Class 12th

Let 'b' and 'x' be the fixed base and equal side of isosceales triangle.

Then,  $\frac{dx}{dt}=-3$ cm/s (Ø decreasing).

This is a short answer type question as classified in NCERT Exemplar

The vapour pressure of a liquid in comparison to air pressure determines its boiling point. At a constant atmospheric pressure, the lower the vapour pressure, the higher the boiling point of a liquid, and vice versa.

Because NaCl is a nonvolatile solute, it reduces the vapour pressure of water when added to it. The boiling point of water rises as a result. Methyl alcohol, on the other hand, is more volatile than water, therefore adding it to the solution raises the overall vapour pressure, lowering the boiling point of water.

Because both components exist in the distillate and the liquid and vapour compositions are the same, this indicates that the liquids have formed an azeotropic combination that cannot be separated at this stage by fractional distillation.

We have, f(x) $\frac{logx}{x},x>0.$

f(x) = $\begin{array}{l}\underset{\_}{x\frac{d}{dx}logx-logx\frac{d}{dx}x}\\ x^2.\end{array}$

$=\frac{x*\frac{1}{x}-logx}{x^2}=\frac{1-logx}{x^2}$

f(x) = $\frac{x^2\frac{d}{dx}(1-logx)-(1-logx)\frac{d}{dx}{x}^2}{x^4}$

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

= $\frac{-1-2+2logx}{x^3}$ = $\frac{2logx-3}{x^3}$

At extreme points, f(x) = 0.

$\Rightarrow \frac{1-logx}{x^2}=0.$

$\Rightarrow logx=1=loge.$

$\Rightarrow x=e.$

At x = e, f"(e) = $\frac{2loge-3}{e^3}=\frac{2-3}{e^3}=-\frac{1}{e^3}<0$

x = e is a point of maximum.

(a) Consider $y=x^{\frac{1}{4}}.Let,x=\frac{16}{81}\&?x=\frac{1}{81}.$

$\begin{array}{l}Then,?y={\left(x+?x\right)}^{\frac{1}{4}}-x^{\frac{1}{4}}\\ ={\left(\frac{17}{81}\right)}^{\frac{1}{4}}-{\left(\frac{16}{81}\right)}^{\frac{1}{4}}\\ ={\left(\frac{17}{81}\right)}^{\frac{1}{4}}-\frac{2}{3}\\ \therefore {\left(\frac{17}{81}\right)}^{\frac{1}{4}}=\frac{2}{3}+?y\end{array}$

Now, $dy$ is approximately equal to $?y$ and is given by,

$\begin{array}{l}dy=\left(\frac{dy}{dx}\right)?x=\frac{1}{4{\left(x\right)}^{\frac{3}{4}}}\left(?x\right)\left(as,y=x^{\frac{1}{4}}\right)\\ =\frac{1}{4{\left(\frac{16}{81}\right)}^{\frac{3}{4}}}\left(\frac{1}{81}\right)=\frac{27}{4*8}*\frac{1}{81}=\frac{1}{32*3}=\frac{1}{96}=0.010\end{array}$

Hence, the approximate value of ${\left(\frac{17}{81}\right)}^{\frac{1}{4}}$ is $\frac{2}{3}+0.010=0.667+0.010$

=0.677

(b) ${(33)}^{\frac{-1}{5}}$

(b) Consider $y=x^{\frac{1}{5}}.Let,x=32\&?x=1$

Now, $dy$ is approximately equal to $?y$ and is given by,

$\begin{array}{l}dy=\left(\frac{dy}{dx}\right)\left(?x\right)=\frac{-1}{5{\left(x\right)}^{\frac{6}{5}}}\left(?x\right)\left(as,y=x^{\frac{1}{5}}\right)\\ =-\frac{-1}{5{\left(x\right)}^6}\left(1\right)=-\frac{1}{320}=-0.003\end{array}$

Hence, the approximate value of ${\mathrm{33}}^{\frac{1}{5}}$
is $\frac{1}{2}+\left(-0.003=0.497\right)$

We have,

$f^{\prime }(x)=[\mathrm{X(x\; -\; 1)}+$
${\mathrm{1]}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},0\le x\le 1$

$f(x)=\frac{1}{3}{\left[x^2-x+1\right]}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{d}{dx}\left(x^2-x+1\right)$

$=\frac{2x-1}{3{\left(x^2-x+1\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

At f(x) = 0.

2x – 1 = 0

$x=\frac{1}{2}\in [0,1]$

$f\left(\frac{1}{2}\right)={\left[\frac{1}{2}\left(\frac{1}{2}-1\right)+1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\left[\frac{1}{4}-\frac{1}{2}+1\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$

$={\left[\frac{1-2+4}{4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

= ${\left[\frac{3}{4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}?0.90$

$f(0)=[\mathrm{0(0\; -\; 1)}$
$\mathrm{+}{1]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1.$

$f(1)=[1(1\; -\; 1)+\mathrm{1]}$
${}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1.$

Option (B) is

Hence maximum value of f(x) = at x = 1 and x = 0.

Option (c) is correct.

We have,

$f(x)=\frac{1-x+x^2}{1+x+x^2}.$

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

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

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

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

At f(x) = 0.

$\Rightarrow \frac{-2(1+x)(1-x)}{{\left(1+x+x^2\right)}^2}=0$

x = 1 and x = -1.

At $x=1,f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$ $x=1,f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$

At x = -1, $f(-1)=\frac{1+1+1}{1-1+1}=3$

The maximum value of f(x) $=\frac{1}{3}x=1$

Hence, option (D) is correct.

The equation of the given curve is

x² = 2y.

Let p (x, y) be a point on the curve.

The distance of p (x, y) from (0, 5) is say S is given by

$s^2=x^2+{(y-5)}^2$

Let z = s² = x² + y² + 25 – 10y = 2y + y² -10y + 25

z = y² – 8y + 25

So,  $\frac{dz}{dy}=2y-8$

$\frac{d^{2}z}{d{y}^2}=2$

At $\frac{dz}{dy}=0\Rightarrow 2y-8=0\Rightarrow y=\frac{8}{2}=4$

At y = 4,  $\frac{d^{2}z}{d{y}^2}=2>0$

y = 4, is point of minimum distance.

So, x² = 2y->x2 = 2 * 4-> x² = 8 $x=\pm 2$

Hence, the point of the nearest distance are $\left(\mathrm{2\surd 2},4\right)$ and

$\left(-\mathrm{2\surd 2},4\right).$

Option (A) is correct.