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Exemplar Solution Class 11th

Find the angle between the lines y = (2 – 3)(x + 5) and y = (2 + 3)(x – 7).

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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:

$\begin{array}{l} T h e g i v e n e q u a t i o n s a r e y = (2 - \sqrt[]{3}) (x + 5) \dots (i) \\ a n d y = (2 + \sqrt[]{3}) (x - 7) \dots (i i) \\ S l o p e o f t h e e q n . (i) m_{1} (s a y) = (2 - \sqrt[]{3}) \\ a n d s l o p e o f t h e e q n . (i i) m_{2} (s a y) = (2 + \sqrt[]{3}) \\ L e t θ b e t h e a n g l e b e t q w e e n t h e t w o g i v e n l i n e s \\ ∴ t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 - \sqrt[]{3} - 2 - \sqrt[]{3}}{1 + (2 - \sqrt[]{3}) (2 + \sqrt[]{3})} | = | \frac{- 2 \sqrt[]{3}}{2} | = | - \sqrt[]{3} | \\ \Rightarrow t a n θ = \sqrt[]{3} o r - \sqrt[]{3} \\ ∴ θ = 6 0^{0} o r 12 0^{0} \\ H e n c e, t h e r e q u i r e d a n g l e i s 6 0^{0} o r 12 0^{0} . \end{array}$

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Exemplar Solution Class 11th

If a circle passes through the points $(0, 0), (a, 0), (0, b)$ , then find the coordinates of its center.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n points a r e (0, 0), (a, 0) a n d (0, b) \\ G e n e r a l e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 2 g x + 2 f y + c = 0 w h e r e t h e c e n t r e i s (- g, - f) a n d r a d i u s = \sqrt[]{g^{2} + f^{2} - c} \\ I f i t p a s s e s t h r o u g h (0, 0) \\ ∴ c = 0 \\ I f i t p a s s e s t h r o u g h (a, 0) a n d (0, b) t h e n \\ a^{2} + 2 g a + c = 0 \Rightarrow a^{2} + 2 g a = 0 [? c = 0] \\ ∴ g = - \frac{a}{2} \\ a n d 0 + b^{2} + 0 + 2 f b + c = 0 \Rightarrow b^{2} + 2 f b = 0 [? c = 0] \\ \Rightarrow f = - \frac{b}{2} \\ H e n c e, t h e c o o r d i n a t e s o f c e n t r e o f t h e c i r c l e a r e (- g, - f) = (\frac{a}{2}, \frac{b}{2}) \end{array}$

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Exemplar Solution Class 11th

Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, –1).

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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:

$S l o p e o f t h e l i n e j o i n i n g t h r o u g h t h e (2, 3) a n d (3, - 1) i s \frac{- 1 - 3}{3 - 2} = - 4 S l o p e o f t h e r e q u i r e d l i n e w h i c h i s p e r p e n d i c u l a r t o i t = \frac{- 1}{- 4} = \frac{1}{4} [$
$m_{1} m_{2} = - 1] E q u a t i o n o f t h e l i n e t h r o u g h t h e (5, 2) i s y - 2 = \frac{1}{4} (x - 5) [$
$\begin{array}{l} y - y_{1} = m (x - x_{1})] \\ \Rightarrow 4 y - 8 = x - 5 \\ \Rightarrow x - 4 y + 3 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x - 4 y + 3 = 0 . \end{array}$

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Exemplar Solution Class 11th

Show that the point $(x, y)$ given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of $t$ , such that $- 1 < t < 1$ , where $a$ is any given real number.

5 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

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

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Exemplar Solution Class 11th

Find the equation of the straight line which passes through the point (1, –2) and cuts off equal intercepts from axes.

3 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:

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

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Exemplar Solution Class 11th

Find the equation of the circle which touches both axes in the first quadrant and whose radius is $a$ .

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} C l e a r l y c e n t r e o f t h e c i r c l e = (a, a) \\ a n d r a d i u s = a \\ E q u a t i o n o f c i r c l e w i t h r a d i u s r a n d c e n t r e (h, k) i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ S o, t h e e q u a t i o n o f r e q u i r e d c i r c l e \\ \Rightarrow {(x - a)}^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} - 2 a x + a^{2} + y^{2} - 2 a y + a^{2} = a^{2} \\ \Rightarrow x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 a x - 2 a y + a^{2} = 0 . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

The third term of a G.P. is , the product of the first five terms is .

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Matching Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t u s c o n s i d e r a G . P, a, a r a n d a r^{2} \\ I f i t i s i n A . P t h e n a r - a \neq a r^{2} - a r \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Kindly consider the following

3 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Matching Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) L e t S_{n} = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} \\ K^{3} - {(K - 1)}^{3} = 3 K^{2} - 3 K + 1 \\ F o r K = 1, 1^{3} - 0^{3} = 3 {(1)}^{2} - 3 (1) + 1 \\ F o r K = 2, 2^{3} - 1^{3} = 3 {(2)}^{2} - 3 (2) + 1 \\ F o r K = 3, 3^{3} - 2^{3} = 3 {(3)}^{2} - 3 (3) + 1 \\ \dots \dots \dots \\ F o r K = n, n^{3} - {(n - 1)}^{3} = 3 {(n)}^{2} - 3 (n) + 1 \\ A d d i n g C o l u m n w i s e, w e g e t \\ n^{3} - 0^{3} = 3 (1^{2} + 2^{2} + 3^{2} + \dots + n^{2}) - 3 (1 + 2 + 3 + \dots + n) + n \\ \Rightarrow n^{3} = 3 . S_{n} - \frac{3 n (n + 1)}{2} + n \\ \Rightarrow n^{3} + \frac{3 n (n + 1)}{2} - n = 3 . S_{n} \\ \Rightarrow \frac{2 n^{3} + 3 n^{2} + 3 n - 2 n}{2} = 3 . S_{n} \\ \Rightarrow 6 . S_{n} = 2 n^{3} + 3 n^{2} + n \\ \Rightarrow 6 . S_{n} = n (2 n^{2} + 3 n + 1) \\ \Rightarrow 6 . S_{n} = n [2 n^{2} + 2 n + n + 1] \\ \Rightarrow 6 . S_{n} = n (n + 1) (2 n + 1) \\ \Rightarrow S_{n} = \frac{n (n + 1) (2 n + 1)}{6} \\ H e r e, (a) \leftrightarrow (i i i) \\ (b) L e t S_{n} = 1^{3} + 2^{3} + 3^{3} + \dots + n^{3} \\ K^{4} - {(K - 1)}^{4} = 4 K^{3} - 6 K^{2} + 4 K - 1 \\ F o r K = 1, 1^{4} - 0^{4} = 4 {(1)}^{3} - 6 {(1)}^{2} + 4 (1) - 1 \\ F o r K = 2, 2^{4} - 1^{4} = 4 {(2)}^{3} - 6 {(2)}^{2} + 4 (2) - 1 \\ F o r K = 3, 3^{4} - 2^{4} = 4 {(3)}^{3} - 6 {(3)}^{2} + 4 (3) - 1 \\ \dots \dots \dots \\ F o r K = n, n^{4} - {(n - 1)}^{4} = 4 {(n)}^{3} - 6 {(n)}^{2} + 4 (3) - 1 \\ A d d i n g C o l u m n w i s e, w e g e t \\ n^{4} - 0^{4} = 4 (1^{3} + 2^{3} + 3^{3} + \dots + n^{3}) - 6 (1^{2} + 2^{2} + 3^{2} + \dots + n^{2}) + 4 (1 + 2 + 3 + \dots + n) - n \\ \Rightarrow n^{4} = 4 . S_{n} - n (n + 1) (2 n + 1) + 2 n (n + 1) - n \\ \Rightarrow n^{4} + n (n + 1) (2 n + 1) - 2 n (n + 1) + n = 4 . S_{n} \\ \Rightarrow n [n^{3} + (n + 1) (2 n + 1) - 2 n - 2 + 1] = 4 . S_{n} \\ \Rightarrow n [n^{3} + 2 n^{2} + 3 n + 1 - 2 n - 1] = 4 . S_{n} \\ \Rightarrow n [n^{3} + 2 n^{2} + n] = 4 . S_{n} \end{array}$

$\begin{array}{l} \Rightarrow \frac{n^{2} (n^{2} + 2 n + 1)}{4} = S_{n} \\ \Rightarrow \frac{n^{2} {(n + 1)}^{2}}{4} = S_{n} \\ ∴ S_{n} = {[\frac{n (n + 1)}{2}]}^{2} \\ H e r e, (b) \leftrightarrow (i) \\ (c) L e t S_{n} = 2 + 4 + 6 + \dots + 2 n \\ = 2 (1 + 2 + 3 + \dots + n) \\ = 2 \frac{n (n + 1)}{2} = n (n + 1) \\ H e r e, (c) \leftrightarrow (i i) \\ (d) L e t S_{n} = 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2} \\ H e r e, (d) \leftrightarrow (i v) \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Kindly consider the following

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 Matching Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) 4, 1, \frac{1}{4}, \frac{1}{16} \\ H e r e, \frac{a_{2}}{a_{1}} = \frac{1}{4}, \frac{a_{3}}{a_{2}} = \frac{1 / 4}{1} = \frac{1}{4} a n d \frac{a_{4}}{a_{3}} = \frac{1 / 16}{1 / 4} = \frac{1}{4} \\ H e n c e, i t i s G . P \\ ∴ (a) \leftrightarrow (i i i) \\ (b) 2, 3, 5, 7 \\ H e r e, a_{2} - a_{1} = 3 - 2 = 1 \\ a_{3} - a_{2} = 5 - 3 = 2 \\ a_{2} - a_{1} \neq a_{3} - a_{2} \\ H e n c e i t i s n o t A . P \\ \frac{a_{2}}{a_{1}} = \frac{3}{2}, \frac{a_{3}}{a_{2}} = \frac{5}{3} \\ S o, \frac{3}{2} \neq \frac{5}{3} \\ S o, i t i s n o t G . P \\ H e n c e, i t i s s e q u e n c e \\ ∴ (b) \leftrightarrow (i i) \\ (c) 13, 8, 3, - 2, - 7 \\ H e r e, a_{2} - a_{1} = 8 - 13 = - 5 \\ a_{3} - a_{2} = 3 - 8 = - 5 \\ a_{2} - a_{1} = a_{3} - a_{2} \\ S o, i t i s a n A . P \\ ∴ (c) \leftrightarrow (i) \end{array}$

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Exemplar Solution Class 11th

How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

C₁ | C₂

(a)   4 letters are used at a time |                                                   (i) 720

(b)   All letters are used at a time | (ii) 240

(c)    All letters are used but the first is a vowel | (iii) 360

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} (a) 4 l e t t e r s a r e u s e d a t a t i m e = {}_{6}{P_{4}} = \frac{6!}{2!} = 360 \\ (b) A l l l e t t e r s a r e u s e d a t a t i m e = {}_{6}{P_{6}} = 6! = 720 \\ (c) A l l l e t t e r s a r e u s e d b u t f i r s t l e t t e r i s v o w e l = 2 \times 5! = 2 \times 120 = 240 \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i), (c) \leftrightarrow (i i) \end{array}$

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Find the angle between the lines y = (2 – 3)(x + 5) and y = (2 + 3)(x – 7).

If a circle passes through the points $(0, 0), (a, 0), (0, b)$ , then find the coordinates of its center.

Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, –1).

Show that the point $(x, y)$ given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of $t$ , such that $- 1 < t < 1$ , where $a$ is any given real number.

Find the equation of the straight line which passes through the point (1, –2) and cuts off equal intercepts from axes.

Find the equation of the circle which touches both axes in the first quadrant and whose radius is $a$ .

The third term of a G.P. is , the product of the first five terms is .

Kindly consider the following

Kindly consider the following