Maths NCERT Exemplar Solutions Class 12th Chapter Six

Let perimeter of $\Delta$ is x and that of square is 22 – x

now area $=\frac{\sqrt[]{3}}{4}{\left(\frac{x}{3}\right)}^2+{\left(\frac{22-x}{4}\right)}^2$

for maximum or minimum,  $\frac{dA}{dx}=0$

x $=\frac{22\sqrt[]{3}}{\sqrt[]{3}+4}$

now side of a $\Delta =\frac{x}{3}$

$=\frac{22\sqrt[]{3}}{3\left(\sqrt[]{3}+4\right)}$

$=\frac{66}{}$$\frac{9}{}$$\frac{+}{}$$\frac{4}{}$

$f_a\left(x\right)=ta{n}^{-1}2x-3ax+7\Rightarrow f_a\text{'}\left(x\right)=\frac{2}{1+4x^2}-3a\Rightarrow f_a\text{'}\left(x\right)\ge 0\Rightarrow 3a\le \frac{2}{1+4x^2}$

$a_{max}=\frac{2}{3}\left(\frac{1}{1+4*\frac{{\pi }^2}{36}}\right)=\frac{6}{9+{\pi }^2}=\overline{a}\therefore f_a\left(\frac{\pi }{8}\right)=ta{n}^{-1}\frac{2\pi }{8}-3\frac{\pi }{8}\frac{6}{9+{\pi }^2}+7=8-\frac{9\pi }{4\left(9+{\pi }^2\right)}$

Let a and b be the roots of the equation  $x^2+\left(3-a\right)x+1=2a$

Therefore a + b = a – 3, ab = 1 – 2a Þ a² + b² = (a – 3)² – 2 (1 – 2a) = a² – 6a + 9 – 2 + 4a = a² – 2a + 7 = (a – 1)² + 6 Þ So,   ${\alpha }^2+{\beta }^2\ge 6$

 $f\left(x\right)=ta{n}^{-1}\left(sinx-cosx\right)$

$f\text{'}\left(x\right)=\frac{cosx+sinx}{{\left(sinx-cosx\right)}^2+1}$ = 0

$\therefore x=\frac{3\pi }{4}$

Sum = tan^-1 $\sqrt[]{2}-\frac{\pi }{4}$

= $co{s}^{-1}\frac{1}{\sqrt[]{3}}-\frac{\pi }{4}$

 $f\left(x\right)=x{e}^{x\left(1-x\right)}$

$f\text{'}\left(x\right)=-e^{x\left(1-x\right)}\left(2x+1\right)\left(x-1\right)$

$f\left(x\right)is\uparrow in\left(\frac{-1}{2},-1\right)$

$(a+\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?x)(a-\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?y)=a^2-b^2$

$\Rightarrow a^2-\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?y+\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?x-2b^2\mathrm{c}\mathrm{o}\mathrm{s}?x\mathrm{c}\mathrm{o}\mathrm{s}?y=a^2-b^2$

Differentiating both sides:

$0-\sqrt[]{2}ab\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\sqrt[]{2}ab(-\mathrm{s}\mathrm{i}\mathrm{n}?x)-2b^2\left[\mathrm{c}\mathrm{o}\mathrm{s}?x\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?y(-\mathrm{s}\mathrm{i}\mathrm{n}?x)\right]=0$

At $\left(\frac{\pi }{4},\frac{\pi }{4}\right)$ :

$\begin{array}{rr}& ab\frac{dy}{dx}-ab-2b^2\left(-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}\right)=0\\ & \Rightarrow \frac{dx}{dy}=\frac{ab+b^2}{ab-b^2}=\frac{a+b}{a-b};a,b>0\end{array}$

 $f\left(2\right)=8,f^{\mathrm{\text{'}}}\left(2\right)=5,f^{\mathrm{\text{'}}}\left(x\right)\ge 1,f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)\ge 4,\forall x\in \left(\mathrm{1,6}\right)$

Using LMVT

$f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)=\frac{f^{\mathrm{\text{'}}}\left(5\right)-f^{\mathrm{\text{'}}}\left(2\right)}{5-2}\ge 4\Rightarrow f^{\mathrm{\text{'}}}\left(5\right)\ge 17$

$f^{\mathrm{\text{'}}}\left(x\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\ge 1\Rightarrow f\left(5\right)\ge 11$

Therefore $f^{\mathrm{\text{'}}}\left(5\right)+f\left(5\right)\ge 28$

 $Mean\frac{{\displaystyle \sum _{i=1}^{15}{x}_i}}{15}=8$

$\Rightarrow {\displaystyle \sum _{i=1}^{15}{x}_i}=120.......\left(i\right)$

S.D = 3

$\Rightarrow {\displaystyle \sum _{i=1}^{15}{x_i}^2}=15\left(9+64\right)=15*73..........\left(ii\right)$

Given 20 has misread as 5

$\therefore$ in new case

${\displaystyle \sum _{q=1}^{15}{x}_{i^2}=15*73-25+400=1470}$

mean in new case

$\overline{x}=\frac{120-5+20}{15}$

$\therefore v$ariance in new case

${\sigma }_{new}^2=\frac{1}{15}\left(1470\right)-81=17$

$f\left(x\right)=3^{{\left(x^2-2\right)}^3+4}=81.3^{{\left(x^2-2\right)}^3}$

$f\text{'}\left(x\right)=81.3^{{\left(x^2-2\right)}^3}.\mathcal{l}n3.3{\left(x^2-2\right)}^2.2x$

From graph : P, Q, R

 $f\left(x\right)=\left|x-1\right|cos\left|x-2\right|sin\left|x-1\right|+\left(x-3\right)\left|x^2-5x+4\right|$

$f\left(x\right)=\left|x-1\right|cos\left(x-2\right)sin\left|x-1\right|+\left(x-3\right)\left|x-1\right|\left|x-4\right|$

x = 1, 4 (doubtful points)

Diff. at x = 1

$\underset{x\to 1}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1}{lim}\frac{\left|x-1\right|\left(sin\left|x-1\right|cos\left(x-2\right)+\left(x-3\right)\left|x-4\right|\right)}{x-1}$

$\begin{array}{l}RHD=\underset{x\to 4^{+}}{lim}\frac{3\left(sin3cos2\right)}{0^{+}}=+\infty \\ LHD=\underset{x\to 4^{-}}{lim}\frac{3\left(sin3cos2\right)}{0^{-}}=-\infty \end{array}\}\Rightarrow$ Not diff. at x = 4