Ncert Solutions Maths class 12th

We have, f (x) = 3x⁴- 8x³ + 12x²- 48x + 25, x ∈  [0, 3].

f (x) = 12x³- 24x² + 24x - 48.

At f (x) = 0.

12x³- 24x² + 24x - 48 = 0.

x³- 2x² + 2x - 4 = 0

x² (x - 2) + 2 (x - 2) = 0

(x - 2) + (x² + 2) = 0

x = 2 ∈ [0, 3] or x = $\pm \surd -2$ which is not possible as

∴f (x) = 3 (2)⁴- 8 (2)³ + 12 (2)²- 4 (2) + 25.

=

We have, p (x) = 41 -f2x - 18x².

P (x) = - 72 - 36x

P (x) = -36

At extreme point,

- 72 - 36x = 0

$\Rightarrow$ $x=\frac{-72}{36}=-2$ .

At x = - 2, p" (x) = - 36 < 0.

∴x = -2 is a point of local maximum and the value of local

Maximum is given by P (2) = 41 - 72 (- 2) - 18 (- 2)²

41 + 144 - 72 = 113 units.

(i) We have,

f(x) = x³ , x ∈ [– 2, 2].

f(x) = 3x².

At, f(x) = 0

3x² = 0

x = 0 <--[-2, 2].

We shall absolute the value of f at x = 0 and points of interval [ -2, 2]. So,

f(0) = 0

f(- 2) = (- 2)³ = 8

f(2) = 2³ = 8.

∴ Absolute maximum value of f(x) = 8 at x = 2

and absolute minimum value of f(x) = -8 at x = -2.

(ii) f (x) =

(i) We have, f (x) = e^x

$\Rightarrow$ f (x) = e^x.

$\Rightarrow f^{\prime }(x)=e^{x.}$

At, extreme points,

f (x) = 0

$\Rightarrow$ e^x = 0 which has no real ‘a’ value

∴f (x) has with maximum or minima

(ii) g (x) = log x

A (ii)

We have, g (x) = log x,

$\Rightarrow$ g (x) = $\frac{1}{x}$

At extreme points,

g (x) = 0

$\to \frac{1}{x}=0.$

$\Rightarrow$ 1 = 0 which is not true.

∴g (x) was value minima or maxima

(iii) h

(i) we have, f(x) = x².

f(x) = 2x.

andf(x) = 2.

AR extreme point, f(x) = 0

2x = 0

x = 0.

When x = 0, f(0) = 2 > 0.

∴x = 0 is a point of local minima and value of local minimum is given by f(0) = 0² = 0.

(ii) g(x) = x³ 3x

A(ii)

we have, g(x) = x³- 3x

$\Rightarrow$ g'(x) = 3x²- 3

$\Rightarrow$ g''(x) = 6x.

At extreme point,

g'(x) = 0

3x²- 3

(i) we have, f(x) = |x + 2| - 1

We know that, for all $x\in ?,|x+2|\ge 0$

$\Rightarrow |x+2|-1\ge -1.$

$\Rightarrow$ f(x)$\ge$- 1.

∴ Minimum value of f(x) = -1 when x + 2 = 0 $\Rightarrow$ x = - 2.

And maximum value of f(x) does not exist.

(ii) $g(x)=-|x+1|+3$

A(ii)

We have, $g(x)=-|x+1|+3$

For all $x\in ,|x+1|\ge 0.$

$\Rightarrow -|x+1|\le 0$

$\Rightarrow -|x+1|+3\le 0+3$

$\Rightarrow$ g(x) $\le \mathrm{}$3.

∴ Maximum value of g(x) = 3 when $|x+1|=0\Rightarrow x=-1.$

And minimum value does not exist.

(iii) h(

(i) We have, f(x) = (2x - 1)² + 3.

For all $x\in ?,(2x-1){}^2\ge 0$

(2x - 1)² + 3 ≥ 3.

f(x) ≥ 3.

∴The minimum value of f(x) = 3. When 2x - 1 = 0--> x = $\frac{1}{2}$

Again as $x\to \infty ,f(x)\to \infty$ as there is vouppa bound to ‘x’ value hence, f(x) has no maximum values.

(ii) $f\left(x\right)=9x^2+12x+2$

A(ii)

We have, f(x) = 9² + 12x + 2.

$f(x)=9\left[x^2+\frac{12x}{9}+\frac{2}{9}\right]$ (Taking 9 common from each te

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

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.