Ncert Solutions Maths class 12th

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New answer posted

4 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

(a) f (x) = x2 + 2x - 5.

f(x) = 2x + 2 = 2 (x + 1).

At, f(x) = 0

2 (x + 1) = 0

x = -1.

At, x (,1),

f(x) = (- ) ve< 0.

So, f (x)is strictly decreasing or (,1).

At x ∈(1,)

f(x) = ( + ve) > 1

f(x) is strictly increasing on (1,).

(b) f(x) = 10 - 6x- 2x2

So, f(x) = - 6 - 4x = - 2 (3 + 2x).

Atf(x) = 0

 2 (3 + 2x) = 0.

 x = 32

At x (,32),

∴f(x) is strictly increasing on (,32)

At x (32,)

f(x) = ( -ve) ( + ve) = ( - ) ve< 0.

∴f (x) is strictly decreasing on (32,)

(c) f (x) = 2x3- 9x2- 12x.

So, f (x) =- 6x2- 18x- 12 = - 6 (x2 + 3x + 2).

= -6 [x2 + x + 2x + 2]

= -6 [x (x + 1) + 2 (x + 1)]

= -6 (x + 1) (x + 2)

At, f (x) = 0.

 6 (x + 1) (x + 2) = 0

x = -1 and x =

...more

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

We have, f (x) = 2x3- 3x2- 36x + 7.

So, f (x) = ddx (2x33x236x+7)=6x26x36.

= 6 (x2-x- 6).

= 6 (x2- 3x + 2x- 6)

= 6 [x (x- 3) + 2 (x- 3)]

= 6 (x- 3) (x + 2).

At, f (x) = 0

6 (x- 3) (x + 2) = 0.

So, when x- 3 = 0 or x + 2 = 0.

x = 3 or x = -2.

Hence we an divide the real line into three disjoint internal

(, 2) II (2, 3)andIII (3, )

At x ∈ (, 2),

f (x) = ( + ve) ( -ve) ( -ve) = ( + ve) > 0.

So, f (x) is strictly increasing in  (, 2)

At, x∈ ( -2,3),

f (x) = ( + ve) ( + ve) ( -ve) = ( -ve) < 0.

So, f (x) is strictly decreasing in ( -2,3).

At, x ∈ (3, )

f (x) = ( + ve) ( + ve) ( + ve) = ( + )ve> 0.

So, f (x) is strictly increasing in  (3, )

∴ (a) f (x) is strictly increasing in  (, 2)and (3, )

(b) f (x) is strictly decr

...more

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

We have, f (x) = 2x2 3x

So, f (x) = ddx (2x23x)=4x3.

Atf (x) = 0.

4x - 3 = 0

i e,  x=34 divides the real line into two

disjoint interval  (, 34) (34, )

(a) Now,

f (x) = 4x - 3 > 0 x (34, )

So, f (x) is strictly increasing in  (34, )

(b) Now, f (x) = 4x - 3 < 0 x (, 34)

So, f (x) is strictly decreasing in  (, 34)

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

We have, f (x) = sin x.

So, f (x) = cosx.

(a) when, x ∈ (0, π2) i e, x in 1st quadrat.

f (x) = cos.x> 0

f (x) is strictly increasing in  (0, π2) .

(b) when, x ∈ (π2, π) in IInd quadrat

f (x) = cosx< 0.

∴f (x) is strictly decreasing (π

(c) When, x ∈ (0, π).

f (x) = cosx is increasing in  (0, π2) and decreasing

in  (π2, π) and f  (π2) = cos π2=0.

∴f (x) is neither increasing not decreasing in (0, π).

New answer posted

4 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

We have, f (x) = e2x

So, f (x) = ddxe2x = e2xddx2x = 2e2x> 0 xR.

∴f (x) is strictly increasing on R.

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

We have, .

f (x) = 3x + 17.

So, f (x) = 3 > 0 xR

∴f (x) is strictly increasing on R.

New answer posted

4 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Given, R (x) = 3x2 + 36x + 5.

Marginal revenue,  ddxR (x)=ddx (3x2+36x+5)

= 3 * 2x + 36

= 6x + 36

When x = 15.

ddxR (x)=6*15+36 = 90 + 36 = 126

Option (D) is correct.

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

The area A of the isle with radius r is given by with respect to radius r A = πr2.

Then, rate of change of area of the circle d·Adx=dπr2dr

= 2πr.

When r = 6 cm

dAdt=2π*6=12π.

Q option (B) is correct.

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Given, R (x) = 13x2 + 26x + 15.

Marginal revenue is the rate of change of total revenue with respect to the number of units sold Marginal revenue (MR) = dR (x)dx

=dd (13x2+26x+15)

= 13 * 2x + 26

= 26x + 26

When x = 7,

MR = 26 * 7 + 26 = 182 + 26 = 208.

Hence, the required marginal reverse = ' 208.

Choose the correct answer for questions 17 and 18.

New answer posted

4 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Given, c (x) = 0.007 x3- 0.003x2 + 15x + 400.

Since the marginal cost is the rate of change of total cost wrt the output we have,

Marginal cost, MC, =dCdx (x)

= 0.007 * 3x2- 0.003 * 2x + 15.

When x = 17,

Then, MC = 0.007 * 2. (17)2 - 0.003 2 (17) + 15.

= 6.069 - 0.102 + 15.

= 20.967

Hence, the required marginal cost = ' 20, 97.

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