Ncert Solutions Maths class 12th

The volume v of a cube with side 'x' metre is v = x³

So,  $dv=\left(\frac{dv}{dx}\right)\Delta x=3x^2\Delta x.$

∴increase in side, Δx = 3% of = $\frac{3x}{100}.$

∴dv = 3x²π $\frac{3x}{100}-0.09x^{3}m{}^3.$

Hence, option (C) is correct.

We have, y = f (x) = 3x² + 15x + 5.

$\frac{dy}{dx}=6x+15$

$\Rightarrow$ dy = (6x + 15) dx

$\Rightarrow$ Δy = (6x + 15) Δx.

Let, x = 3 and Δx = 0.02 then,

Δy = f (x + Δx) - f (x)

$\Rightarrow$ f (x + Δx) = f (x) + Δy = f (x) + (6x + 5) Δx.

$\Rightarrow$ f (3 + 0.02) = 3 (3)² + 15 (3) + 5 + (6 * 3 + 15) (0.02).

$\Rightarrow$ f (3.02) = 27 + 45 + 5 + (18 + 15) (0.02).

= 77 + 0.66

= 77.66

∴ Option (D) is correct.

Let be the radius of the sphere &r be the error in measuring the radius.

Then, π = 9m and Δr = 0.03m.

Now, surface area S of the sphere is

S = 4πr²

So,  $\frac{ds}{dr}=8\pi r.$

∴e, this = $\left(\frac{ds}{dr}\right)$ Δr = 8πr.Δr = 8π * 9 * 0.03

= 2.16πm³.

Appropriate error in calculating the surface area is 2.16πm³.

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

$V=\frac{4}{3}\pi r^3.$

So,  $\frac{\mathrm{dU}}{dx}=4\pi r^2$

$\Rightarrow dv=\left(\frac{dv}{dr}\right)\cdot \Delta r=4\pi r^2(\Delta \pi )$

$\Rightarrow$ dv = 4π (7)² (.0.02) = 3.92 πm3

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

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

So,  $dS=\frac{dS}{dx}\Delta x=12x\cdot \Delta x.$

Given decrease in side,  $\Delta x=-1\mathrm{\%}x=-\frac{x}{100}.$

$\therefore dS=12x\left(-\frac{x}{100}\right)=-0.12x^2 m^2.$

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

So,  $d=\left(\frac{d}{dx}\right)\cdot \Delta x=3x^2\cdot \Delta x.$

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

$\Rightarrow \Delta x=\frac{x}{100}$

$\therefore dv=3x^2\left(\frac{x}{100}\right)=0.03x^{3}m{}^3.$

Given, y = f (x) = x³- 7x² + 15.

So,  $\frac{dy}{dx}=f^{\prime }(x)=3x^2-14x.$

$\Rightarrow$ dy = (3x²- 14x) dx.

$\Rightarrow$ Δy = (3x²- 14x) Δx.

Let, x = 5 and Δx = 0.001. Then,

Δy = f (x + Δx) f (x).

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

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

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

= -35 + 0.005 = - 34.995.

Given, y = f (x) = 4x² + 5x + 2.

So, f (x) = 8x + 5. $\frac{dy}{dx}$ = 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.