y i \\ ∴ | z | = \sqrt[]{x^{2} + y^{2}} \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 2 \Rightarrow x^{2} + y^{2} = 4 \dots (i) \\ Since z = x + y i \\ z^{2} = x^{2} + y^{2} i^{2} + 2 x y i \Rightarrow z^{2} = x^{2} - y^{2} + 2 x y i \\ ∴ R e (z^{2}) = x^{2} - y^{2} \\ \Rightarrow x^{2} - y^{2} = 0 \dots (i i) \\ F r o m e q n . (i) a n d (i i), w e g e t \\ x^{2} + y^{2} = 4 \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt[]{2} a n d y = \pm \sqrt[]{2} \\ H e n c e, z = \pm \sqrt[]{2} \pm i \sqrt[]{2}, - \sqrt[]{2} \pm i \sqrt[]{2} . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

. $c o s \frac{π}{15} \cdot c o s \frac{4 π}{15} \cdot c o s \frac{8 π}{15} \cdot c o s \frac{16 π}{15} = \frac{1}{16}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . c o s \frac{2 π}{15} . c o s \frac{4 π}{15} . c o s \frac{8 π}{15} . c o s \frac{16 π}{15} \\ = c o s 2 4^{0} . c o s 4 8^{0} . c o s 9 6^{0} . c o s 19 2^{0} \\ = \frac{1}{16 s i n 2 4^{0}} [(2 s i n 2 4^{0} c o s 2 4^{0}) (2 c o s 4 8^{0}) (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 4 8^{0} . 2 c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 4 8^{0} c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 9 6^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 9 6^{0} c o s 9 6^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 19 2^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 19 2^{0} c o s 19 2^{0}] \\ = \frac{1}{16 s i n 2 4^{0}} s i n 38 4^{0} = \frac{1}{16 s i n 2 4^{0}} s i n (36 0^{0} + 2 4^{0}) \\ = \frac{1}{16 s i n 2 4^{0}} * s i n 2 4^{0} [? s i n (36 0^{0} + θ) = s i n θ] \\ = \frac{1}{16} R . H . S . \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

$s i n 1 0^{\circ}$ is greater than $c o s 1 0^{\circ}$ .

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f s i n 1 0^{0} > c o s 1 0^{0} \\ \Rightarrow s i n 1 0^{0} > c o s (9 0^{0} - 8 0^{0}) \\ \Rightarrow s i n 1 0^{0} > s i n 8 0^{0} w h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The equality $s i n A + s i n 2 A + s i n 3 A = 3$ holds for some real value of $A$ .

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n A + s i n 2 A + s i n 3 A = 3 \\ Since t h e maximum v a l u e o f s i n A i s 1 b u t f o r s i n 2 A a n d s i n 3 A i t i s n o t e q u a l t o 1 . \\ S o i t i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $t a n A = \frac{1 - c o s B}{s i n B}$ , then $t a n 2 A = t a n B$ .

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n A = \frac{1 - c o s B}{s i n B} = \frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2} = t a n \frac{B}{2} \\ t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A} = \frac{2 t a n B / 2}{1 - t a n^{2} B / 2} \\ ∴ t a n 2 A = t a n B \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The maximum distance of a point on the graph of the function $y = 3 s i n x + c o s x$ from the $x$ -axis is ___.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : y = \sqrt[]{3} s i n x + c o s x \dots (i) \\ ∴ T h e maximum distance f r o m a point o n t h e g r a p h o f e q n . (i) f r o m x - a x i s \\ = \sqrt[]{{(\sqrt[]{3})}^{2} + {(1)}^{2}} = \sqrt[]{3 + 1} = 2 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 2 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Given $x > 0$ , the values of $f (x) = - 3 c o s 2 x + \frac{3}{x}$ lie in the interval ___.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : f (x) = - 3 c o s \sqrt[]{3 + x + x^{2}} \\ P u t \sqrt[]{3 + x + x^{2}} = y \\ ∴ f (x) = - 3 c o s y \\ ? - 1 \leq c o s y \leq 1 \\ 3 \geq - 3 c o s y \geq - 3 \\ \Rightarrow - 3 \leq - 3 c o s y \leq 3 \\ \Rightarrow - 3 \leq - 3 c o s \sqrt[]{3 + x + x^{2}} \leq 3, x > 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s [- 3, 3] . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

$3 {(s i n x - c o s x)}^{4} + 6 {(s i n x + c o s x)}^{2} + 4 (s i n^{6} x + c o s^{6} x) =$ ___.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s \\ 3 {(s i n x - c o s x)}^{4} + 6 {(s i n x + c o s x)}^{2} + 4 (s i n^{6} x + c o s^{6} x) \\ = 3 {[s i n^{2} x + c o s^{2} x - 2 s i n x c o s x]}^{2} + 6 [s i n^{2} x + c o s^{2} x + 2 s i n x c o s x] + 4 [{(s i n^{2} x)}^{3} + {(c o s^{2} x)}^{3}] \\ = 3 {[1 - 2 s i n x c o s x]}^{2} + 6 [1 + 2 s i n x c o s x] + 4 [{(s i n^{2} x + c o s^{2} x)}^{3} - 3 s i n^{2} x c o s^{2} x (s i n^{2} x + c o s^{2} x)] \\ = 3 [1 + 4 s i n^{2} x c o s^{2} x - 4 s i n x c o s x] + 6 (1 + 2 s i n x c o s x) + 4 [1 - 3 s i n^{2} x c o s^{2} x] \\ = 3 + 12 s i n^{2} x c o s^{2} x - 12 s i n x c o s x + 6 + 12 s i n x c o s x + 4 - 12 s i n^{2} x c o s^{2} x \\ = 3 + 6 + 4 = 13 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 13 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If for complex numbers $z_{1}$ and $z_{2}$ , $arg (z_{1}) - arg (z_{2}) = 0$ , then show that $? z_{1} - z_{2} ? = ? z_{1} ? ? z_{2} ?$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t f o r z_{1} a n d z_{2}, a r g (z_{1}) - a r g (z_{2}) = 0 \\ L e t u s r e p r e s e n t z_{1} a n d z_{2} i n p o l a r f o r m \\ z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ a r g (z_{1}) = θ_{1} a n d a r g (z_{2}) = θ_{2} \\ Since a r g (z_{1}) - a r g (z_{2}) = 0 \\ \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2} \\ N o w z_{1} - z_{2} = r_{1} (c o s θ_{1} - i s i n θ_{1}) - r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} c o s θ_{1} - i r_{1} s i n θ_{1} - r_{2} c o s θ_{1} + i r_{2} s i n θ_{1} [? θ_{1} = θ_{2}] \\ = (r_{1} c o s θ_{1} - r_{2} c o s θ_{1}) + i (r_{1} s i n θ_{1} - r_{2} s i n θ_{1}) \\ ∴ | z_{1} - z_{2} | = \sqrt[]{{(r_{1} c o s θ_{1} - r_{2} c o s θ_{1})}^{2} + {(r_{1} s i n θ_{1} - r_{2} s i n θ_{1})}^{2}} \\ = \sqrt[]{(r_{1}^{2} c o s^{2} θ_{1} + r_{2}^{2} c o s^{2} θ_{1} - 2 r_{1} r_{2} c o s^{2} θ_{1} + r_{1}^{2} s i n^{2} θ_{1} + r_{2}^{2} s i n^{2} θ_{1} - 2 r_{1} r_{2} s i n^{2} θ_{1})} \\ = \sqrt[]{r_{1}^{2} (c o s^{2} θ_{1} + s i n^{2} θ_{1}) + r_{2}^{2} (c o s^{2} θ_{1} + s i n^{2} θ_{1}) - 2 r_{1} r_{2} (c o s^{2} θ_{1} + s i n^{2} θ_{1})} \\ = \sqrt[]{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2}} = \sqrt[]{{(r_{1} - r_{2})}^{2}} = r_{1} - r_{2} = | z_{1} | - | z_{2} | \\ H e n c e, | z_{1} - z_{2} | = | z_{1} | - | z_{2} | \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $? z_{1} ? = ? z_{2} ? = . . . = ? z_{n} ? = 1$ , then show that $? z_{1} + z_{2} + z_{3} + . . . + \frac{1_{}}{z_{1}} = \frac{z_{2}}{z_{2}} + \frac{z_{3}}{z_{3}} + \frac{z_{4}}{z_{3}} + . . . + \frac{z_{1}}{z_{n}}$ .

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} W e h a v e | z_{1} | = | z_{2} | = \dots = | z_{n} | = 1 \\ \Rightarrow {| z_{1} |}^{2} = {| z_{2} |}^{2} = \dots = {| z_{n} |}^{2} = 1 \dots (i) \\ \Rightarrow z_{1} {\bar{z}}_{1} = z_{2} {\bar{z}}_{2} = \dots = z_{n} {\bar{z}}_{n} = 1 [? z \bar{z} = {| z |}^{2}] \\ \Rightarrow z_{1} = \frac{1}{{\bar{z}}_{1}}, z_{2} = \frac{1}{{\bar{z}}_{2}} = \dots = z_{n} = \frac{1}{{\bar{z}}_{n}} \\ L . H . S . | z_{1} + z_{2} + z_{3} + \dots + z_{n} | \\ = | \frac{z_{1} {\bar{z}}_{1}}{{\bar{z}}_{1}} + \frac{z_{2} {\bar{z}}_{2}}{{\bar{z}}_{2}} + \frac{z_{3} {\bar{z}}_{3}}{{\bar{z}}_{3}} + \dots + \frac{z_{n} {\bar{z}}_{n}}{{\bar{z}}_{n}} | \\ = | \frac{{| z_{1} |}^{2}}{{\bar{z}}_{1}} + \frac{{| z_{2} |}^{2}}{{\bar{z}}_{2}} + \frac{{| z_{3} |}^{2}}{{\bar{z}}_{3}} + \dots + \frac{{| z_{n} |}^{2}}{{\bar{z}}_{n}} | [? z \bar{z} = {| z |}^{2}] \\ = | \frac{1}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}} + \dots + \frac{1}{{\bar{z}}_{n}} | [Using (i)] \\ = | \bar{\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + \dots + \frac{1}{z_{n}}} | [? {\bar{z}}_{1} + {\bar{z}}_{2} = \bar{z_{1} + z_{2}}] \\ = | \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + \dots + \frac{1}{z_{n}} | [? | z | = | {\bar{z}}_{1} |] \\ L . H . S . - R . H . S . H e n c e p r o v e d . \end{array}$

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