Ncert Solutions Maths class 12th

Side of the tin square piece = 18 cm

Let x cm be the thought of the square to be cut from each corner.

The volume v of the box after cutting

$\Rightarrow$ v = length × breadth × height

= (18 - 2x) (18 - 2x)  x x

= (18 - 2x)²x

= (324 + 4x²- 72x) x

= 4x³- 72x² + 324x

So,  $\frac{d}{dx}=12x^2-144x+324$

$\frac{d^{2}v}{d{x}^2}=24x-144$

As $\frac{d}{dx}=0\to 12x^2-144x+324=0$

$\Rightarrow$ x²- 12x + 27 = 0

$\Rightarrow$x²- 9x- 3x + 27

Let, x and y be the two positive number

Then, x + y = 16 y = 16 - x

Let p be the sun of the cubes then

p = x³ + y³ = x³ + (16 -x)³ = x³ + (16)³-x³- 48x (16 -x)

p = 16³ + 48x²- 76 8x

So,  $\frac{dp}{dx}=96x-768.$

$\Rightarrow \frac{d^{2}p}{d{x}^2}=96.$

At $\frac{dp}{dx}=0$

96x - 768 = 0

$\Rightarrow x=\frac{768}{96}=8.$

∴at x = 8,  $\frac{d^{2}p}{d{x}^2}=96\cdot >0$

So, x = 8 is a point of local minima

So, y = 16 - 8 = 8

Hence, x = 8, y = 8

We have, x + y = 35.

y = 35 - x

Let the product, P =x² y⁵

$\Rightarrow$ P = x² (35 -x)⁵

So,  $\frac{dp}{dx}$ = x² 5 (35 -x)⁴ (1) + (35 -x)⁵ 2x

= x (35 -x)⁴ [ - 5x + (35 -x) 2]

= x (35 -x)⁴ [ - 5x + 70 - 2x]

= x (35 -x)⁴ (70 - 7x)

= 7x (35 -x)⁴ (10 -x)

At $\frac{dp}{dx}=0$

$\Rightarrow$7x (35 -x)⁴ (10 -x) = 0

$\Rightarrow$ x = 0, 35, 10

As x is a (+) ve number we

We have, x + y = 60.where x, y > 0

x = 60 - y.

Let the product P = xy³ = (60 -y) y³ = 60y³-y⁴

$\Rightarrow \frac{dp}{dy}=180y^2-4y^3$

= 4y² (45 -y)

At $\frac{dp}{dy}=0$

$\Rightarrow$ 4y² (45 -y) = 0

$\Rightarrow$ y = 0 and y = 45

As y > 0, y = 45

When, y > 45,  $\frac{dp}{dy}=(+ve)(-ve)$

= ve < 0

Ad y < 45,  $\frac{dp}{dy}=(+ve)(+ve)$

= (+ ve) > 0

∴p is maximum when y = 45 from + ve to- ve or y increases through 45.

So, x = 60 - y = 6Ø - 4

Let ‘x’ and ‘y’ be the two number

Then, x + y = 24 $\Rightarrow$ y = 24 - x

Let ‘P’ be their product then,

P = xy = x (24 - x) = 24x -x²

P (x) = 24x -x²

$\Rightarrow \frac{dp}{dx}=24-2x.$

$\Rightarrow \frac{d^{2}p}{d{x}^2}=-2.$

At $\frac{dp}{dx}=0$

24 - 2x = 0

$\Rightarrow x=\frac{24}{2}=12.$

So, P (12) $\frac{d^{2}p}{d{x}^2}=-2<0.$

x = 12 is a point of local maxima

Hence, y = 24 - 12 = 12.

The uqdtwno (x, y) is (12, 12).

We have, f(x) = x + sin 2x ,x ∈ [0, 2π].

f(x) = 1 + 2cos 2x

At f(x) = 0

1 + 2 cos2x = 0

$\Rightarrow cos2x=-\frac{1}{2}=-cos\frac{\pi }{3}=cos\pi -\frac{\pi }{3}=cos\frac{2\pi }{3}$

$\therefore 2x=2n\pi \pm \frac{2\pi }{3},n=1,2,3.$

$\Rightarrow x=n\pi \pm \frac{2\pi }{3}$

$n=0,x=\pm \frac{\pi }{3}\to x=\frac{\pi }{3}\in \left[0,2\pi \right].$

$n=1,x=\pi \pm \frac{\pi }{3}\to x=\pi +\frac{\pi }{3}and\pi -\frac{\pi }{3}$

$=\frac{4\pi }{3}ad\frac{2\pi }{3}\in [0,2\pi ]$

$n=2,x=2\pi \pm \frac{\pi }{3}\to x=2\pi +\frac{\pi }{3}ad2\pi -\frac{\pi }{3}.$

$\Rightarrow x=\frac{51}{3}\in [0,2\pi ].$

Hence, $x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3}and\frac{5\pi }{3}$

Missing

At $x=\frac{\pi }{3},f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+sin\frac{2\pi }{3}=1.05+$ $sin\left(\pi -\frac{\pi }{3}\right)=1.05+sin\frac{\pi }{3}$

$=1.05+\surd 3/2$

= 1.05 + 0.87

= 1.92

At $x=\frac{2\pi }{3},f\left(\frac{2\pi }{3}\right)=\frac{2\pi }{3}+sin2\times \frac{2\pi }{3}$ $=2.10+sin\left(\pi +\frac{\pi }{3}\right)$

$=2.10-sin\frac{\pi }{3}=2,10-0.87$

= 1.23

At $x=\frac{4\pi }{3},f\left(\frac{4\pi }{3}\right)=\frac{4\pi }{3}+sin2\times \frac{4\pi }{3}=4\cdot 2+sin\left(3x-\frac{\pi }{3}\right)$

$=4.2+sin\frac{\pi }{3}=4.2+0.87.$

=5.07.

At $x=\frac{5\pi }{3},f\left(\frac{5\pi }{3}\right)=\frac{5\pi }{3}+sin2\times \frac{5\pi }{3}=5.25+sin\left(3x+\frac{\pi }{3}\right)$

$=5\cdot 25-sin\frac{1}{3}$

= 5.25 - 0.87 = 4.38

At and points,

f(0) = 0 + sin2 × 0 = 0

f(2π) = 2π + sin 2 × 2π = 6.2 + 0 = 6.28

∴Maximum value of f(x) =

We have, f (x) = x⁴- 62x² + ax + 9, x∈  [0, 2].

$\Rightarrow$ f (x) = 4x³- 124x + a

∴f (x) active its maxⁿ value at x = 1∈ [0, 2]

∴f (1) = 0.

4 (1)³- 124 (1) + a = 0

$\Rightarrow$ a = 124 - 4 = 120.

∴a = 120

We have, f (x) =2x³- 24x + 107, x [1,3]

f (x) = 6x²- 24.

At f (x) = 0

6x²- 24= 0

$\Rightarrow x^2=\frac{24}{6}=4$

x = ±2. ->x = 2 ∈ [1, 3].

So, f (2) = 2 (2)³- 24 (2) + 107 = 16 - 48 + 107 = 75.

f (1) = 2 (1)³- 24 (1) + 107 = 2 - 24 + 107 = 85.

f (3) = 2 (3)³- 24 (3) + 107 = 54 - 72 + 107 = 89

∴ Maximum value of f (x) in interval [1, 3]

We have, f (x) = sin x + cos x.

f (x) = cos x - sin x.

At f (x) = 0

cosx - sin x = 0

sinx = cos x

$\Rightarrow \frac{sinx}{cosx}=1.$

$\Rightarrow tanx=1=tan\frac{\pi }{4}$

$\Rightarrow x=\frac{\pi }{4}orx=n\pi +\frac{\pi }{4}.$

At $x=\frac{\pi }{4}+n\pi$ ,

$f\left(n\pi +\frac{\pi }{4}\right)=sin\left(n\pi +\frac{\pi }{4}\right)+cos\left(nx+\frac{\pi }{4}\right).$

$={(-1)}^{x}sin\frac{\pi }{4}+{(-1)}^{n}sin\frac{\pi }{4}.$

We have, f(x) = sin 2x, x ∈ [0, 2π],

f(x) = 2cos 2x.

At f(x) = 0.

2 cos 2x = 0

cos 2x = 0

$\Rightarrow 2x=(2x+1)\frac{\pi }{2},x=0,1,2,3.$

$\Rightarrow x=(2x+1)\frac{\pi }{4}.$

$x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\in \left[0,2\pi \right]$

$\therefore f\left(\frac{\pi }{4}\right)=sin\frac{2\pi }{4}=sin\frac{\pi }{2}=1$ .

$f\left(\frac{3\pi }{4}\right)=sin2\left(\frac{3\pi }{4}\right)=$ $sin\frac{3\pi }{2}$ $f\left(\frac{7\pi }{4}\right)$

$=sin\left(\pi +\frac{\pi }{2}\right)=$ $=sin2\times \frac{7\pi }{4}$

$=-sin\frac{\pi }{2}$ $=sin\frac{7\pi }{2}$

= 1. $=sin3\pi +\frac{\pi }{2}$

$f\left(\frac{5\pi }{4}\right)=sin2\times \left(\frac{5\pi }{4}\right)=sin\frac{5\pi }{2}=sin\left(2x+\frac{\pi }{4}\right)$ $=-sin\frac{\pi }{2}$

$=sin\frac{\pi }{4}=1$ = 1.

f(0) = sin 2(0) = sin 0 = 0

f(2π) = sin 2(2π) = sin 4π = 0

Hence, the points of maximum $xf\mathrm{x\; are.}$

$(\frac{\pi }{4},2)and\left(\frac{5\pi }{4},1\right).$