Maths

23. x² – x + 2 = 0

Comparing the given equation with ax² + bx + c = 0

We have, a = 1, b = –1 andc = 2

Hence, discriminant of the equation is

b² – 4ac = (-1)² – 4 * 1 * 2 = 1 – 8 = –7

Therefore, the solution of the quadratic equation is

21. –x² + x – 2 = 0

Comparing the given equation with ax² + bx + c = 0

We have, a = –1, b = 1 and c = –2

Hence, discriminant of the equation is

b² – 4ac = 1² – 4 * (-1)* (-2) = 1 – 8 = –7

Therefore, the solution of the quadratic equation is

20. x² + 3x + 9 = 0

Comparing the given equation with ax² + bx + c = 0

We have, a = 1, b = 3 and c = 9

Hence, discriminant of the equation is

b² – 4ac = 3² – 4 * 1* 9 = 9 – 36 = –27

Therefore, the solution of the quadratic equation is

18. x² + 3 = 0

=>x² = –3

=>x = $\pm \surd 3$

=>x= ± √3$i$ [since, √-1 = i]

=>x = 0 ± √3$i$

81. Given, yx = xy

Taking log,

x log y .log x

Differentiating w r t 'x' we get,

$\Rightarrow x\frac{d}{dx}logy+logy\frac{dx}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}$

$\Rightarrow \frac{x}{y}·\frac{dy}{dx}+logy=\frac{y}{x}+logx\frac{dy}{dx}$

$\Rightarrow logx\frac{dy}{dx}-\frac{x}{y}\frac{dy}{dx}=logy-\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}\left[\frac{ylogx-x}{y}\right]=\frac{xlogy-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y(xlogy-y)}{x(ylogx-x)}.$

80. Given, xy + yx = 1

Let 4 = xy and v =., we have,

u + v = 1.

$\Rightarrow \frac{dy}{dx}+\frac{dv}{dy}=0$ ___ (1)

So, u = xy

= log u = y log x(taking log)

Now, differentiating w r t 'x',

$\frac{1}{4}\frac{dy}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}.$

$\frac{dy}{dx}=4\left[\frac{y}{x}+logx\frac{dy}{dx}\right]$

$\Rightarrow x^y·\frac{y}{x}+x^{y}logx·\frac{dy}{dx}$

= xy- 1y + xy log x $\frac{dy}{dx}.$

And v = yx.

log v = x log y.

Differentiating w r t 'x',

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{d}{dx}logy+logy\frac{dx}{dx}$

$=\frac{x}{y}\frac{dy}{dx}+logy$

$\Rightarrow \frac{dv}{dx}=v\left[\frac{x}{y}\frac{dy}{dx}+logy\right]$

$=y^x·\frac{x}{y}·\frac{dy}{dx}+y^{x}log·y$

= yx- 1. $x\frac{dy}{dx}$ + yx log y.

So, eqn (1) becomes

xy- 1y + xy log x $\frac{dy}{dx}$ + yx - 1 $\frac{dy}{dx}$ + yx log y = 0

$\Rightarrow \frac{dy}{dx}\left(x^{y}logx+y^{x+1}·x\right)$ = - (xy- 1y + yx log y)

$\Rightarrow \frac{dy}{dx}=\frac{-\left(x^{y-1}·y+y^{x}logy\right)}{\left(x^{y}logx+y^{x-1}·x\right).}$

78. Let y = x^x cos x  $\frac{a^2+1}{x^2-1}$

Putting  4 = x^x cos x and v = $\frac{x^2+1}{x^2-1}$ we have,

y = u + v

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dx}+\frac{dv}{dx}$ ____ (1)

As u $\left|=\right|$ x^x cos x.

Taking log,

Log u = x cos x log x

Differentiating w r t 'x',

$\frac{1}{u}\frac{dy}{dx}=x\frac{d}{dx}$ [cos x log x] + cos x log x $\frac{dx}{dx}$

= x $\left\{cotx\frac{d}{dx}logx+logx\frac{d}{dx}cosx\right\}$ + cos x log x.

$=x\left\{cosx·\frac{1}{x}-sinx·logx\right\}$ + cos x log x.

= cos x- sin x. log x + cos x log x.

$\frac{dy}{dx}=u$ [cosx + cos x log x- sin x log x]

= x^x cos x [cos x + cos x log x-x sin x log x]

And v = $\frac{x^2+1}{x^2-1}$

So, $\frac{dv}{dx}=\frac{\left(x^2-1\right)\frac{d}{dx}\left(x^2+1\right)-\left(x^2+1\right)\frac{d}{dx}(x-1)}{{\left(x^2-1\right)}^2}$

$=\frac{\left(x^2-1\right)(2x)-\left(x^2+1\right)(2x)}{{\left(x^2-1\right)}^2}$

$=\frac{2x^3-2x-2x^3-2x}{{\left(x^2-1\right)}^2}$

$=\frac{-4x}{{\left(x^2-1\right)}^2}.$

Hence, eqn (1) becomes,

$\frac{dy}{dx}$ x^{xcos x} [cos x + cos x log x-x sin x log x] $\frac{-4x}{{\left(x^2-1\right)}^2}$

17. Kindly go through the solution

16. Kindly go through the solution