Ncert Solutions Maths class 12th

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New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Let x be the radius of the sphere & Δπ be the error in measuring the radius.

Then, π = 7m and Δr = 0.02m.

Now, volume v of sphere is

V=43πr3.

So,  dUdx=4πr2

dv= (dvdr)Δr=4πr2 (Δπ)

dv = 4π (7)2 (.0.02) = 3.92 πm3

∴The appropriate error is calculating the volume is 3.92πm3.

New answer posted

10 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

We know that, the surface area 5 of a 'x' when length cube, is S = 6x2.

So,  dS=dSdxΔx=12xΔx.

Given decrease in side,  Δx=1%x=x100.

dS=12x (x100)=0.12x2 m2.

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

We know that, the volume v of side 'a' mete of cube is v = x3.

So,  d= (ddx)Δx=3x2Δx.

Given that, increase in side = 1% of x.

Δx=x100

dv=3x2 (x100)=0.03x3 m3.

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Given, y = f (x) = x3- 7x2 + 15.

So,  dydx=f (x)=3x214x.

dy = (3x2- 14x) dx.

Δy = (3x2- 14x) Δx.

Let, x = 5 and Δx = 0.001. Then,

Δy = f (x + Δx) f (x).

f (x + Δx) = f (x) + Δy = f (x) + (3x2- 4x) Δx.

f (5 + 0.001) = 53- 7 (5)2 + 15 + [3 (5)2 - 14 (5)]. (0.001).

f (5.001) = 125 - 175 + 15 + (75 - 70) (0.001)

= -35 + 0.005 = - 34.995.

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Given, y = f (x) = 4x2 + 5x + 2.

So, f (x) = 8x + 5. dydx = 8x + 5 dy = (8x + 5) dx.

Let x = 2 and Δx = 0.01.Then,

f (x + Δx) = f (2 + 0.01) = f (2.01).

Δy = f. (x + Δx) f (Δx).

f (x +Δx) = f (x) +Δy.

= f (x) + dy = f (x) + (8x + 5) dx.

= f (2.01) = f (2) + (8 x 2 + 5). Δx {∴dx = Δx}

= 4 (2)2 + 5 (2) + 2 + 21 (0.01)

= 16 + 10 + 2 + 0.21 = 28.21.

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

(i) Let y= ?x : Let x = 25 and x = 0. 3.

Then, ?y = ?x+?x?x

=?25.3?25

=?25.3?5

?25.3=5+?y?????????????????y~dy.

= 5 + dy

5+(dydx)Ax.

=5+12?x?x.

=5+12?25?0.3

=5+0.32*5

= 5 + 0.03

?25.3=5.030

(ii) ?49.5

A.(ii)

Let y = ?x. Let x = 49 and x = 0.5.

Then, ?y=?x+?x?x.

=?49.5?49

?49.5=7+?y=7+(dydx)?x

=7+12?x?x

=7+12?49*(0.5)

=7+0.514

= 7 + 0.0357.

?49.5=7?035

(iii) ?0.6

A.(iii)

Let y = ?x. Let x = 0.64 and ?x = 0.04.

Then, ?y=?x+Ax?x

=?0.64-0.04?0.64.

=?0.6?0.8

?0.6=0.8+?y=0.8+(dydx)?x

=0.8+12?x?x

=0.8+12?0.64*(??0.04)

=0.8?0.042*0.8

= 0.8 - 0.025.

= 0.775.

(iv) (0.009)13

A.(iv)

Let y=x13 Let x = 0.008 and ?x = 0.00 1.

Then, ?y = [x+?x]13?[x]13

=(0.009)13?(0.008)13

(0.009)13=0.2+?y=0.2+dydx??x

=0.2+13(x13)2?x

=0.2+13[(0.008)13]2*(0.001)

=0.2+0.0013*(0.0)2=0.2+0.0010.12

= 0.2 + 0.0083.

= 0.208.

(v) (0.999)110

A.(v)

Let y=x110. Let x = 1 and ?x = -0.001

Then, ?y=(x+?x)110?x110

=(0.999)110?(1)110

?1(0.999)110=1+Ay

=1+dydx?x.

=1+110x910*?x

=1?0.00110(.1910)=1?0.00110=1?0.0001

= 0.999.

(vi) (15)14

A.(vi)

Let y=x14. Then, x = 16 and ?x = 1.

Then, ?y=(x+Ax)14?x14.

=(15)14?(16)14

?(15)14=2+Ay=2+dydxAx

=2+14x34?(?1)

=4?14(16)34

=2?14*8

=4?132

=64?132=6332=1.968

(vii) (26)13

A.(vii)

Let y=x13. Let x = 27 and ?x = 1.

Then, ?y=(x+?x)13?x13

=(26)13?(27)13

?(26)13=3+4y=3+dydxAx=3+13x23Ax,

=3?13(2F)23

=3?127

=81?127=8027=2.962.

(viii) (255)14

A.(viii)

Let y=x14. Let x = 256 and ?x = 1.

Then, ?y=(x+?x)14?x14.

=(255)14?(256)14

?(255)14=(256)14+4y=4+dydxAx.

=4+14x34?4x

=4?14(256)34

=4?1256

=1024?1256

=1023256=3.996

(ix) (82)14

A.(ix)

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