a = -2, b = 0, c = -3

Getting \(\lambda =-9\)
and \(\mu=27\)
Alternative Method :
\((\overset\wedge i+3\overset\wedge j+9\overset\wedge k)\times(3\overset\wedge i-\lambda \overset\wedge j+\mu\overset\wedge k)=0\)
\(\Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & 3 & 9 \\ 3 & -\lambda & \mu \end{matrix} \right| \)
\(\Rightarrow \overset\wedge i(3\mu+9\lambda)-\overset\wedge j(\mu-27)+\overset\wedge k(-\lambda-9)=0\)
\(\Rightarrow 3\mu+9\lambda=0 ....(i)\)
\(\Rightarrow \mu-27=0 ...(ii)\)
\(\Rightarrow -\lambda-9=0...(iii)\)
from eqn.(ii) and (iii),
\(\mu=27\)
and \(\lambda=-9\)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

We have, y = tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)
Put x = cos2\(\theta\)
\(\Rightarrow\)2\(\theta\) = cos^-1x
\(\Rightarrow\)\(\theta\) = 1/2cos^-1x
y = tan^-1\(\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
y = \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \)
(\(\because\)cos2\(\theta\) = 2cos²\(\theta\)-1 = 1-2sin²\(\theta\)
y = tan^-1(tan \(\theta\))
y = \(\theta\) = 1/2cos^-1x
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)​​​​​​​

P(E) = \(\frac { 7 }{ 13 } \)
P(F) = \(\frac { 9 }{ 13 } \)
and P(E\(\cap\)F) = \(\frac { 4 }{ 13 } \)
(a) \(P(\bar { E } /F)=\frac { P(\bar { E } \cap F) }{ P(F) } \)
\(=\frac { P(F)-P(E\cap F)) }{ P(F) } \)
\(=1-\frac { P(E\cap F) }{ P(F) } =1-\frac { \frac { 4 }{ 13 } }{ \frac { 9 }{ 13 } } \)
\(\Rightarrow P(\bar { E } /F)=1-\frac { 4 }{ 9 } =\frac { 5 }{ 9 } \)
(b) \(P(\bar { E } /\bar { F } )=\frac { P(\bar { E } \cap \bar { F } ) }{ P(\bar { F } ) } \)
\(=\frac { P\overline { (E\cup F) } }{ P(\bar { F } ) } \)
\(=\frac { 1-P(E\cup F) }{ 1-P(F) } \)
\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
\(=\frac { 7 }{ 13 } +\frac { 9 }{ 13 } -\frac { 4 }{ 13 } \)
\(=\frac { 12 }{ 13 } \)
\(P(E/F)=\frac { 1-\frac { 12 }{ 13 } }{ 1-\frac { 9 }{ 13 } } \)
\(=\frac { \frac { 1 }{ 13 } }{ \frac { 4 }{ 13 } } \)
\(=\frac { 1 }{ 4 } \)

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=(2\overset\wedge i+\overset\wedge j+3\overset\wedge k)+(\overset\wedge i+2\overset\wedge j-\overset\wedge k)\)
\(=3\overset\wedge i+3\overset\wedge j+2\overset\wedge k\)
\(\overset\wedge c=\frac{\overset\rightarrow c}{\left|c \right| }\)
\(=\frac{2i+3j+2k}{\sqrt{9+9+4}}\)
\(=\frac{3}{\sqrt{22}}\overset\wedge i+\frac{3}{\sqrt{22}}\overset\wedge j+\frac{2}{\sqrt{22}}\overset\wedge k\)

\(\overset\rightarrow {AB} \) = Position vector of B-Position vector of A
= (−5\(\hat{i}\) +7\(\hat{j}\)) − (2\(\hat{i}\)+1\(\hat{j}\))
= (−5−2)\(\hat{i}\)+ (7−1)\(\hat{j}\)
= −7\(\hat{i}\)+ 6\(\hat{j}\)
∴ The scalar components are (-7,6,0).

(i) f(g) = f(x)g(x)
\(fg=(\sqrt { x+4 } )(\sqrt { x-4 } )=\sqrt { x^{ 2 }-4 } \)
(ii) \(\frac { f }{ g } =\frac { f(x) }{ g(x) } =\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \times \frac { \sqrt { x-4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x^{ 2 }-16 } }{ x-4 } \)

\(\int { \left[ \frac { 2sinxcosx }{ 2 } \right] ^{ 2 } } dx\)
\(=\frac { 1 }{ 4 } \int { (sin2x)^{ 2 } } dx\)
\(=\frac { 1 }{ 4 } \int { { sin }^{ 2 }2x } dx\)
\(\frac { 1 }{ 4 } \int { \frac { 1-cos4x }{ 2 } } dx=\frac { 1 }{ 8 } \int { 1dx } -\frac { 1 }{ 8 } \int { cos4xdx } \)
 \(=\frac { x }{ 8 } -\frac { 1 }{ 8 } \frac { sin4x }{ 4 } \)
\(=\frac { 1 }{ 8 } \left( x-\frac { 1 }{ 4 } sin4x \right) +c\)

\(y^2 =8x + 3\)
\(2y\frac{dy}{dt}=8\frac{dx}{dt}\)
\(\frac{dy}{dt}=8\frac{dx}{dt}\)
\(2y\cdot8\frac{dx}{dt}=8\frac{dx}{dt}\)
\(\Rightarrow y=\frac{8}{16}=\frac{1}{2}\)
\(y=\frac{1}{2},y^2=8x+3\)
\((\frac{1}{2})^2=8x+3\)
\(\frac{1}{4}-3=8x\)
\(\Rightarrow x=\frac{1-12}{4}\times \frac{1}{8}=-\frac{11}{32}\)
Points  \((-\frac{11}{32},\frac{1}{2})\)

We have, \({ A }_{ 3\times x }\) and \({ B }_{ 4\times y }\) = \({ AB }_{ 3\times 3 }\)
(i) \(\left( { A }_{ 3\times x }, \right) \left( { B }_{ 4\times y } \right) =\left( { AB }_{ 3\times 3 } \right) \)
\(\therefore\)   x = 4 and y = 3
(ii) \(\left( { A }_{ x\times 2 } \right) \left( { B }_{ y\times 4 } \right) =\left( { AB }_{ 3\times 4 } \right) \)
\(\therefore\)  y = 2, x = 3

We have, \(X+Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] ,X-Y=\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(\left( X+Y \right) +\left( X-Y \right) =\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] +\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(2X=\left[ \begin{matrix} 8 & 8 \\ 12 & 4 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(\therefore \ X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
and \(Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -2 & -1 \\ -1 & -1 \end{matrix} \right] \)

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

\(\text { Let } \vec{a}=\sqrt{2} \hat{i}+\hat{j}+\hat{k} \)
\(\text { then } \hat{a}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k} \)
\(\text { Cosine of an angle with } y \text { -axis is coefficient of } \hat{j} \text { , }\)
\(cos\beta =\frac { 1 }{ 2 } \)

\(\text { We have }\int \tan ^{-1}(\cot x) d x=\int \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x=\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} x-\frac{x^{2}}{2}+C \)

\(\int \frac{1+\tan x}{1-\tan x} d x =\int \frac{\cos x+\sin x}{\cos x-\sin x} d x \)
\(=-\int \frac{1}{t} d t \)
\(=-\log |t|+C \)
\(=-\log |\cos x-\sin x|+C\)

Consider : \(\int({\sqrt x+{1\over \sqrt x}})^2dx=\int({x+{1\over x+2}+2})dx={x^2\over2}+log |x| + 2x + c\)

\(\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\(\Rightarrow 2+y=5,2x+2=8\Rightarrow x=3,y=3 \)
\(\therefore \ x+y=3+3=6\)

\( f(x)=27 x^3 and g(x)=x^{1 / 3}\\ \operatorname{gof}(x)=g[f(x)]=g\left[27 x^3\right]=\left[27 x^3\right]^{1 / 3} \\ =\left[(3 x)^3\right]^{1 / 3}=3 x \therefore \operatorname{gof}(x)=3 x \)

Let \(I= \int _{ 0 }^{ \pi /2 }{ \frac { x\sin { x\cos { x } } }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x } } } dx\)
\(\therefore I= \int _{ 0 }^{ \pi /2 }{ \frac { \left( \frac { \pi }{ 2 } -x \right) \sin { \left( \frac { \pi }{ 2 } -x \right) \cos { \left( \frac { \pi }{ 2 } -x \right) } } }{ \sin ^{ 4 }{ \left( \frac { \pi }{ 2 } -x \right) } +\cos ^{ 4 }{ \left( \frac { \pi }{ 2 } -x \right) } } } dx\)
\(\Rightarrow I= \int _{ 0 }^{ \pi /2 }{ \frac { \left( \frac { \pi }{ 2 } -x \right) \cos { x } \sin { x } }{ \cos ^{ 4 }{ x } +\sin ^{ 4 }{ x } } } dx\)
\(=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } \cos { x } }{ \cos ^{ 4 }{ x } +\sin ^{ 4 }{ x } } } dx-1\)
\(\Rightarrow 2I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } \cos { x } }{ \cos ^{ 4 }{ x } +\sin ^{ 4 }{ x } } } dx\)
\(\Rightarrow I= \frac { \pi }{ 4 } \int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } \cos { x } }{ \cos ^{ 4 }{ x } +\sin ^{ 4 }{ x } } } dx.\)
\(Put\sin ^{ 2 }{ x } =t\) so that  \(2\sin { x } \cos { x } dx=dt\)
\(i.e.\sin { x\cos { x } } dx=\frac { 1 }{ 2 } dt.\)
When \(x=0,t=0.\)  when 
\(\therefore I= \frac { \pi }{ 4 } \int _{ 0 }^{ 1 }{ \frac { \frac { 1 }{ 2 } dt }{ (1{ -t) }^{ 2 }+{ t }^{ 2 } } } =\frac { \pi }{ 8 } \int _{ 0 }^{ 1 }{ \frac { dt }{ 2{ t }^{ 2 }-2t+1 } }\)
\(=\frac { \pi }{ 16 } \int _{ 0 }^{ 1 }{ \frac { dt }{ { t }^{ 2 }-t+\frac { 1 }{ 2 } } } =\frac { \pi }{ 16 } \int _{ 0 }^{ 1 }{ \frac { dt }{ \left( t-\frac { 1 }{ 2 } \right) ^{ 2 }+\left( \frac { 1 }{ 2 } \right) ^{ 2 } } } \)
\(=\frac { \pi }{ 16 } \frac { 1 }{ 1/2 } { \left[ \tan ^{ -1 }{ \left( \frac { 1-1/2 }{ 1/2 } \right) } \right] }_{ 0 }^{ 1 }\)
\(=\frac { \pi }{ 8 } { \left[ \tan ^{ -1 }{ \left( 2t-1 \right) } \right] }_{ 0 }^{ 1 }\)
\(=\frac { \pi }{ 8 } { \left[ \tan ^{ -1 }{ \left( 1 \right) - } \tan ^{ -1 }{ \left( -1 \right) } \right] }_{ 0 }^{ 1 }\)
\(=\frac { \pi }{ 8 } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\frac { { \pi }^{ 2 } }{ 16 } .\)

We have: \(f(x)=x^{ 100 }+sinx-1\)
\(f'(x)=100x^{ 99 }+cosx.\)
At(-1, 1)    \(f'(-1)=100(-1)^{ 99 }+cos(-1)\)
=-100 + cos1 < 0
Hence 'f' is strictly decreasing on (-1, 14)
Yes, strictness is not always good in life.

We have :\(f(x)=sin^{ 2 }x-cosx\)
\(f'(x)=2sinxcosx+sinx\)
\(=sinx(2cosx+1)\)
Now \(f'(x)=0\Rightarrow sinx(2cosx+1)=0\)
\(\Rightarrow sinx=0orcosx=\frac { -1 }{ 2 } \)
\(\Rightarrow x=0,\pi orx=\frac { 2\pi }{ 3 } \)
Now \(f(0)=0-1=-1.f\left( \frac { 2\pi }{ 3 } \right) =sin^{ 2 }\frac { 2\pi }{ 3 } -cos\frac { 2\pi }{ 3 } \)           

We have : {1, 2, 3, 4, 5, 6}
Here 'X' is the larger of the two numbers obtained>
Now  X = 2, 3, 4, 5, 6
\(P(X=2)=\frac { 1 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 1 }{ 15 } \)
\(P(X=3)=\frac { 2 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 2 }{ 15 } \)
\(P(X=4)=\frac { 3 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 3 }{ 15 } \)
\(P(X=5)=\frac { 4 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 4 }{ 15 } \)
\(P(X=6)=\frac { 5 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 5 }{ 15 } \)
Hence, probability distribution is :       

(ii) we have

(I) Mean \(\mu =\sum { { p }_{ i }-{ x }_{ i }=\frac { 70 }{ 15 } =\frac { 14 }{ 3 } } \)
(II) Variance \({ \sigma }^{ 2 }=\sum { { p }_{ i } } { x }_{ i }^{ 2 }-{ \mu }^{ 2 }\)
\(=\frac { 350 }{ 15 } -{ \left( \frac { 14 }{ 3 } \right) }^{ 2 }=\frac { 350 }{ 15 } -\frac { 196 }{ 9 } =\frac { 1050-980 }{ 45 } =\frac { 70 }{ 45 } =\frac { 14 }{ 9 } \)

Let y = \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) \)
Put \(x=tan\theta \)
\(y={ tan }^{ -1 }\frac { \sqrt { 1+{ tan }^{ 2 }\theta } -1 }{ tan\theta } \)
\(={ tan }^{ -1 }\left( \frac { sec\theta -1 }{ tan\theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { 1-cos\theta }{ sin\theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { 2{ sin }^{ 2 }\frac { \theta }{ 2 } }{ 2{ sin }\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { { sin }\frac { \theta }{ 2 } }{ cos\frac { \theta }{ 2 } } \right) ={ tan }^{ -1 }\left( { tan }\frac { \theta }{ 2 } \right) \)
\(=\frac { \theta }{ 2 } =\frac { 1 }{ 2 } { tan }^{ -1 }x\)
Hence \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \frac { 1 }{ 1+{ x }^{ 2 } } =\frac { 1 }{ 2(1+{ x }^{ 2 }) } \)

\(\text { We have }|\vec{a}|=\sqrt{1^{2}+2^{2}}=\sqrt{5} \text { and }|\vec{b}|=\sqrt{2^{2}+1^{2}}=\sqrt{5}\)
So, | \(\overset { \rightarrow }{ a } \) | =| \(\overset { \rightarrow }{ b } \) |. But, the two vectors are not equal since their corresponding components are distinct

(i) Let \(x_{ 1 },x_{ 2 }y\in R\).
Now  \(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \)  \(3-4_{ x1 }=3-4_{ x2 }\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\Rightarrow f\) is one-one.
Let \(y\in R\). Let \(y=f(x_{ 0 })\).
Then \(3-4x_{ 0 }=y\Rightarrow x_{ 0 }=\frac { 3-y }{ 4 } \).
Now \(y\in R\Rightarrow \frac { 3-y }{ 4 } \in R\Rightarrow x_{ 0 }\in R\)
\(f(x_{ 0 })=3-4x_{ 0 }=3-4\frac { 3-y }{ 4 } =3-3+y=y\).
\(\because \) For each \(y\in R\) , there exists \(x_{ 0 }\in R\) such that 
\(f(x_{ 0 })=y\)
\(\because \)  \(f\) is onto
Hence, 'f' is ne-one and onto or bijective.
(ii) Here  f(1) = 1 + 1 = 2,
f(-1) = 1 + 1 = 2.
Now \(1\neq -1\) but f(1) = f(-1)
\(\because \)   \(f\) is onto
Also range of \(f\) is \([1,\infty )\neq R\)
\(\because \)  \(f\) is onto.
Hence, 'f' os not bijective

Let 'x' and 'y' the number of packets of food P and Q respectively.
We have:  \(x\ge 0\)  ...(1)
\(y\ge 0\)  ....(2)
\(12x+3y\ge 240\quad i.e.4x+y\ge 80\) .....(3)
\(4x+20y\ge 460\quad i.e.x+5y\ge 115\) .....(4)
\(6x+4y\le 300\quad i.e.3x+2y\le 150\) .....(5)
The mathematical problem is as below:
Minimize Z=6x+3y subject to (1)-(5)
Draw the lines
x = 0, 4x + y = 80, x + 5y = 80, x + 5y = 115 and 3x + 2y = 150.
The lines 3x + 2y = 150 and 4x + y = 80 meet at L (2, 72)
The lines 4x + y = 80 and x + 3y = 115 meet at M (15, 20)
The lines 3x + 2y = 150 and x + 5y = 115 meet at N(40,15).

The shaded portion represents the feasible region, which is bounded.
Apply Corner Point Method, we have:

Hence, minimum amount of vitamin A = 150 units.

(i) The number of trials is finite. When the drawing is done with replacement, the probability of success (say, red ball) is p = \(\frac{7}{16}\)which is same for all six trials (draws). Hence, the drawing of balls with replacements are Bernoulli trials.
(ii) When the drawing is done without replacement, the probability of success (i.e., red ball) in first trial is \(\frac{7}{16}\)in 2nd trial is \(\frac{6}{15}\) if the first ball drawn is red or \(\frac{7}{15}\) if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trials, hence the trials are not Bernoulli trials

Note that f '(x) = – sin x
(a) Since for each x \(\in\) (0, \(\pi\)), sin x > 0, we have f '(x) < 0 and so f is decreasing in (0, \(\pi\)).
(b) Since for each x \(\in\) (\(\pi\), 2\(\pi\)), sin x < 0, we have f '(x) > 0 and so f is increasing in (\(\pi\), 2\(\pi\)).
(c) Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2\(\pi\)).

\(We\quad have:f(x)={ x }^{ 3 }+{ x }^{ 2 }-1\)
Which is polynomial function
\(and\ { D }_{ f }=R\)
\(Let\quad c\in { D }_{ f }\)
\(Then\ \lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ ({ x }^{ 3 }+{ x }^{ 2 }-1) } \)
\(={ c }^{ 3 }+{ c }^{ 2 }-1=f(c)\)
\(\Rightarrow \) f is continuous at x = c.
But c is arbitrary.
Hence, f is continuous at each of its domains.

\( f(x)=\begin{cases} -x,if\quad x<0 \\ x,\quad if\quad x\ge 0. \end{cases}\)
Clearly the function is defined at 0 and f(0) = 0. Left hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (-x) } =0\)
Similarly, the right hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (x) } =0\)
Thus, the left hand limit, right hand limit and the value of the function coincide at x = 0.
Hence, f is continuous at x = 0.

Let events B₁, B₂, B₃ be the following :
B₁ : the bolt is manufactured by machine A
B₂ : the bolt is manufactured by machine B
B₃ : the bolt is manufactured by machine C
Clearly, B₁, B₂, B₃ are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space.
Let the event E be ‘the bolt is defective’.
The event E occurs with B₁ or with B₂ or with B₃. Given that,
P(B₁) = 25% = 0.25, P (B₂) = 0.35 and P(B₃) = 0.40
Again P(E|B₁) = Probability that the bolt drawn is defective given that it is manufactured by machine A = 5% = 0.05
Similarly, P(E|B₂) = 0.04, P(E|B₃) = 0.02.
Hence, by Bayes' Theorem, we have
\(\mathrm{P}\left(\mathrm{B}_{2} \mid \mathrm{E}\right) =\frac{\mathrm{P}\left(\mathrm{B}_{2}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{2}\right)}{\mathrm{P}\left(\mathrm{B}_{1}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{1}\right)+\mathrm{P}\left(\mathrm{B}_{2}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{2}\right)+\mathrm{P}\left(\mathrm{B}_{3}\right) \mathrm{P}\left(\mathrm{E} \mid \mathrm{B}_{3}\right)} \)
\(=\frac{0.35 \times 0.04}{0.25 \times 0.05+0.35 \times 0.04+0.40 \times 0.02} \)
\(=\frac{0.0140}{0.0345}=\frac{28}{69} \)

\(\overrightarrow { AL } +\overrightarrow { BM } +\overrightarrow { CN } \)=\(\lambda \)\(\overrightarrow { CK } \)

Since marginal cost is the rate of change of total cost with respect to the output, we have
Marginal \(\operatorname{cost}(\mathrm{MC})=\frac{d C}{d x}=0.005\left(3 x^{2}\right)-0.02(2 x)+30\)
When x = 3, MC = 0.015(3² ) − 0.04(3) + 30
= 0.135 – 0.12 + 30 = 30.015
Hence, the required marginal cost is Rs. 30.02 (nearly).

\(\text { Consider } \frac{1}{2} \sin x(2 \sin 3 x \sin 2 x) =\frac{1}{2} \sin x(\cos x-\cos 5 x) \)
\(=\frac{1}{4}(2 \sin x \cos x-2 \cos 5 x \sin x)=\frac{1}{4}(\sin 2 x-\sin 6 x+\sin 4 x) \)
\(\therefore \int \sin x \sin 2 x \sin 3 x d x =\frac{1}{4} \int(\sin 2 x-\sin 6 x+\sin 4 x) d x \)
\(=\frac{1}{4}\left[\frac{-\cos 2 x}{2}+\frac{\cos 6 x}{6}-\frac{\cos 4 x}{4}\right]+C\)

Let f be continuous at a critical point c in I. Then
(i) If f ′(x) changes sign from positive to negative as x increases through c, i.e., if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.
(ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e., if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.
(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection

Let numbers be x, y, z then
x + y + z = -1
2y + 3z = 5
x + y – z = -1
\(x=-\frac { 7 }{ 2 } \), y = \(\frac { 5 }{ 2 } \), z = 0

\(\int { \frac { \sqrt { x^{ 2 }+1 } \left\{ log({ x }^{ 2 }+1)-2logx \right\} }{ x^{ 4 } } } dx\)
\(=\sqrt { 1+\frac { 1 }{ { x }^{ 2 } } } \left( log\left( 1+\frac { 1 }{ x^{ 2 } } \right) \right) \frac { 1 }{ x^{ 3 } } dx\)
Let, \(1+\frac { 1 }{ { x }^{ 2 } } ={ t }^{ 2 }\)
\(\Rightarrow \frac { -2 }{ { x }^{ 3 } } dx=2tdt\)
\(\Rightarrow \frac { 1 }{ { x }^{ 3 } } dx=-tdt\)
\(=-\int { t(2logt)t\quad dt+ } -2\int { logt.{ t }^{ 2 }dt } \)
\(=-2logt.\frac { { t }^{ 3 } }{ 3 } +\int { 2\frac { 1 }{ t } .\frac { t^{ 3 } }{ 3 } dt } \)
\(=-\frac { 2 }{ 3 } logt.{ t }^{ 3 }+\frac { 2 }{ 9 } { t }^{ 3 }+C\)
\(=\frac { 2 }{ 3 } \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) ^{ 3 }\left( -log\left( 1+\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ 3 } \right) +C\)

E₁ = Event that 1 occur;
E₂ = Event that 1 does not occur;
A = Event that the man reports that 1 occur.
\(P({ E }_{ 1 })=\frac { 1 }{ 6 } ;P({ E }_{ 2 })=\frac { 5 }{ 6 } ;\)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 5 } ;P(A/{ E }_{ 2 })=\frac { 2 }{ 5 } ;\)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 6 } .\frac { 3 }{ 5 } }{ \frac { 1 }{ 6 } .\frac { 3 }{ 5 } +\frac { 5 }{ 6 } .\frac { 2 }{ 5 } } =\frac { 3 }{ 13 } \)

Let P(x, y) be the nearest point
\(\therefore \ D=\sqrt{(x-11a)^2+y^2}\)
\(S=(x-11a)^2+y^2=(x-11a)^2+4ax\)
\(\frac{dS}{dx}=2(x-11a)+4a\)
\(\frac{dS}{dx}=0\Rightarrow x=9a\)
\(\therefore\ y=pm6a\)
\(\Rightarrow\ \frac{d^2S}{dx^2}=2>0\)
For minimum distance, coordinates are \(p(9a,\pm6a)\)

We know that A = I₃A
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow 2{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 2 & 4 & 4 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow 4{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 8 & 4 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 0 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2}\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }+{ 3R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ 4R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }-\frac { 4 }{ 3 } { R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 16 }{ 3 } & -\frac { 8 }{ 3 } \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] \)
Applying \({ R }_{ 1 }\rightarrow \frac { 1 }{ 8 } { R }_{ 1 },{ R }_{ 2 }\rightarrow \frac { 1 }{ 3 } { R }_{ 2 }\) and R₃ = - R₃
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] A\)
\({ A }^{ -1 }=\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \)
Matrix representation of given linear equation is
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \)  AX = B
\(\Rightarrow \)  X = A^-1B
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 1 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0+\frac { 10 }{ 3 } -\frac { 7 }{ 3 } \\ 19-\frac { 65 }{ 3 } +\frac { 14 }{ 3 } \\ -19+20+0 \end{matrix} \right] \)
\(\therefore \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right] \)
\(\therefore \)  x = 1, y = 2, z = 1

Use A=lA. Proceed. [Refer 2]

\(\left[ \begin{matrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{matrix} \right] \)

We have 1 + x² > x²
\(\Rightarrow \sqrt { 1+{ x }^{ 2 } } >x\) for x ∈ [1, 2]
\(\Rightarrow \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } <\frac { 1 }{ x } \)
\(\Rightarrow \int _{ 1 }^{ 2 }{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } dx } <\int _{ 1 }^{ 2 }{ \frac { dx }{ x } } \Rightarrow { I }_{ 1 }<{ I }_{ 2 }\)

As y’ = 3x² – 3, for a point of absolute maximum or minimum y’=0 ⇒ x = ± 1.
y]_x=0 = 2,
y]_x=1 = 1 – 3 + 2 = 0,
y]_x=-1 = -1 +3+ 2 = 4,
y]_x=2 = 8 – 6 + 2 = 4

y = tan^-1 \(\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \) 
= tan^-1 \(\left( \frac { \pi }{ 4 } \right) -{ tan }^{ -1 }{ x }^{ 2 }\)
\({ y }^{ ' }=0-\frac { -1 }{ 1+{ x }^{ 4 } } 2x=\frac { -2x }{ 1+{ x }^{ 4 } } \)

As \(\lim _{ x\rightarrow 0 }{ \left( \frac { sinx }{ x } +cosx \right) } \) 
= 1 + 1 = 2 2k
⇒ k = 1

As diag (3, -1) is a diagonal matrix. Its order is 2 × 2 with diagonal elements 3 and (-1).