Maths

64. Let x litre of water is required to added.

Then, total mixture =(x+1125) litre.

As, 0 % of acid is contained in pure mater (neutral).

$\therefore$ 0% of x + 45% of 1125% > 25% of (x+1125)

$\Rightarrow \frac{0*x}{100}+\frac{45}{100}*1125>\frac{25}{100}(x+1125)$ (multiplying by 100 throughout)

$\Rightarrow$

$\Rightarrow \frac{1125\cdot (45-25)}{25}>\frac{25x}{25}$ (dividing by 25 throughout)

$\Rightarrow \frac{1125*20}{25}>x$

$\Rightarrow 900>x$ ……(1)

And 0% of x+ 45% of 1125<30 of (x+1125)

$\Rightarrow \frac{0}{100}*x+\frac{45}{100}*1125<\frac{30}{100}(x+1125)$

$\Rightarrow 45*1125<30(x+1125)$ {multiplying by 100 throughout)

$\Rightarrow 45*1125<30x+30*1125$

$\Rightarrow$ 45 * 1125 – 30 * 1125 < 30x

$\Rightarrow \frac{1125(45-30)}{30}<\frac{30x}{30}$ {dividing by 30 throughout}

$\Rightarrow \frac{1125*15}{30}<x$

$\Rightarrow$ 562.5

From (1) and (2),

562.5

Thus, the water to be added has to be more than 562.5 litre but less than 900 litres.

63. Let x litre of 20 % boric acid solution is required to added. Since, we have 640 litres of 8 % boric acid solution.

Total mixture will have (x + 640) litre.

$\therefore 2\mathrm{\%of}x+8\mathrm{\%of}640>4\mathrm{\%of}(x+640).$

$\begin{array}{l}\Rightarrow \frac{2x}{100}+\frac{8*640}{100}>\frac{4*(x+640)}{100}\\ \end{array}$

$$ $\Rightarrow \frac{2x}{100}*\frac{100}{2}+\frac{8}{100}*640*\frac{100}{2}>\frac{4}{100}*(x+640)*\frac{100}{2}$ (multiplying throughout by $\frac{100}{2}$ )

$\Rightarrow x+2560>2(x+640)$

$\Rightarrow 2x+2560>2x+1280$

$\Rightarrow 2560-1280>2x-x$

$\Rightarrow 1280>x$ (1)

And 2 % of x + 8 % of 640 < 6 % of (x+ 640)

$\Rightarrow \frac{2x}{100}+\frac{8}{100}*640<\frac{6}{100}(x+640)$

$\Rightarrow 2x+8*640<6x+6*64\mathrm{0(multiplyingby}10\mathrm{0throughout)}$

$\Rightarrow 8*640-6*640<6x-2x$

$\Rightarrow 640(8-6)<4x$

$\begin{array}{l}\Rightarrow \frac{640*2}{4}<\frac{4x}{4}\mathrm{4\; (dividing\; by\; 4\; throughout)}\\ \Rightarrow 320<\mathrm{x(2)}\end{array}$

So, from (1) and (2) we get,

320

Thus, the number of litre of 2 % boric acid solution will have to be more than 320 litres but less than 1280 litres.

62. It is given that

$68<\; F<77.$

Putting $f=\frac{9}{5}c+32$ we get,

$68<\frac{9}{5}c+32<77.$

$\Rightarrow 68-32<\frac{9}{5}c+32-32<77-32.$  (Subtracting 32 throughout)

$\Rightarrow 36<\frac{9}{5}c<45$

Multiplying by $\frac{5}{9}$ throughout,

$\Rightarrow 36*\frac{5}{9}<\frac{9}{5}*c*\frac{5}{9}<45*\frac{5}{9}$

$\begin{array}{l}\\ \Rightarrow 20<c<25\end{array}$

Hence, the required range of temperature is between $20^{°}25^{°}.$

19. Given f (x) = x2 sin x + 5.

At x = .

$f(\pi )={\pi }^2-sin\pi +5={\pi }^2-0+5={\pi }^2+5$

$\underset{x\to \pi }{lim}$ f (x) = $\underset{x\to \pi }{lim}$  [x2 sin x + 5]

If x = $\pi +h$ then as x, h 0, so,

$\underset{x\to \pi }{lim}$ f (x) = $\underset{x\to 0}{lim}$  [ ( + h)2 sin ( + h) + 5]

= ( + 0)2 $\underset{h\to 0}{lim}$ $\begin{array}{rr}\hfill [sin\pi cosh+cos\pi \cdot sina]& \hfill +5.\end{array}$

= 2 $\underset{h\to 0}{lim}$ sin cos h $\underset{h\to 0}{lim}$ cos sin h + 5

= x2 0 * (1) ( 1) 0 + 5.

= 2 + 5 = f (x)

So, f is continuous at x = .