Show that the function \(f(x)=\left\{\begin{aligned} -x^{2}, & x<0 \\ 0, & x=0 \\ x^{2}, & x>0 \end{aligned}\right.\) 
Derivable at x =0

2 a³b³e³

Given, 
\(f(t)=\left| \begin{matrix} \cos { t } & t & 1 \\ 2\sin { t } & t & 2t \\ \sin { t } & t & t \end{matrix} \right| =f(t)=\left| \begin{matrix} \cos { t } & t & 1 \\ 0 & -t & 0 \\ \sin { t } & t & t \end{matrix} \right|\)
\( [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }]\)
\(=t\left| \begin{matrix} \cos { t } & 1 & 1 \\ 0 & -1 & 0 \\ \sin { t } & 1 & 1 \end{matrix} \right| \)
Expanding an along R₂, we get
t[(-1)(tcost-sint)]
= -t²cost+sint
\(\lim _{ t\rightarrow 0 }{ \frac { f(t) }{ { t }^{ 2 } } } =\lim _{ t\rightarrow 0 }{ \frac { -{ t }^{ 2 }cost+tsint }{ { t }^{ 2 } } } \)
\(=\lim _{ t\rightarrow 0 }{ \left( \frac { -{ t }^{ 2 }cost }{ { t }^{ 2 } } +\frac { tsint }{ { t }^{ 2 } } \right) } \)
\(=\lim _{ t\rightarrow 0 }{ \left( -cost+\frac { sint }{ t } \right) } \)
\(=-1+\lim _{ t\rightarrow 0 }{ \frac { sint }{ t } } \)
= -1+1 = 0

(1-sin x)² y"=cos x

\(\frac { dy }{ dx } =-2{ cos }^{ 3 }y\quad sin \ \ y\)

\(\text { Let } E=\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right.\)
\(\text { Now, } 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right) \) 
\(\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right] \)
\(\Rightarrow \quad 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 / 3}{8 / 9}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \)
\(\text { Put } \tan ^{-1}\left(\frac{3}{4}\right)=\theta \Rightarrow \tan \theta=\frac{3}{4} \) 
\(\text { Then, } \sin \theta=\frac{3}{5} \Rightarrow \theta=\sin ^{-1}\left(\frac{3}{5}\right) \)
\(\Rightarrow \tan ^{-1}\left(\frac{3}{4}\right)=\sin ^{-1}\left(\frac{3}{4}\right)\) 
From Eqs. (i) and (ii), we get
2\( \tan ^{-1}\left(\frac{1}{3}\right)=\sin ^{-1}\left(\frac{3}{5}\right)\)
Now, put \(\tan ^{-1} 2 \sqrt{2}=\phi \Rightarrow \tan \phi=2 \sqrt{2}\)
Then, \(\cos \phi=\frac{1}{3} \Rightarrow \phi=\cos ^{-1}\left(\frac{1}{3}\right)\)
Now, \(\tan ^{-1} 2 \sqrt{2}=\cos ^{-1}\left(\frac{1}{3}\right)\)
\(\tan ^{-1}\left(\frac{3}{4}\right)=\sin ^{-1}\left(\frac{3}{5}\right)\) 
\( \therefore E =\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right) \)
\(=\sin \left(\sin ^{-1} \frac{3}{5}\right)+\cos \left(\cos ^{-1} \frac{1}{3}\right) \)
\(=\frac{3}{5}+\frac{1}{3}=\frac{9+5}{15}=\frac{14}{15} \)

We know tan x is increasing for \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)\(\left[y=\tan x \Rightarrow y^{\prime}=\sec ^{2} x>0\right]\)
\(\text {Also } 1>\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}>\tan \frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1} >1>\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}>1>\tan ^{-1}(1) . \)

For \(x_{ 1 },x_{ 2 }\in R,f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(\frac { x_{ 1 } }{ x_{ 1 }^{ 2 }+1 } =\frac { x_{ 2 } }{ x_{ 2 }^{ 2 }+1 } \)
\(\Rightarrow \) \(x_{ 1 }x_{ 2 }^{ 2 }+x_{ 1 }=x_{ 2 }x_{ 1 }^{ 2 }+x_{ 2 }\)
\(\Rightarrow \) \(x_{ 1 }x_{ 2 }(x_{ 1 }-x_{ 2 })=x_{ 2 }-x_{ 1 }\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\quad or\quad x_{ 1 }x_{ 2 }=1.\)
Here there are points \(x_{ 1 }\) and \(x_{ 2 }\) with \(x_{ 1 }\neq x_{ 2 }\) and \(f(x_{ 1 })=f(x_{ 2 })\).
\(\left[ For\ Ex.\ Take\ x_{ 1 }=2,x_{ 2 }=\frac { 1 }{ 2 } \right] \)
then \(f(x_{ 1 })=\frac { 2 }{ 4+1 } =\frac { 2 }{ 5 } \) and \(f(x_{ 2 })=\frac { 1/2 }{ 1+1/4 } =\frac { 2 }{ 5 } \)
Thus \(f(x_{ 1 })=f(x_{ 2 })=\frac { 2 }{ 5 } \) but \(x_{ 1 }\neq x_{ 2 }\quad i.e.f(x)=\frac { 1 }{ 2 } ]\)
Hence 'f' is not one-one.
Also 'f' is not onto.
For if so, then for \(1\in R\) , there exists \(x\in R\) such that f(x) = 1,
Which gives \(\frac { x_{ 2 } }{ x_{ 2 }^{ 2 }+1 } =1\)
But there is no such x in the domain R.
[ ∵ x² − x + 1 = 0 does not give any real value of x]   

Since, X = Number of four seen
On tossing two dice, X = 0, 1, 2
Also, \(P(4)=\frac{1}{10} \text { and } P(\operatorname{not} 4)=\frac{9}{10}\) 
So, \(P(X=0)=P(\operatorname{not} 4) \cdot P(\operatorname{not} 4)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}\) 
\(P(X=1)=P(\operatorname{not} 4) \cdot P(4)+P(4) \cdot P(\operatorname{not} 4)\) 
\(=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100}\) 
\(P(X=2)=P(4) \cdot P(4)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}\) 
Thus, we get the following table
\(\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & 81 / 100 & 18 / 100 & 1 / 100 \\ \hline \end{array}\) 
\(\begin{array}{c|c|c|c} \hline X P(X) & 0 & 18 / 100 & 2 / 100 \\ \hline X^{2} P(X) & 0 & 18 / 100 & 4 / 100 \\ \hline \end{array}\) 
\(\therefore \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}=\Sigma X^{2} P(X)-\left[\sum X P(X)\right]^{2}\) 
\(=\left(0+\frac{18}{100}+\frac{4}{100}\right)-\left(0+\frac{18}{100}+\frac{2}{100}\right)^{2}\) 
\(=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\) 
\(=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18\)

Let E_I = Bag I is selected
E ₂ = Bag II is selected
and A = Black ball is drawn
Then,\(P\left(E_{1}\right)=P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{A}{E_{1}}\right)=\frac{3}{5}, P\left(\frac{A}{E_{2}}\right)=\frac{2}{6}\) 
\(\therefore\) Required probability
\(P(A)=P\left(E_{1}\right) \times P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \times P\left(\frac{A}{E_{2}}\right)\) 
= \(\frac{7}{15}\) 

The equation of a plane through the line of intersection of the planes
\(x+2 y+3 z-4=0 \text { and } 2 x+y-z+5=0\) is 
\((x+2 y+3 z+4)+\lambda(2 x+y-z+5)=0\) 
As, this is perpendicular to the plane
\(5 x+3 y+6 z+8=0\) 
\(\therefore \quad 5(1+2 \lambda)+3(2+\lambda)+6(3-\lambda)=0\) 
\(\left[\because a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\right]\) 
\(\Rightarrow 5+10 \lambda+6+3 \lambda+18-6 \lambda=0\) 
\(\lambda=-\frac{29}{7}\) 
Then, from Eq. (i), we get
\(x\left[1+2\left(\frac{-29}{7}\right)\right]+y\left(2-\frac{29}{7}\right)+z\left(\frac{29}{7}+3\right)-4+5\left(\frac{-29}{7}\right)=0\) 
\(\Rightarrow x(7-58)+y(14-29)+z(29+21)-28-145=0\) 
\(\Rightarrow -51 x-15 y+50 z-173=0\) 
Hence, the required equation of plane is
\(51 x+15 y-50 z+173=0\)

Given,\(|\vec{r}|=14\) and if \(\vec{r}=a \hat{i}+b \hat{j}+c \hat{k}\) then \(a=2 \lambda, b=3 \lambda\)  \(\text { and } c=-6 \lambda \text { for some } \lambda \neq 0 \text { . }\)
\(\therefore \text { Direction cosines } l, m \text { and } n \text { are } l=\frac{a}{|\vec{r}|}=\frac{2 \lambda}{14}=\frac{\lambda}{7}\)
\(m=\frac{b}{|\vec{r}|}=\frac{3 \lambda}{14} \text { and } n=\frac{c}{|\vec{r}|}=\frac{-6 \lambda}{14}=\frac{-3 \lambda}{7}\) 
Also, we know that \(l^{2}+m^{2}+n^{2}=1\) 
\( \therefore \frac{\lambda^{2}}{49}+\frac{9 \lambda^{2}}{196}+\frac{9 \lambda^{2}}{49}=1 \)
\(\Rightarrow \frac{4 \lambda^{2}+9 \lambda^{2}+36 \lambda^{2}}{196}=1 \)
\(\Rightarrow 49 \lambda^{2}=196 \Rightarrow \lambda^{2}=\frac{196}{49} \)
\(\lambda^{2}=4 \Rightarrow \lambda=\pm 2 \) 
So, the direction cosines l,m and n are \(\frac{2}{7}, \frac{3}{7} \text { and } \frac{-6}{7}\) .
[\(\because \vec{r} \)makes an acute angle with X -axis, so we will take positive value of \(\lambda\)]
\(\because \ \vec{r}=\hat{r} \cdot|\vec{r}|\) 
\(\therefore \vec{r}=(l \hat{i}+m \hat{j}+n \hat{k}) \cdot|\vec{r}|=\left(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right) \cdot 14\) 
\(=4 \hat{i}+6 \hat{j}-12 \hat{k}\) 
 Thus, the components of \(\vec{r} \text { are } 4 \hat{i}, 6 \hat{j} \text { and }-12 \hat{k}\).

\(\frac{d y}{d x}=y+\frac{y}{x}\) 
\(y=K x e^{x}\)

Let the events be As:
A : At least one girl is chosen
B : At least 2 girls are chosen
We requireP(B/A)
\(P(\overset { - }{ A } )\)= P(No girl is chosen)
\(=\frac { ^{ 8 }{ C }_{ 4 } }{ ^{ 12 }{ C }_{ 4 } } =\frac { 70 }{ 495 } =\frac { 14 }{ 99 } \)
\(P(A)=1-P(\overset { - }{ A } )=1-\frac { 14 }{ 99 } =\frac { 85 }{ 99 } \)
And \(P(A\cap B)\)= P(2 boys and 2 girls) = \(\frac { ^{ 8 }{ C }_{ 2 }.^{ 4 }{ C }_{ 2 } }{ ^{ 12 }{ C }_{ 4 } } \)
\(=\frac { 6\times 28 }{ 495 } =\frac { 56 }{ 165 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 56 }{ 165 } \times \frac { 99 }{ 85 } =\frac { 168 }{ 425 } \)

Let E_I = Letter has come from CALCUITA
E₂ = Letter has come from TATANAGAR
and E = Two consecutive letters (i.e. alphabets) TA are visible on envelope
\(\therefore P\left(E_{1}\right)=\frac{1}{2}, P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{E}{E_{1}}\right)=\frac{n\left(E \cap E_{1}\right)}{n\left(E_{1}\right)}=\frac{1}{7}\) 
[\(\therefore\) pairs of consecutive letters are CA, AL, LC, CU, UT, TT, TA]
and \(P\left(\frac{E}{E_{2}}\right)=\frac{n\left(E \cap E_{2}\right)}{n\left(E_{2}\right)}=\frac{2}{8}\) 
[\(\therefore\)8 pairs of consecutive letters are TA, AT, TA;AN, NA, AG, GA, AR]
\(\therefore P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\) 
[using Baye's theorem]
\(=\frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{1}{7}+\frac{1}{2} \times \frac{2}{8}}=\frac{\frac{2}{16}}{\frac{8+14}{2 \times 7 \times 8}}=\frac{2 \times 7}{22}=\frac{7}{11}\) 
Hence, the probability that the letter TA came from TATANAGAR is \(\frac{7}{11}\).

m =cos\(\pi\over4\) = \(\frac { 1 }{ \sqrt { 2 } } \) and  n = cos\(\pi\over2\)=0
Now \({ i }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1\) gives \({ i }^{ 2 }+\frac { 1 }{ 2 } +0=1\)
\(\Rightarrow { i }^{ 2 }=\frac { 1 }{ 2 } \quad \Rightarrow i=\pm \frac { 1 }{ \sqrt { 2 } } \) 
Hence, the reqd.vector is \(\overset { \wedge }{ r } =3\sqrt { 2 } (l\overset { \wedge }{ i } +m\overset { \wedge }{ j } +n\overset { \wedge }{ k } )\) 
\(i.e\quad \overset { \rightarrow }{ r } =3\sqrt { 2 } \left( \pm \frac { 1 }{ \sqrt { 2 } } \overset { \wedge }{ i } +\frac { 1 }{ \sqrt { 2 } } \overset { \wedge }{ j } +0\overset { \wedge }{ k } \right) i.e\overset { \rightarrow }{ r } =\pm 3\overset { \wedge }{ i } +3\overset { \wedge }{ j } \)

\((\overrightarrow { a } -\overrightarrow { d } )\) x \((\overrightarrow { b } -\overrightarrow { c } )\) 
\(=\overrightarrow { a } *\overrightarrow { b } -\overrightarrow { a } *\overrightarrow { c } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } *\overrightarrow { c } \)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } -\overrightarrow { c } \)
\([\because \overrightarrow { a } *\overrightarrow { b } =\overrightarrow { c } -\overrightarrow { d } and\overrightarrow { a } *\overrightarrow { c } =\overrightarrow { b } *\overrightarrow { d } ]\)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } +\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { c } *\overrightarrow { d } =\overrightarrow { 0 } \) 
Hence, \((\overrightarrow { a } -\overrightarrow { d } )\) is parallel to \((\overrightarrow { b } -\overrightarrow { c } )\)

\(0.018\pi cm^3\)

The area A₁ of square of side x is given by A₁= x²
and area A₂ of square of side y is given by
\(A_{2}=y^{2}=\left(x-x^{2}\right)^{2} \)
\(\frac{d A_{1}}{d x} =2 x, \frac{d A_{2}}{d x}=2\left(x-x^{2}\right)(1-2 x) \)
\(\frac{d A_{2}}{d A_{1}} =\frac{d A_{2}}{d x}= \frac{d A_{1}}{d x}=\frac{2\left(x-x^{2}\right)(1-2 x)}{2 x} \)
\(=(1-x)(1-2 x)=1-3 x+2 x^{2} \)

\({ds\over dt}\infty {1\over x}\)