Maths

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

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

From graph : P, Q, R

S = {1,2, 3, …., n, 2022}

HCF (n, 2022) = 1

2022 = 2 * 1011 ->3 * 337

2022 = 2 * 3 * 337 (prime factorization)

Let n (A) = no members divisible by 2 = 1011

Let n (B) = no members divisible by 3 = 674

Let n (C) = no members divisible by 337 = 6

$n\left(A\cup B\cup C\right)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(A\cap B\right)-n\left(B\cap C\right)-n\left(C\cap A\right)+n\left(A\cap B\cap C\right)$

= 1011 + 674 + 6 – 337 – 2 – 3 + 1

= 1350

$n\left(A\cap B\right)=337$

$n\left(\left(A\cup B\cup C\right)\text{'}\right)=2022-1350=672$

Prob. $=\frac{672}{2022}=\frac{336}{1011}=\frac{112}{337}$

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

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

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

L : y = mx + c, m > 0

y = m (x – 1)

$\frac{x^2}{4}-\frac{y^2}{4}=1$

$y=mx\pm \sqrt[]{4m^2-4}$

$\pm \sqrt[]{4m^2-4}=-m,m>0$

$\Rightarrow \sqrt[]{4m^2-4}=m$

$M\left(\frac{5+\sqrt[]{21}}{2},\sqrt[]{3}+\sqrt[]{7}\right),N\left(\frac{5-\sqrt[]{21}}{2},\sqrt[]{3}-\sqrt[]{7}\right)$

$=$$\text{exception 25:}$$\left(2\text{exception 25:}\right)$$=$$2$

 $\left|x-1\right|\le y\le \sqrt[]{5-x^2}$

$y=\left|x-1\right|,y^2+x^2=5,y\ge 0y=\sqrt[]{5-x^2}$

${\left(x-1\right)}^2+x^2=5\Rightarrow 2x^2-2x-4=0\Rightarrow x^2-x-2=0$

$\left(x-2\right)\left(x+1\right)=0\to x=-1,2$

$A={\displaystyle \underset{-1}{\overset{2}{\int }}\left(\sqrt[]{5-x^2}-\left|x-1\right|\right)dx}$

$Area={\displaystyle \underset{-1}{\overset{2}{\int }}\left(\sqrt[]{5-x^2}dx-\left|x-1\right|dx\right)}$

$={\displaystyle \underset{-1}{\overset{2}{\int }}\sqrt[]{5-x^2}dx+{\displaystyle \underset{-1}{\overset{1}{\int }}\left(x-1\right)dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(x-1\right)dx}}}$

$=\frac{5}{2}si{n}^{-1}\left(1\right)-\frac{1}{2}=\frac{5\pi }{4}-\frac{1}{2}$

 $f\left(\alpha \right)={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{lo{g}_{10}t}{1+t}}dt$

$f\left(\frac{1}{\alpha }\right)={\displaystyle \underset{1}{\overset{\frac{1}{\alpha }}{\int }}\frac{lo{g}_{10}t}{1+t}dt={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{-lo{g}_{10}z}{1+\frac{1}{z}}-\frac{1}{z^2}dz}}$

$={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{lo{g}_{10}z}{z\left(z+1\right)}dz}$

$={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{lo{g}_{10}t}{t}}dt=\frac{1}{\mathcal{l}n10}{\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{\mathcal{l}nt}{t}dt=\frac{1}{\mathcal{l}n10}{\left[\frac{{\left(\mathcal{l}nt\right)}^2}{2}\right]}_1^{\alpha }=\frac{{\left(\mathcal{l}n\alpha \right)}^2}{2\mathcal{l}n10}}$

$f\left(e^3\right)+f\left(e^{-3}\right)=\frac{{\left(\mathcal{l}n{e}^3\right)}^2}{2\mathcal{l}n10}=\frac{9}{2\mathcal{l}n10}$

$\left(\widehat{a}+\widehat{b}\right).\left(\widehat{a}+2\widehat{b}+2\left(\widehat{a}*\widehat{b}\right)\right),\widehat{a}.\widehat{b}=\frac{1}{\sqrt[]{2}}$

= 1 + $\frac{1}{\sqrt[]{2}}+\sqrt[]{2}$ + 2 + 0 + 0

$=3+\frac{3}{\sqrt[]{2}}$

$\left|\widehat{a}+2\widehat{b}+2{\left(\widehat{a}*\widehat{b}\right)}^2\right|=1+4+4\left(\frac{1}{2}\right)+2\sqrt[]{2}+0=7+2\sqrt[]{2}$

$\Rightarrow \left(2+\sqrt[]{2}\right)\left(7+2\sqrt[]{2}\right)co{s}^2\theta =9+\frac{9}{2}+9\sqrt[]{2}=\frac{27+18\sqrt[]{2}}{2}$

$\Rightarrow co{s}^2\theta =\frac{27+18\sqrt[]{2}}{2}*\frac{1}{18+11\sqrt[]{2}}=\frac{\left(27+18\sqrt[]{2}\right)\left(18-11\sqrt[]{2}\right)}{2\left(324-242\right)}$

$=\frac{486+324\sqrt[]{2}-297\sqrt[]{2}-396}{2*82}$

164 cos² (θ)= 90 + $2$$7$

 ${\left(a-1\right)}^2+4={\left(b-1\right)}^2+16$

$={\left(a-1\right)}^2+{\left(b-1\right)}^2$

${\left(b-1\right)}^2=4\&{\left(a-1\right)}^2=16$

$\Rightarrow b=1\pm 2a=1\pm 4$

= 3, 1= 5, 3

x + 2y = 3 …… (i)

3x – y = 8…… (ii)

x + 2y = 3

3x – y = 83 * 2

$\oplus \to 7x=-13\Rightarrow x=\frac{-13}{7}$

$y=\frac{-39}{7}+8=\frac{17}{7}$

$k_1+k_2=x+y=\frac{4}{7}$

 $\left(p\wedge q\right)\to \left(p\wedge r\right)$

$\equiv \sim \left(p\wedge q\right)\vee \left(p\wedge r\right)$

$\left(A\right)\left(\sim q\right)\vee \left(p\wedge r\right)\left(B\right)\left(\sim p\right)\vee \left(p\wedge r\right)$

$\left(C\right)\sim \left(p\wedge r\right)\vee \left(p\wedge q\right)\left(D\right)\sim \left(p\wedge q\right)\vee r$

Option (D) is correct

$100co{t}^2\alpha =15^2-9^2=144$

$\frac{OT}{15}=\frac{2}{3}$

$cot\alpha =\frac{6}{5}OT=10$

$\Delta AOT,OA=OTcot\alpha =10cot\alpha$