Tags Maths

Maths

Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths

Follow Ask Question

6.5k

Questions

Discussions

Active Users

Followers

New question posted

10 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Sol:

$\begin{array}{l} G i v e n t h a t \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \\ = \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \times \frac{\sqrt[]{x + h} + \sqrt[]{x}}{\sqrt[]{x + h} + \sqrt[]{x}} [R a t i o n a l i z i n g t h e] \\ = \underset{h \to 0}{l i m} \frac{x + h - x}{h [\sqrt[]{x + h} + \sqrt[]{x}]} = \underset{h \to 0}{l i m} \frac{h}{h [\sqrt[]{x + h} + \sqrt[]{x}]} \\ = \underset{h \to 0}{l i m} \frac{1}{\sqrt[]{x + h} + \sqrt[]{x}} \\ T a k i n g, w e h a v e \\ \frac{1}{\sqrt[]{x} + \sqrt[]{x}} = \frac{1}{2 \sqrt[]{x}} . \end{array}$

0 Follower 2 Views

New answer posted

10 months ago

Maths Exemplar Solution

Find the equation of a straight line on which the length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of the x-axis.

(Hint: Use normal form, where ω = 30°.)

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : \\ O M = 4 u n i t s \\ ∠ B A X = 12 0^{0} \\ ∴ ∠ B A O = 18 0^{0} - 12 0^{0} o r ∠ M O A + ∠ M A O = 9 0^{0} [? O M ⊥ A B] \\ θ + 6 0^{0} = 9 0^{0} \Rightarrow θ = 9 0^{0} - 6 0^{0} \\ ∴ θ = 3 0^{0} \\ S o, e q u a t i o n o f A B i n i t s n o r m a l f o r m \\ x c o s θ + y s i n θ = p \\ \Rightarrow x c o s 3 0^{0} + y s i n 3 0^{0} = 4 \\ \Rightarrow x * \frac{\sqrt[]{3}}{2} + y * \frac{1}{2} = 4 \Rightarrow \sqrt[]{3} x + y = 8 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \sqrt[]{3} x + y = 8 . \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

If the intercept of a line between the coordinate axes is divided by the point (–5, 4) in the ratio 1: 2, then find the equation of the line.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t a a n d b b e t h e o n t h e g i v e n l i n e . \\ ∴ C o o r d i n a t e s o f A a n d B a r e (a, 0) a n d (0, b) r e s p e c t i v e l y \\ ∴ - 5 = \frac{1 * 0 + 2 * a}{1 + 2} \\ \Rightarrow 2 a = - 15 \Rightarrow a = \frac{- 15}{2} [\begin{array}{l} ? X = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ a n d Y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{array}] \\ ∴ A = (\frac{- 15}{2}, 0) \\ a n d 4 = \frac{1 * b + 0 * 2}{1 + 2} \\ \Rightarrow 4 = \frac{b}{3} \Rightarrow b = 12 \\ ∴ B = (0, 12) \\ S o, t h e e q u a t i o n o f l i n e A B i s \\ y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ \Rightarrow y - 0 = \frac{12 - 0}{0 + \frac{15}{2}} (x + \frac{15}{2}) \\ \Rightarrow y = \frac{12 * 2}{15} (x + \frac{15}{2}) \\ \Rightarrow y = \frac{8}{5} (x + \frac{15}{2}) \\ \Rightarrow 5 y = 8 x + 60 \\ \Rightarrow 8 x - 5 y + 60 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 8 x - 5 y + 60 = 0 . \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n a r e a x + b y + 8 = 0 \dots (i) \\ a n d 2 x - 3 y + 6 = 0 \dots (i i) \\ F r o m e q n . (i) w e g e t, \\ a x + b y + 8 = 0 \Rightarrow a x + b y = - 8 \\ \Rightarrow \frac{a}{- 8} x + \frac{b}{- 8} y = 1 \Rightarrow \frac{x}{\frac{- 8}{a}} + \frac{y}{\frac{- 8}{b}} = 1 \\ S o, t h e o n t h e a x e s a r e \frac{- 8}{a} a n d \frac{- 8}{b} \\ F r o m e q n . (i i) w e g e t, \\ 2 x - 3 y + 6 = 0 \Rightarrow 2 x - 3 y = - 6 \\ \Rightarrow \frac{2 x}{- 6} - \frac{3 y}{- 6} = 1 \Rightarrow \frac{x}{- 3} + \frac{y}{2} = 1 \\ S o, t h e o n t h e a x e s a r e - 3 a n d 2 . \\ A c c o r d i n g t o t h e q u e s t i o n \\ \frac{- 8}{a} = + 3 \Rightarrow a = - \frac{8}{3} \\ a n d \frac{- 8}{b} = - 2 \Rightarrow b = + 4 \\ H e n c e, t h e r e q u i r e d v a l u e s o f a a n d b a r e - \frac{8}{3} a n d 4 . \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$(a) G i v e n$

$\begin{array}{l} t h a t : 2 x + y = 5 \dots (i) \\ x + 3 y + 8 = 0 \dots (i i) \\ 3 x + 4 y = 7 \dots (i i i) \\ E q u a t i o n o f a n y l i n e t h r o u g h t h e o f o f e q n . (i) a n d e q n . (i i) i s \\ (2 x + y - 5) + λ (x + 3 y + 8) = 0 \dots (i v) [λ =] \\ \Rightarrow 2 x + y - 5 + λ x + 3 λ y + 8 λ = 0 \\ \Rightarrow (2 + λ) x + (1 + 3 λ) y - 5 + 8 λ = 0 \\ S l o p e o f l i n e m_{1} (s a y) = - \frac{(2 + λ)}{1 + 3 λ} [? m = \frac{- a}{b}] \\ S l o p e o f l i n e 3 x + 4 y = 7 i s \\ m_{2} (s a y) = - \frac{3}{4} \\ I f e q n . (i i i) i s p a r a l l e l t o e q n . (i v) t h e n \\ m_{1} = m_{2} \\ ∴ - \frac{(2 + λ)}{1 + 3 λ} = - \frac{3}{4} \\ \Rightarrow \frac{2 + λ}{1 + 3 λ} = \frac{3}{4} \Rightarrow 8 + 4 λ = 3 + 9 λ \\ \Rightarrow 9 λ - 4 λ = 5 \Rightarrow 5 λ = 5 \Rightarrow λ = 1 \\ O n p u t t i n g t h e v a l u e o f λ i n e q n . (i v) w e g e t \\ (2 x + y - 5) + 1 (x + 3 y + 8) = 0 \\ \Rightarrow 2 x + y - 5 + x + 3 y + 8 = 0 \\ \Rightarrow 3 x + 4 y + 3 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 3 x + 4 y + 3 = 0 . \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

Find the equation of lines passing through (1, 2) and making an angle 30° with the y-axis.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t t h e l i n e m a k e s a n g l e 3 0^{0} w i t h y a x i s \\ ∴ A n g l e m a d e b y t h e l i n e w i t h x a x i s i s 6 0^{0} \\ ∴ S l o p e o f t h e l i n e \\ m = t a n 6 0^{0} \\ \Rightarrow m = \sqrt[]{3} \\ S o, t h e e q u a t i o n o f t h e l i n e t h r o u g h t h e (1, 2) a n d s l o p e \sqrt[]{3} i s \\ y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 2 = \sqrt[]{3} (x - 1) \\ \Rightarrow y - 2 = \sqrt[]{3} x - \sqrt[]{3} \\ \Rightarrow y - \sqrt[]{3} x - \sqrt[]{3} - 2 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e i s y - \sqrt[]{3} x - \sqrt[]{3} - 2 = 0 . \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

Show that the tangent of an angle between the lines x/a + y/b = 1 and x/a − y/b = 1 is 2ab / (a² − b²).

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

10 months ago

Maths Exemplar Solution

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e a n y l y i n g i n t h e e q u a t i o n x + y = 4 \\ ∴ x_{1} + y_{1} = 4 \dots (i) \\ o f t h e (x_{1}, y_{1}) f r o m t h e e q u a t i o n 4 x + 3 y = 10 \\ \frac{4 x_{1} + 3 y_{1} - 10}{\sqrt[]{{(4)}^{2} + {(3)}^{2}}} = 1 \\ | \frac{4 x_{1} + 3 y_{1} - 10}{5} | = 1 \\ 4 x_{1} + 3 y_{1} - 10 = \pm 5 \\ T a k i n g (+) s i g n 4 x_{1} + 3 y_{1} - 10 = 5 \\ \Rightarrow 4 x_{1} + 3 y_{1} = 15 \dots (i i) \\ F r o m e q n . (i) w e g e t y_{1} = 4 - x_{1} \\ P u t t i n g t h e v a l u e o f y_{1} i n e q n . (i i) w e g e t \\ 4 x_{1} + 3 (4 - x_{1}) = 15 \\ \Rightarrow 4 x_{1} + 12 - 3 x_{1} = 15 \\ \Rightarrow x_{1} + 12 = 15 \\ \Rightarrow x_{1} = 3 a n d y_{1} = 4 - 3 = 1 \\ S o, t h e r e q u i r e d i s (3, 1) \\ N o w t a k i n g (-) s i g n, 4 x_{1} + 3 y_{1} - 10 = - 5 \\ \Rightarrow 4 x_{1} + 3 y_{1} = 5 \dots (i i i) \\ F r o m e q n . (i) w e g e t y_{1} = 4 - x_{1} \\ 4 x_{1} + 3 (4 - x_{1}) = 5 \\ \Rightarrow 4 x_{1} + 12 - 3 x_{1} = 5 \\ \Rightarrow x_{1} + 12 = 5 \\ \Rightarrow x_{1} = - 7 a n d y_{1} = 4 - (- 7) = 11 \\ S o, t h e r e q u i r e d i s (- 7, 11) \\ H e n c e, t h e r e q u i r e d o n t h e g i v e n l i n e a r e (3, 1) a n d (- 7, 11) . \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e h a v i n g a a n d b o n t h e a x i s i s \\ \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ G i v e n t h a t a + b = 14 \Rightarrow b = 14 - a \\ \Rightarrow \frac{x}{a} + \frac{y}{14 - a} = 1 \dots (i i) \\ I f e q n . (i i) p a s s e s t h r o u g h t h e (3, 4) t h e n \\ \frac{3}{a} + \frac{4}{14 - a} = 1 \\ \Rightarrow \frac{3 (14 - a) + 4 a}{a (14 - a)} = 1 \\ \Rightarrow 42 + a = 14 a - a^{2} \\ \Rightarrow a^{2} + a - 14 a + 42 = 0 \\ \Rightarrow a^{2} - 13 a + 42 = 0 \\ \Rightarrow a^{2} - 7 a - 6 a + 42 = 0 \\ \Rightarrow a (a - 7) - 6 (a - 7) = 0 \\ \Rightarrow (a - 6) (a - 7) = 0 \Rightarrow a = 6, 7 \\ ∴ b = 14 - 6 = 8, b = 14 - 7 = 7 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e s a r e \\ \frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4 x + 3 y = 24 \\ a n d \frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7 \end{array}$

0 0

New answer posted

10 months ago

Maths Exemplar Solution

Find the equation of the circle which touches the $x$ -axis and whose center is $(1, 2)$ .

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} S i n c e t h e c i r c l e w h o s e c e n t r e i s (1, 2) t o u c h x - a x i s \\ ∴ r = 2 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ \Rightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = {(2)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 4 \\ \Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 x - 4 y + 1 = 0 . \end{array}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
691k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Maths

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience