Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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New question posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

13 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} N o w p u t t i n g t h e v a l u e o f K i n e q . (i) w e g e t \\ 6 x^{2} + 4 π r^{2} = x^{2} (π + 6) \\ \Rightarrow 6 x^{2} + 4 π r^{2} = π x^{2} + 6 x^{2} \Rightarrow 4 π r^{2} = π x^{2} \Rightarrow 4 r^{2} = x^{2} \\ ∴ 2 r = x \\ ∴ x : 2 r = 1 : 1 \\ N o w d i f f e r e n t i a t i n g e q . (i i) w . r . t ., x, w e h a v e \\ \frac{d^{2} V}{d x^{2}} = 6 x - \frac{3}{} \frac{d}{d x} [x {(K - 6 x^{2})}^{1 / 2}] \\ = 6 x - \frac{3}{} [x . \frac{1}{2} \times (- 12 x) + {(K - 6 x^{2})}^{1 / 2} . 1] \\ = 6 x - \frac{3}{} [\frac{- 6 x^{2}}{} +] \\ = 6 x - \frac{3}{} [\frac{- 6 x^{2} + K - 6 x^{2}}{}] = 6 x + \frac{3}{} [\frac{12 x^{2} - K}{}] \\ = 6 + \frac{3}{} [\frac{\frac{12 K}{π + 6} - K}{}] [P u t x =] \\ = 6 + \frac{3}{} [\frac{12 K - π K - 6 K}{}] = 6 + \frac{3}{} [\frac{6 K - π K}{}] \\ = 6 + \frac{3}{π} [(6 K - π K)] > 0 s o i t i s minima . \\ H e n c e, t h e r e q u i r e d r a t i o i s 1 : 1 w h e n t h e c o m b i n e d v o l u m e i s minimum . \end{array}$

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New question posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

12 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides. Also, find the maximum volume.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

An open box with a square base is to be made of a given quantity of cardboard of area c². Show that the maximum volume of the box is $\frac{c ³}{6 \sqrt 3}$ cubic units.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

If the straight-line x cosα + y sinα = p touches the curve, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then prove that a² cos²α + b² sin²α =p².

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription, and it is believed that for every increase of Re 1/-, one subscriber will discontinue the service. Find what increase will bring maximum profit.

27 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t u s c o n s i d e r t h a t t h e c o m p a n y i n c r e a s e s t h e a n n u a l s u b s c r i p t i o n b y R s . x \\ S o, x i s t h e n u m b e r o f s u b s c r i b e r s w h o d i s c o n t i n u e t h e s e r v i c e s . \\ ∴ T o t a l r e v e n u e, R (x) = (500 - x) (300 + x) \\ = 150000 + 500 x - 300 x - x^{2} \\ = - x^{2} + 200 x + 150000 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t R^{'} (x) = - 2 x + 200 \\ F o r l o c a l maxima a n d l o c a l minima, R^{'} (x) = 0 \\ - 2 x + 200 = 0 \Rightarrow x = 100 \\ R^{''} (x) = - 2 < 0 M a x i m a \\ S o, R (x) i s maximum a t x = 100 \\ H e n c e, i n o r d e r t o g e t maximum profit, the company should increase its a n n u a l s u b s c r i p t i o n \\ b y R s . 100 . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

Find the points of local maxima, local minima, and the points of inflection of the function f(x) = $x^{5} - 5 x ? + 5 x ³ - 1$ Also, find the corresponding local maximum and local minimum values.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{5} - 5 x^{4} + 5 x^{3} - 1 \\ f^{'} (x) = 5 x^{4} - 20 x^{3} + 15 x^{2} \\ F o r l o c a l maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 5 x^{4} - 20 x^{3} + 15 x^{2} = 0 \Rightarrow 5 x^{2} (x^{2} - 4 x + 3) = 0 \\ \Rightarrow 5 x^{2} (x^{2} - 3 x - x + 3) = 0 \Rightarrow x^{2} (x - 3) (x - 1) = 0 \\ ∴ x = 0, x = 1 a n d x = 3 \\ N o w f^{''} (x) = 20 x^{3} - 60 x^{2} + 30 x \\ \Rightarrow f^{''} {(x)}_{a t x = 0} = 20 {(0)}^{3} - 60 {(0)}^{2} + 30 (0) = 0 w h i c h i s n e i t h e r M a x i m a n o r M i n i m a . \\ ∴ f (x) h a s t h e point o f inflextion a t x = 0 \\ \Rightarrow f^{''} {(x)}_{a t x = 1} = 20 {(1)}^{3} - 60 {(1)}^{2} + 30 (1) \\ = 20 - 60 + 30 = - 10 < 0 M a x i m a \\ \Rightarrow f^{''} {(x)}_{a t x = 3} = 20 {(3)}^{3} - 60 {(3)}^{2} + 30 (3) \\ = 540 - 540 + 90 = 90 > 0 M i n i m a \\ T h e m a x i m u m v a l u e o f t h e f u n c t i o n a t x = 1 \\ f (x) = {(1)}^{5} - 5 {(1)}^{4} + 5 {(1)}^{3} - 1 \\ = 1 - 5 + 5 - 1 = 0 \\ T h e minimum v a l u e o f t h e f u n c t i o n a t x = 3 \\ f (x) = {(3)}^{5} - 5 {(3)}^{4} + 5 {(3)}^{3} - 1 \\ &thins \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Application of Derivatives

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{π}{3}$ .

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

Upvote

Ncert Solutions Maths class 12th

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AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.

<p><strong> AB is a diameter of a circle and C is any point on the circle. Show that the area of ?ABC is maximum when it is isosceles.</strong></p>

If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

<p><strong>If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?</strong></p>

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides. Also, find the maximum volume.

An open box with a square base is to be made of a given quantity of cardboard of area c². Show that the maximum volume of the box is $\frac{c ³}{6 \sqrt 3}$ cubic units.

If the straight-line x cosα + y sinα = p touches the curve, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then prove that a² cos²α + b² sin²α =p².

Find the points of local maxima, local minima, and the points of inflection of the function f(x) = $x^{5} - 5 x ? + 5 x ³ - 1$ Also, find the corresponding local maximum and local minimum values.

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{π}{3}$ .