Ncert Solutions Maths class 12th

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

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)

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

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

$=?25.3?25$

$=?25.3?5$

$?25.3=5+?y\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}?y~dy.$

= 5 + dy

= $5+\left(\frac{dy}{dx}\right)Ax.$

$=5+\frac{1}{\mathrm{2?x}}?x.$

$=5+\frac{1}{\mathrm{2?25}}?0.3$

$=5+\frac{0.3}{2\times 5}$

= 5 + 0.03

$?25.3=5.030$

(ii) ?49.5$$

A.(ii)

Let y = ?x$.$ Let x = 49 and x = 0.5.

Then, $?y=?x+?x?x.$

$=?49.5?49$

$?49.5=7+?y=7+{\left(\frac{dy}{dx}\right)}_{}^{?x}$

$=7+\frac{1}{\mathrm{2?x}}?x$

$=7+\frac{1}{\mathrm{2?49}}\times (0.5)$

$=7+\frac{0.5}{14}$

= 7 + 0.0357.

$?49.5=7?035$

(iii) ?0.6$$

A.(iii)

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

Then, $?y=?x+Ax?x$

$=?0.64-0.04?0.64.$

$=?0.6?0.8$

$?0.6=0.8+?y=0.8+\left(\frac{dy}{dx}\right)?x$

$=0.8+\frac{1}{\mathrm{2?x}}?x$

$=0.8+\frac{1}{\mathrm{2?0.64}}\times (?\text{?}0.04)$

$=0.8\frac{?0.04}{2\times 0.8}$

= 0.8 - 0.025.

= 0.775.

(iv) ${(0.009)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

A.(iv)

Let $y=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ Let x = 0.0

The given eqn of curve is $y^2=4x$ .

Then, differentiating wrt x we get,

$2y\frac{dy}{dx}=4$

$\Rightarrow \frac{dy}{dx}=\frac{2}{y}$ which is the slope of the tangent to the curve.

The line $y=x+1$ compared to $y=mx+c$ gives slope of line = 1.

Since, tangent is the line we have,

$\frac{2}{y}=1.$

$\Rightarrow y=2$

Putting y = 2 in $y^2=4x$ we get,  $4x=2^2\Rightarrow x=1.$

Hence, the required point is

Given,  $y=2x^2+3sinx$

Slope of tangent,  $\frac{dy}{dx}=4x+3cosx$

${\frac{dy}{dx}|}_{x=0}=4\times 0+3\times cos0=3$

So, slope of normal $=\frac{-1}{3}$

Option (D) is correct.

The given eqⁿ of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ______(1)

Differentiating eqn (1) wrt ‘x’ we get,

$\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$

$\Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{2x}{a^2}$

$\Rightarrow \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$

${\frac{dy}{dx}|}_{\left(x,y\right)}=\left(x_0,y_0\right)\frac{b^2{x}_0}{a^2{y}_0}$ is the reqd slope of tangent to the curve

So, eqⁿ of tangent at point $\left(x_0,y_0\right)$ is

$y-y_0=\frac{b^2{x}_0}{a^2{y}_0}\left(x-x_0\right).$

$\Rightarrow \frac{y{y}_0}{b^2}-\frac{{y_0}^2}{b^2}=\frac{x{x}_0}{a^2}-\frac{{x_0}^2}{a^2}$

$\Rightarrow \frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=\frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}$

As $\left(x_0,y_0\right)$ lies on the parabola given by eqn (1) we write,

$\frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}=1$

Hence, $\frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1$