Ncert Solutions Maths class 12th

(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) = sin x + cos x , x ∈ [0, π]

A.(ii)

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

f(x) = cos x - sin x.

atf(x) = 0

$\Rightarrow$ cosx - sin x = 0

$\Rightarrow$ sinx = cos x

$\Rightarrow \frac{si{n}^{°}x}{corx}=1tanx=1\Rightarrow tanx=tan\frac{\pi }{4}$

$\Rightarrow x=\frac{\pi }{4}\in \left[0,\pi \right]$

(iii) f(x) = 4x $-\frac{1}{2}{x}^2,x\in \left[-2,\frac{9}{2}\right]$

A.(iii)

We have, f(x) = 4x $-\frac{1}{2}{x}^2,x\in \left[-2,\frac{9}{2}\right]$

f(x) = 4 - x

atf(x) = 0

$\Rightarrow$ 4-  x = 0

$\Rightarrow$ x = 4 $\in$$\left[-2,\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right]$

$\therefore f(4)=4(4)-\frac{1}{2}{(4)}^2=16-8=8.$

$f(-2)=4(-2)-\frac{1}{2}{(-2)}^2=-8-2=-10.$

$f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}{\left(\frac{9}{2}\right)}^2=18-\frac{81}{8}=\frac{94-81}{8}=\frac{63}{8}.$

= 7.87.5

Hence, absolute maximum value of f(x) = 8 at x = 4

and absolute minimum value of f(x) = -10 at x = -2.

(iv) f(x) = (x - 1)² + 3, x∈[ -3, 1]

A.(iv)

We have, f(x) = (x - 1)² + 3, x∈[ -3, 1]

f(x) = 2(x -1)

atf(x) = 0

2(x - 1) = 0

x = 1∈ [ -3, 1]

∴f(1) = (1 - 1)² + 3 = 3

f(- 3) = (- 3 - 1)² + 3 = 16 + 3 = 1

∴ Absolute maximum value of f(x) = 19 at x = -3.

And absolute minimum value of f(x) = 3 at x = 1.

(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 = 0.

3(x²- 1) = 0 ⇒ 3(x - 1)(x + 1) = 0.

x = 1 or x = -1.

At x = 1, g"(1) = 6.1 = 6 > 0.

So, x = 1 is a point of local minima and value of local minimum is given by g(1) = 1³- 3.1 = 1 - 3 = - 2.

And at x = -1, g"( -1) = 6 ( -1) = 6 < 0.

So, x = -1 is a point of local minima and value of local minimum is given by

g(- 1) = (- 1)³- 3(- 1) = 1 + 3 = 2.

(iii) $h\left(x\right)=sinx+cosx.0<x<\frac{\pi }{2}$

A(iii)

we have, h(x) = sin x + cos x.

$\Rightarrow$ h'(x) = cos x - sin x

$\Rightarrow$ h'(x) = -sin x - cos x.

At extreme points,

h'(x) = 0

cosx - sin x = 0

cosx = sin x

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

$\Rightarrow$ 1 = tan x

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

$\Rightarrow x=\frac{\pi }{4}.$

(iv) $f\left(x\right)=sinx-cosx,0<x<2\pi$

A(iv)

we have, f(x) = sin x cos x.

$\Rightarrow$ f(x) = cos x + sin x

$\Rightarrow$ f(x) = sin x + cos x.

At extreme points,

f(x) = 0

$\Rightarrow$ cosx + sin x = 0.

$\Rightarrow \frac{sinx}{cosx}=\frac{-cosx}{cosx}$

$\Rightarrow tanx=-1=-tan\frac{\pi }{4}=tan\left(\pi -\frac{\pi }{4}\right)=tan\left(2\pi -\frac{\pi }{4}\right)$

$\therefore x=\left(\pi -\frac{\pi }{4}\right)$ IInd quadrate or $\left(2\pi -\frac{\pi }{4}\right)$ IIIth quadrate

$\Rightarrow x=\frac{4\pi -\pi }{4}\frac{8x-\pi }{4}$

$\Rightarrow x=\frac{3\pi }{4}orx=\frac{f\pi }{4}.$

$Atx=\frac{3\pi }{4},f^{\prime \prime }\left(\frac{3\pi }{4}\right)=-sin\frac{3\pi }{4}+cos\frac{3\pi }{4}$

$=-sin\left(\pi -\frac{\pi }{4}\right)+cos\left(\pi -\frac{\pi }{4}\right)$

$=-sin\frac{\pi }{4}-cot\frac{\pi }{4}$

(v) f(x) = x³- 6x² + 9x + 15.

A(v)

we have, f(x) = x³- 6x² + 9x + 15.

$\Rightarrow$ f(x) = 3x²- 12x + 9.

$\Rightarrow$ f'(x) = 6x - 12.

At extreme point, f'(x) = 0.

$\Rightarrow$ 3x²- 12x + 9 = 0.

$\Rightarrow$ x²- 4x + 3 = 0

$\Rightarrow$ x²-x - 3x + 3 = 0

$\Rightarrow$ x(x - 1) -3(x - 1) = 0

$\Rightarrow$ (x - 1)(x - 3) = 0

$\Rightarrow$ x = 1 or x = 3.

At, x = 1,f"(1) = 6 * 1 - 12 = 6 - 12 = - 6 < 0

∴x = 1 is a point of local maxima and the value of local

Maximum is given by f(1) = 1³- 6(1)² + 9(1) + 15

= 1 - 6 + 9 + 15

= 19.

And at x = 3,f"(3) = 6 * 3 - 12 = 18 - 12 = 6 > 0

∴x = 3 is a point of local minima and the value of

local minimum is given by f (3) = 3³- 6(3)² + 9(3) + 15.

= 27 - 54 + 27 + 15.

= 15.

(vi) $g(x)=\frac{x}{2}+\frac{2}{x},x>0$

A(vi)

We have, $g(x)=\frac{x}{2}+\frac{2}{x},x>0$

$\Rightarrow g^{\prime }(x)=\frac{1}{2}-\frac{2}{x^2}$

$\Rightarrow g{g}^{\prime z}(x)=\frac{x^2-4}{2x^2}=\frac{(x-2)(x+2)}{2x^2}$

$g^{\prime \prime }(x)=\frac{4}{x^3}.$

At, extreme point, g'(x) = 0

$\Rightarrow \frac{(x+2)(x-2)}{x^2}=0$

$\Rightarrow$ x = -2 or x = 2

Given, that x > 0, hence we have x = 2.

At, x = 2, ${\stackrel{?}{g}}^{\prime }(x)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}>0$

∴x = 2 is a point of local minima and value of local

Minimum is given by g(2) = $\frac{2}{2}+\frac{2}{2}=1+1=2.$

(vii) $f(x)=\frac{1}{x^2+2}$

A(vii)

Given, $f(x)=\frac{1}{x^2+2}$

$\Rightarrow g^{\prime }(x)=-\frac{1}{{\left(x^2+2\right)}^2}\frac{d}{dx}\left(x^2+2\right)=\frac{-2x}{{\left(x^2+2\right)}^2}.$

$\Rightarrow g^{\prime \prime }(x)=-\left[\frac{{\left(x^2+2\right)}^2\frac{d}{dx}2x-2x\frac{d}{dx}{\left(x^2+2\right)}^2}{{\left(x^2+2\right)}^4}\right]$

$=\frac{-\left[2{\left(x^2+2\right)}^2-2x\cdot 2\left(x^2+2\right)\cdot 2x\right]}{{\left(x^2+2\right)}^4}.$

$=\frac{-\left[2{\left(x^2+2\right)}^2\left[x^2+2-4x^2\right]\right]}{{\left(x^2+2\right)}^4}$

$=\frac{-2\left(x^2+2\right)\left(2-3x^2\right)}{{\left(x^2+2\right)}^4}=-\frac{2\left(2-3x^2\right)}{{\left(x^2+2\right)}^3}.$

At extreme points, g'(x) = 0.

$\Rightarrow \frac{-2x}{{\left(x^2+2\right)}^2}=0.$

$\Rightarrow$ x = 0.

At, x = 0, $g^{\prime \prime }(x)=\frac{-2\left(2-2*0^2\right)}{{\left(0^2+2\right)}^3}=\frac{-2(2-0)}{2^3}=-\frac{1}{2}<0$

∴x = 0 is a point of local maxima and value of local maximum is given by $g(0)=\frac{1}{0^2+2}=\frac{1}{2}.$

(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(x) = sin (2x) + 5.

A(iii)

we have, h(x) = sin (2x) + 5.

For all $x\in ?,-1\le sin2x\le 1.$ {range of sine function is [-1, 1]}

$\Rightarrow$ -1 + 5 $\le$sin 2x + 5 $\le$1 + 5.

$\Rightarrow$ 4 $\le$h(x) $\le$6.

∴ Maximum value of h(x) = 6.

Minimum value of h(x) = 4.

(iv) $f(x)=|sin4x+3|.$

A(iv)

we have, $f(x)=|sin4x+3|.$

As for all $x\in ?,-1\le sin4x\le 1$

$\Rightarrow$ -1 + 3 $\le$sin 4x + 3$\le$ 1 + 3

$\Rightarrow \left|2\right|\le |sin4x+3|\le \left|4\right|.$

$\Rightarrow$ 2 $\le$f(x) $\le$4.

∴ Maximum value of f(x) = 4.

Minimum value of f(x) = 2.

(v) h(x) = x + 1.,x∈ ( -1, 1).

A(v)

we have, h(x) = x + 1.,x∈ ( -1, 1).

Given, -1

- 1 + 1 <x + 1 < 1 + 1

0 <h (x) < 2.

∴ Maximum value and minimum value of h(x) does not exist.