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Exemplar Solution Class 11th

If $θ$ lies in the first quadrant and $c o s θ = \frac{8}{17}$ , then find the value of $c o s (3 0^{\circ} + θ) + c o s (4 5^{\circ} - θ) + c o s (12 0^{\circ} - θ) .$

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 Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s θ = \frac{8}{17} \\ ∴ s i n θ = \sqrt[]{1 - {(\frac{8}{17})}^{2}} = \sqrt[]{1 - \frac{64}{289}} = \sqrt[]{\frac{289 - 64}{289}} = \sqrt[]{\frac{225}{289}} = \frac{15}{17} \\ B u t θ l i e s i n I q u a d r a n t . \\ ∴ s i n θ = \frac{15}{17} \\ N o w c o s (3 0^{0} + θ) + c o s (4 5^{0} - θ) + c o s (12 0^{0} - θ) \\ = c o s 3 0^{0} c o s θ - s i n 3 0^{0} s i n θ + c o s 4 5^{0} c o s θ + s i n 4 5^{0} s i n θ + c o s 12 0^{0} c o s θ + s i n 12 0^{0} s i n θ \\ = \frac{\sqrt[]{3}}{2} c o s θ - \frac{1}{2} s i n θ + \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ - \frac{1}{2} c o s θ + \frac{\sqrt[]{3}}{2} s i n θ \\ = (\frac{\sqrt[]{3}}{2} c o s θ + \frac{\sqrt[]{3}}{2} s i n θ) - \frac{1}{2} (s i n θ + c o s θ) + \frac{1}{\sqrt[]{2}} (c o s θ + s i n θ) \\ = \frac{\sqrt[]{3}}{2} (c o s θ + s i n θ) - \frac{1}{2} (s i n θ + c o s θ) + \frac{1}{\sqrt[]{2}} (c o s θ + s i n θ) \\ = (\frac{\sqrt[]{3}}{2} - \frac{1}{2} + \frac{1}{\sqrt[]{2}}) (c o s θ + s i n θ) = (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) (\frac{8}{17} + \frac{15}{17}) \\ = (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) (\frac{23}{17}) \\ = \frac{23}{17} (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) . \\ H e n c e, t h e r e q u i r e d s o l u t i o n = \frac{23}{17} (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) . \end{array}$

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Exemplar Solution Class 11th

Give an example of a statement $P (n)$ which is true for all $n$ . Justify your answer.

4 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e r e q u i r e d s t a t e m e n t i s P (n) : 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2} \\ J u s t i f i c a t i o n : P (1) : 1 = \frac{1 (1 + 1)}{2} \\ P (k) : 1 + 2 + 3 + \dots + k = \frac{k (k + 1)}{2} . L e t i t b e t r u e . \\ P (k + 1) : 1 + 2 + 3 + \dots + k + (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{(k + 1) (k + 2)}{2} \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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Exemplar Solution Class 11th

Give an example of a statement $P (n)$ which is true for all $n \geq 4$ but $P (1)$ , $P (2)$ , and $P (3)$ are not true. Justify your answer.

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e r e q u i r e d s t a t e m e n t i s P (n) = 2 n < n! \\ J u s t i f i c a t i o n : P (n) : 2 n < n! \\ P (1) : 2.1 < 1! \Rightarrow 2 < 1 n o t t r u e \\ P (2) : 2.2 < 2! \Rightarrow 4 < 2.1 \Rightarrow 4 < 2 n o t t r u e \\ P (3) : 2.3 < 3! \Rightarrow 6 < 3.2.1 \Rightarrow 6 < 6 n o t t r u e \\ P (4) : 2.4 < 4! \Rightarrow 8 < 4.3.2.1 \Rightarrow 8 < 24 T r u e \\ P (5) : 2.5 < 5! \Rightarrow 10 < 5.4.3.2.1 \Rightarrow 10 < 120 T r u e \\ H e n c e, P (n) : 2 n < n! i s n o t t r u e f o r P (1), P (2) a n d P (3) b u t i t i s t r u e f o r a l l v a l u e s o f n \geq 4 . \end{array}$

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Exemplar Solution Class 11th

Domain of  $\sqrt[]{a^{2} - x^{2}}$ ( $a > 0$ ) is

(a) $(- a, a)$  

(b) $[- a, a]$  

(c) $[0, a]$  

(d) $(- a, 0]$

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t f (x) = \sqrt[]{a^{2} - x^{2}} \\ f (x) i s d e f i n e d i f a^{2} - x^{2} \geq 0 \\ \Rightarrow x^{2} - a^{2} \leq 0 \Rightarrow x^{2} \leq a^{2} \\ \Rightarrow x \leq \pm a \Rightarrow - a \leq x \leq a \\ ∴ D o m a i n o f f (x) = [- a, a] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

If <math> <mrow> <mi>x</mi> <mo>=</mo> <mi>s</mi> <mi>e</mi> <mi>c</mi> <mi>?</mi> <mo>−</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>?</mi> </mrow> </math> and <math> <mrow> <mi>y</mi> <mo>=</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>e</mi> <mi>c</mi> <mi>?</mi> <mo>+</mo> <mi>c</mi> <mi>o</mi> <mi>t</mi> <mi>?</mi> </mrow> </math> , then show that <math> <mrow> <mi>x</mi> <mi>y</mi> <mo>+</mo> <mi>x</mi> <mo>−</mo> <mi>y</mi> <mo>+</mo> <mn>1</mn> <mo>=</mo> <mn>0</mn> <mo>.</mo> </mrow> </math>    [Hint: Find <math> <mrow> <mi>x</mi> <mi>y</mi> <mo>+</mo> <mn>1</mn> </mrow> </math> and then show <math> <mrow> <mi>x</mi> <mo>−</mo> <mi>y</mi> <mo>=</mo> <mo>−</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mi>y</mi> <mo>+</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> .]

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Exemplar Solution Class 11th

Let  $f (x) = \sqrt[]{1 + x^{2}}$ then

(a) $f (x y) = f (x) \cdot f (y)$  

(b) $f (x y) \geq f (x) \cdot f (y)$  

(c) $f (x y) \leq f (x) \cdot f (y)$  

(d) None of these

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Class 11th

If $a c o s 2 θ + b s i n 2 θ = c,$ has $α$ and $β$ as its roots, then prove that $t a n α + t a n β = \frac{2 b}{a + c} .$ 

[Hint: Use the identities for $c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}$ and $s i n 2 θ$ = $\frac{2 t a n θ}{1 + t a n^{2} θ}$ ]

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 a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : a c o s 2 θ + b s i n 2 θ = c \\ \Rightarrow a [\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}] + b [\frac{2 t a n θ}{1 + t a n^{2} θ}] = c \\ \Rightarrow a - a t a n^{2} θ + 2 b t a n θ = c (1 + t a n^{2} θ) [? c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}, s i n 2 θ = \frac{2 t a n θ}{1 + t a n^{2} θ}] \\ \Rightarrow a - a t a n^{2} θ + 2 b t a n θ - c t a n^{2} θ - c = 0 \\ \Rightarrow - (a + c) t a n^{2} θ + 2 b t a n θ + (a - c) = 0 \\ \Rightarrow (a + c) t a n^{2} θ - 2 b t a n θ + (c - a) = 0 \\ Since α a n d β a r e t h e r o o t s o f t h i s e q u a t i o n \\ \Rightarrow t a n α + t a n β = \frac{- (- 2 b)}{a + c} \\ \Rightarrow t a n α + t a n β = \frac{2 b}{a + c} . H e n c e p r o v e d . \end{array}$

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Exemplar Solution Class 11th

Range of  $f (x) = \frac{1}{1 - 2 c o s x}$ is

(a) $[\frac{1}{3}, 1]$  

(b) $[- 1, \frac{1}{3}]$  

(c) $(- \infty, - 1] \cup [\frac{1}{3}, \infty)$  

(d) $[- \frac{1}{3}, 1]$

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{1}{1 - 2 c o s x} \\ W e k n o w t h a t - 1 \leq c o s x \leq 1 \\ \Rightarrow 1 \geq c o s x \geq - 1 \Rightarrow - 1 \leq - c o s x \leq 1 \\ \Rightarrow - 2 \leq - 2 c o s x \leq 2 \Rightarrow - 2 + 1 \leq 1 - 2 c o s x \leq 2 + 1 \\ \Rightarrow - 1 \leq 1 - 2 c o s x \leq 3 \Rightarrow - 1 \leq \frac{1}{1 - 2 c o s x} \leq \frac{1}{3} \\ \Rightarrow - 1 \leq f (x) \leq \frac{1}{3} \\ S o, t h e r a n g e o f f (x) = [- 1, \frac{1}{3}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

Find the value of the expression $3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (2 π / 2 + α) + s i n^{6} (5 π - α)] .$

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 a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \\ 3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)] \\ = 3 [c o s^{4} α + s i n^{4} (π + α)] - 2 [c o s^{6} α + s i n^{6} (π - α)] \\ = 3 [c o s^{4} α + s i n^{4} α] - 2 [c o s^{6} α + s i n^{6} α] \\ = 3 [c o s^{4} α + s i n^{4} α + 2 s i n^{2} α c o s^{2} α - 2 s i n^{2} α c o s^{2} α] - 2 [\begin{array}{l} {(c o s^{2} α + s i n^{2} α)}^{3} \\ - 3 c o s^{2} α s i n^{2} α (c o s^{2} α + s i n^{2} α) \end{array}] \\ = 3 [{(c o s^{2} α + s i n^{2} α)}^{2} - 2 s i n^{2} α c o s^{2} α] - 2 [1 - 3 c o s^{2} α s i n^{2} α] \\ = 3 [1 - 2 s i n^{2} α c o s^{2} α] - 2 [1 - 3 c o s^{2} α s i n^{2} α] \\ = 3 - 6 s i n^{2} α c o s^{2} α - 2 + 6 c o s^{2} α s i n^{2} α \\ = 3 - 2 = 1 \\ H e n c e, t h e v a l u e o f t h e g i v e n expression i s 1 . \end{array}$

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Exemplar Solution Class 11th

If $P = {1, 2}$ , then $P \times P \times P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}$ .

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : P = {1, 2} \\ ∴ P \times P = {1, 2} \times {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} \\ P \times P \times P = {(1, 1), (1, 2), (2, 1), (2, 2)} \times {1, 2} \\ = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)} \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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If $θ$ lies in the first quadrant and $c o s θ = \frac{8}{17}$ , then find the value of $c o s (3 0^{\circ} + θ) + c o s (4 5^{\circ} - θ) + c o s (12 0^{\circ} - θ) .$

Give an example of a statement $P (n)$ which is true for all $n$ . Justify your answer.

Give an example of a statement $P (n)$ which is true for all $n \geq 4$ but $P (1)$ , $P (2)$ , and $P (3)$ are not true. Justify your answer.

Find the value of the expression $3 [s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)] - 2 [s i n^{6} (2 π / 2 + α) + s i n^{6} (5 π - α)] .$

If $P = {1, 2}$ , then $P \times P \times P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}$ .