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Exemplar Solution Class 11th

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

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 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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Exemplar Solution Class 11th

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

3 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 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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Exemplar Solution Class 11th

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

2 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

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

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Exemplar Solution Class 11th

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

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

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

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Exemplar Solution Class 11th

In a triangle $△ A B C$ with $∠ C = 9 0^{\circ}$ , the equation whose roots are $t a n A$ and $t a n B$ is _.

Hint :If $A + B = 9 0^{0} \Rightarrow t a n A t a n B = 1$ , then $t a n A + t a n B = \frac{2}{S i n A}$ _.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n a Δ A B C w i t h ∠ C = 9 0^{0} \\ S o, t h e e q u a t i o n w h o s e r o o t s a r e t a n A a n d t a n B i s \\ x^{2} - (t a n A + t a n B) x + t a n A . t a n B = 0 \\ A + B = 9 0^{0} [? ∠ C = 9 0^{0}] \\ \Rightarrow t a n (A + B) = t a n 9 0^{0} \\ \Rightarrow \frac{t a n A + t a n B}{1 - t a n A . t a n B} = \frac{1}{0} \\ \Rightarrow 1 - t a n A . t a n B = 0 \\ \Rightarrow t a n A . t a n B = 1 \dots (i) \\ N o w t a n A + t a n B = \frac{s i n A}{c o s A} + \frac{s i n B}{c o s B} \\ = \frac{s i n A c o s B + c o s A s i n B}{c o s A c o s B} \\ = \frac{s i n (A + B)}{c o s A c o s B} = \frac{s i n 9 0^{0}}{c o s A . c o s (9 0^{0} - A)} \\ = \frac{1}{c o s A s i n A} \\ ∴ t a n A + t a n B = \frac{2}{2 c o s A s i n A} = \frac{2}{s i n 2 A} \dots (i i) \\ F r o m (i) a n d (i i) w e g e t \\ x^{2} - (\frac{2}{s i n 2 A}) x + 1 = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s x^{2} - (\frac{2}{s i n 2 A}) x + 1 = 0 . \end{array}$

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Exemplar Solution Class 11th

If $s i n x + c o s x = a$ , then:

i. $s i n^{6} x + c o s^{6} x =$ _
ii. $∣ s i n x - c o s x ∣ =$ _

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 Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n x + c o s x = a \\ {(s i n x + c o s x)}^{2} = a^{2} \\ \Rightarrow s i n^{2} x + c o s^{2} x + 2 s i n x c o s x = a^{2} \\ \Rightarrow 1 + 2 s i n x c o s x = a^{2} \\ \Rightarrow s i n x c o s x = \frac{a^{2} - 1}{2} \dots (i) \\ (i) s i n^{6} x + c o s^{6} x = {(s i n^{2} x)}^{3} + {(c o s^{2} x)}^{3} \\ = {(s i n^{2} x + c o s^{2} x)}^{3} - 3 s i n^{2} x c o s^{2} x (s i n^{2} x + c o s^{2} x) \\ = {(1)}^{3} - 3 {(\frac{a^{2} - 1}{2})}^{2} . 1 = 1 - \frac{3 {(a^{2} - 1)}^{2}}{4} = \frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}] \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}] . \\ (i i) {| s i n x - c o s x |}^{2} = s i n^{2} x + c o s^{2} x - 2 s i n x c o s x \\ = 1 - 2 (\frac{a^{2} - 1}{2}) = 1 - (a^{2} - 1) = 1 - a^{2} + 1 \\ = 2 - a^{2} \\ ∴ | s i n x - c o s x | = \sqrt[]{2 - a^{2}} [? | s i n x - c o s x | > 0] \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \sqrt[]{2 - a^{2}} . \end{array}$

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Exemplar Solution Class 11th

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

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 Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n A = \frac{1 - c o s B}{s i n B} \\ t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A} = \frac{2 (\frac{1 - c o s B}{s i n B})}{1 - {(\frac{1 - c o s B}{s i n B})}^{2}} \\ = \frac{2 (\frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2})}{1 - {(\frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2})}^{2}} [\begin{array}{l} ? 1 - c o s B = 2 s i n^{2} B / 2 \\ s i n B = 2 s i n B / 2 c o s B / 2 \end{array}] \\ = \frac{2 (\frac{s i n B / 2}{c o s B / 2})}{1 - {(\frac{s i n B / 2}{c o s B / 2})}^{2}} = \frac{2 t a n B / 2}{1 - t a n^{2} B / 2} = t a n B \\ S o, t a n 2 A = t a n B \\ H e n c e, t h e v a l u e o f f i l l e r i s t a n B . \end{array}$

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Exemplar Solution Class 11th

If $k = s i n \frac{π}{18} s i n \frac{5 π}{18} s i n \frac{7 π}{18}$ , then the numerical value of $k$ is ___.

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 Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : k = s i n (\frac{π}{18}) s i n (\frac{5 π}{18}) s i n (\frac{7 π}{18}) \\ \Rightarrow k = s i n 1 0^{0} . s i n 5 0^{0} . s i n 7 0^{0} \\ \Rightarrow k = s i n 1 0^{0} . s i n (9 0^{0} - 4 0^{0}) . s i n (9 0^{0} - 2 0^{0}) \\ \Rightarrow k = s i n 1 0^{0} . c o s 4 0^{0} . c o s 2 0^{0} \\ \Rightarrow k = s i n 1 0^{0} . \frac{1}{2} [2 c o s 4 0^{0} . c o s 2 0^{0}] \\ \Rightarrow k = s i n 1 0^{0} . \frac{1}{2} [c o s (4 0^{0} + 2 0^{0}) + c o s (4 0^{0} - 2 0^{0})] \\ \Rightarrow k = \frac{1}{2} s i n 1 0^{0} [c o s 6 0^{0} + c o s 2 0^{0}] \\ \Rightarrow k = \frac{1}{2} s i n 1 0^{0} [\frac{1}{2} + c o s 2 0^{0}] \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{2} s i n 1 0^{0} c o s 2 0^{0} \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} (2 s i n 1 0^{0} c o s 2 0^{0}) \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} [s i n (1 0^{0} + 2 0^{0}) + s i n (1 0^{0} - 2 0^{0})] \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} [s i n 3 0^{0} + s i n (- 1 0^{0})] \\ \Rightarrow k = \frac{1}{4} s i n 1 0^{0} + \frac{1}{4} s i n 3 0^{0} - \frac{1}{4} s i n 1 0^{0} \\ \Rightarrow k = \frac{1}{4} s i n 3 0^{0} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} . \\ H e n c e, t h e v a l u e o f f i l l e r i s \frac{1}{8} . \end{array}$

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Exemplar Solution Class 11th

The value of $\frac{s i n 5 0^{\circ}}{s i n 13 0^{\circ}}$ is ___.

4 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 Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \frac{s i n 5 0^{0}}{s i n 13 0^{0}} = \frac{s i n 5 0^{0}}{s i n (18 0^{0} - 5 0^{0})} = \frac{s i n 5 0^{0}}{s i n 5 0^{0}} = 1 \\ H e n c e, t h e v a l u e o f f i l l e r i s 1 . \end{array}$

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Exemplar Solution Class 11th

If  $z_{1}, z_{2}$ and  $z_{3}, z_{4}$ are two pairs of conjugate complex numbers, then find $arg (\frac{z_{1}}{z_{4}}) + arg (\frac{z_{2}}{z_{3}})$ .

2 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

$\begin{array}{l} L e t t h e p o l a r f o r m o f z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) \\ ∴ z_{2} = {\bar{z}}_{1} = r_{1} (c o s θ_{1} - i s i n θ_{1}) = r_{1} [c o s (- θ_{1}) + i s i n (- θ_{1})] \\ S i m i l a r l y, z_{3} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ ∴ z_{4} = {\bar{z}}_{3} = r_{1} (c o s θ_{2} - i s i n θ_{2}) = r_{2} [c o s (- θ_{2}) + i s i n (- θ_{2})] \\ a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}}) = a r g (z_{1}) - a r g (z_{4}) + a r g (z_{2}) - a r g (z_{3}) \\ = θ_{1} - (- θ_{2}) + (- θ_{1}) - θ_{2} \\ = θ_{1} + θ_{2} - θ_{1} - θ_{2} = 0 \\ H e n c e, a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}}) = 0 \end{array}$

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

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

In a triangle $△ A B C$ with $∠ C = 9 0^{\circ}$ , the equation whose roots are $t a n A$ and $t a n B$ is _.

Hint :If $A + B = 9 0^{0} \Rightarrow t a n A t a n B = 1$ , then $t a n A + t a n B = \frac{2}{S i n A}$ _.

If $s i n x + c o s x = a$ , then:

i. $s i n^{6} x + c o s^{6} x =$ _
ii. $∣ s i n x - c o s x ∣ =$ _

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

If $k = s i n \frac{π}{18} s i n \frac{5 π}{18} s i n \frac{7 π}{18}$ , then the numerical value of $k$ is ___.

The value of $\frac{s i n 5 0^{\circ}}{s i n 13 0^{\circ}}$ is ___.

If  $z_{1}, z_{2}$ and  $z_{3}, z_{4}$ are two pairs of conjugate complex numbers, then find $arg (\frac{z_{1}}{z_{4}}) + arg (\frac{z_{2}}{z_{3}})$ .