Let E₁ and E₂ be the events of choosing a boy and a girl respectively from the class.
Given that the number of boys to the number of girls = 1: 2
\(\therefore P({ E }_{ 1 })=\frac { 1 }{ 1+2 } =\frac { 1 }{ 3 } \) and P(E₂) = \(\frac { 2 }{ 1+12 } =\frac { 2 }{ 3 } \)
Let A be the event that a student chosen will get first class
Given P(A/E₁) = 0.28 and P(A/E₂) = 0.25
\(\therefore\)  By theorem of total probability, 
P(A) = P(E₁).P(A/E₁) + P(E₂).P(A/E₂)
\(\Rightarrow \quad =\frac { 1 }{ 3 } \times 0.28+\frac { 2 }{ 3 } \times 0.25\)
\(=\frac { 28 }{ 300 } +\frac { 50 }{ 300 } =\frac { 78 }{ 300 } =0.26\) 

\(\lim _{ x\rightarrow 0 }{ \frac { { x }^{ \frac { 2 }{ 3 } }-9 }{ x-27 } } =\lim _{ x\rightarrow 0 }{ \frac { { x }^{ \frac { 2 }{ 3 } }-{ (27) }^{ \frac { 2 }{ 3 } } }{ x-27 } } =\frac { 2 }{ 3 } { (27) }^{ \frac { 2 }{ 3 } -1 }=\frac { 2 }{ 3 } { (27) }^{ \frac { -1 }{ 3 } }\)
\(=\frac { 2 }{ 3(27)^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3({ 3 }^{ 3 })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(3) } =\frac { 2 }{ 9 } \)

The greatest integer function \(f(x)=\lfloor x\rfloor\) is not continuous at every integer point n, since \(\left.\lim _{x \rightarrow n^{-}} \mid x\right\rfloor=n-1\) and \(\lim _{x \rightarrow n^{+}}\lfloor x\rfloor=n .\) Thus f'(n) does not exist.

Given \(\overrightarrow{a}=2\hat{i}+2\hat{j}-\hat{k}\)
\(\overrightarrow{b}=6\hat{i}-3\hat{j}+2\hat{k}\)
\(\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}+2\hat{j}-\hat{k})(6\hat{i}-3\hat{j}+2\hat{k})=12-6-2=12-8=4\)

For 3-digit even numbers, the unit's place can be occupied by one of the 3 digits 2, 4 or 6. The remaining 5 digits can be arranged int he remaining 2 places in ₅P² ways.
\(\therefore\) By the multiplication rule, the required number of 3-digit even number is 3 \(\times\) ⁵P₂ = 3 \(\times\) 5 \(\times\) 4 = 60.

In the expansion of (1+a)^m+n
Coefficient of a^m = ^m+nC_m = \(\frac{(m+n)!}{m!(m+n-m)!}=\frac{(m+n)!}{m!n!}\)   --- (1)
Coefficient of aⁿ = ^m+nC_n = \(\frac{(m+n)!}{n!(m+n-m)!}=\frac{(m+n)!}{n!m!}\)  --- (2)
(1) = (2) ⇒ Coefficient of a^m = Coefficient of aⁿ.

In this example, it is easy to see that whenever a circle C₁ is related to C₂ then C₂ is also related to C₁. Hence the relation is symmetric, whereas in example 1 it does not hold as whenever 2 divides 4 then it does not imply 4 divides 2 as 4 does not divide 2.

\(\sin { \left( \pi +\theta \right) } =-\sin { \theta } \)
\(\sin { \left( \pi +\theta \right) } =\sin { \pi } \cos { \theta } +\cos { \pi } \sin { \theta } \)
\(=\left( 0 \right) \cos { \theta } +\left( -1 \right) \sin { \theta } \)
\(=0-\sin { \theta } \)
\(\sin { \left( \pi +\theta \right) } =-\sin { \theta } \)

Let I = \(\int { \frac { 1 }{ { x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { 1 }{ 3 } } } } \)dx
Here, the exponents of x are \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 3 } \) and the LCM of their denominator is 6.
So, to remove fractional exponents put x = t⁶ \(\Rightarrow\) dx = 6t⁵ dt.
I = \(\int { \frac { { 6t }^{ 5 } }{ { t }^{ 3 }+{ t }^{ 2 } } } dt=6\int { \frac { { t }^{ 5 }dt }{ { t }^{ 2 }\left( t+1 \right) } } =6\int { \frac { { t }^{ 3 } }{ t+1 } } dr=6\int { \frac { \left( { t }^{ 3 }+1 \right) -1 }{ t+1 } } dt\)
= \(\int { \frac { \left( t+1 \right) ^{ 3 }-3t\left( t+1 \right) -1 }{ t+1 } } dt=6\) \(\int { \left[ \frac { \left( t+1 \right) ^{ 3 } }{ t+1 } -\frac { 3t(t+1) }{ t+1 } -\frac { 1 }{ t+1 } \right] dt } \)
= \(6\left[ f\left( t+1 \right) ^{ 2 }-3t-\frac { 1 }{ t+1 } dt \right] 6\left[ f{ t }^{ 2 }+2t+1-3t-\frac { 1 }{ t+1 } dt \right] =6\left( f{ t }^{ 2 }-t+1-\frac { 1 }{ t+1 } \right) dt\)
= \(6\left( \frac { { t }^{ 3 } }{ 3 } -\frac { { t }^{ 2 } }{ 2 } +t-log|t+1| \right) +c\)
= \(2.\sqrt { x } -3x^{ \frac { 1 }{ 3 } }+6x^{ \frac { 1 }{ 6 } }-6log|{ x }^{ \frac { 1 }{ 6 } }+1|+c\)

Consider the following events
E₁: Company insured scooter driver
E₂: Company insured a car driver
E₃: Company insured a truck driver
\(\therefore P({ E }_{ 1 })=\frac { 2000 }{ 2000+4000+6000 } =\frac { 2000 }{ 12000 } =\frac { 1 }{ 6 } \)
\(P({ E }_{ 2 })=\frac { 4000 }{ 12000 } =\frac { 1 }{ 3 } ,\)
\(P({ E }_{ 3 })=\frac { 6000 }{ 12000 } =\frac { 1 }{ 2 } \)
It is given that
P(A/E₁) = 0.01, P(A/E₁) = 0.03 and P(A/E₃) = 0.15
By Bayes' theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 6 } \times \frac { 1 }{ 100 } }{ \frac { 1 }{ 6 } \times \frac { 1 }{ 100 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 100 } +\frac { 1 }{ 2 } \times \frac { 15 }{ 100 } } =\frac { \frac { 1 }{ 600 } }{ \frac { 1 }{ 600 } +\frac { 1 }{ 100 } +\frac { 15 }{ 200 } } =\frac { \frac { 1 }{ 600 } }{ \frac { 52 }{ 600 } } \)
\(=\frac { 1 }{ 600 } \times \frac { 600 }{ 52 } =\frac { 1 }{ 52 } \)

Given \(x=4{ z }^{ 2 }+5\)
\(\Rightarrow \frac { dx }{ dz } =8z\)
\( \Rightarrow y=6{ z }^{ 2 }+7z+3\)
\(\Rightarrow \frac { dy }{ dz } =12z+7\)
\(\therefore \frac { dy }{ dx } =\frac { dy }{ dz } /\frac { dx }{ dz } =\frac { 12z+7 }{ 8z } \)
Differentiating again with respect to 'x', we get
\(\frac { d^{ 2 }y }{ dx^{ 2 } } =\frac { 8z(12)-(12z+7)8 }{ 64{ z }^{ 2 } } .\frac { dz }{ dx } \)
\(=\frac { 96z-96z-56 }{ 64{ z }^{ 2 } } \times \frac { 1 }{ 8z } =\frac { -56 }{ 64{ z }^{ 2 }\times 8 } =\frac { -7 }{ 64{ z }^{ 3 } } \)
\(\therefore \frac { d^{ 2 }y }{ dx^{ 2 } } =\frac { -7 }{ 64{ z }^{ 3 } } \)

Let \(\triangle\)=\(\left| \begin{matrix} x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \\ { x }^{ 3 } & { y }^{ 3 } & { z }^{ 3 } \end{matrix} \right| \)
Taking x, y, z common from C₁, C₂ and, C₃ repectively.
\(\triangle =xyz\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \)
Applying C₂ ⟶ C₂ - C₁ and C₃ ⟶ C₃ - C₁ we get,
\(\triangle =xyz\left| \begin{matrix} 1 & 0 & 1 \\ x & y-x & z-x \\ { x }^{ 2 } & { y }^{ 2 }-{ x }^{ 2 } & { z }^{ 2 }-{ x }^{ 2 } \end{matrix} \right| \)
Taking (y - x) and (z - x) common from C₂ & C₃ respectively|
\(\triangle\) = xyz(y - x)(z - x)\(\left| \begin{matrix} 1 & 0 & 0 \\ x & 1 & 1 \\ { x }^{ 2 } & y+x & z+x \end{matrix} \right| \)
Expanding along R₁ we get,
\(\triangle\) = xyz(y - x)(z - x)[z + x - y - x]
\(\triangle\) = xyz(y - x)(z - x)[z - y]
\(\triangle\) = xyz(x - y)(y - z)(z - x)
Hence proved.

Let A₁, A₂ and A₃ be the event of A, B, C becoming managers of the company respectively. Let X be the event that the office canteen will be improved.
Then, \(P\left(A_1\right)=\frac{5}{10}=0.5 \)
\(P\left(A_2\right)=\frac{3}{10}=0.3 \)
\(P\left(A_3\right)=\frac{2}{10}=0.2 \)
\(P\left(X / A_1\right)=0.4 \)
\(P\left(X / A_2\right)=0.5\)
\(P\left(X / A_3\right)=0.3\)
\(P\left(A_2 / X\right)=\frac{P\left(A_2\right) P\left(X / A_2\right)}{P\left(A_1\right) P\left(X / A_1\right)+P\left(A_2\right) P\left(X / A_2\right)}+P\left(A_3\right) P\left(X / A_3\right)\)
\(=\frac{0.3(0.5)}{0.5(0.4)+0.3(0.5)+0.2(0.3)} \)
\(=\frac{0.15}{0.2+0.15+0.06} \)
\(=\frac{0.15}{0.41}=\frac{15}{41}\)

\( \int\left[(x+4)^5+\frac{5}{(2-5 x)^4}-\operatorname{cosec}^2(3 x-1)\right] d x =\int(x+4)^5 d x+5 \int \frac{1}{(2-5 x)^4} d x\)
\( =\frac{(x+4)^6}{6}+5\left(\frac{-1}{3(2-5 x)^3 \times(-5)}\right) -\left(\frac{-\cot (3 x-1)}{3}\right)+c \)
\(=\frac{(x+4)^6}{6}+\frac{1}{3(2-5 x)^3}+\frac{\operatorname{cosec} 2(3 x-1) d x}{-\left(\frac{-\cot (3 x-1)}{3}\right)+c} \)

Given \(f(x)={x^2-2x-8\over x+2}\)|
f(x) does not exist at x = - 2
\(\therefore\) It has a removable discontinuity at x = - 2
\(lim_{x\rightarrow -2}f(x)=lim_{x\rightarrow -2}{x^2-2x-8\over x+2}\)
\(lim_{x\rightarrow -2}f(x)=lim_{x\rightarrow -2}{(x-4)(x+2)\over (x+2)}=lim_{x\rightarrow-2}x-4=-2-4=-6\)
\(\therefore\) The continuous function g(x) can be written as
g(x) = {\(\begin{matrix} \frac { { x }^{ 2 }-2x-8 }{ x+2 } & if\ x\ \neq -2 \\ 6& if\quad x=-2 \end{matrix}\)

\(lim_{x-1}{3\sqrt{7+x^3}-\sqrt{3+x^2}\over x-1}\)\(=lim_{x-1}{({7+x^3})^{1\over3}-8^{1\over3}\over 7+x^3-8} \times{x^3-1\over x-1}-{(3+x^2)^{1\over2}-4^{1\over2}\over 3+x^2-4}\times {x^2-1\over x-1}\)|

[\(\because\) when x\(\rightarrow\) 1, x³ \(\rightarrow\) 1 and 7 + x³\(\rightarrow\) 7+ 1 = 8, also when x \(\rightarrow\) 1, x² \(\rightarrow\) 1 and 3 + x²\(\rightarrow\) 4]\(=lim_{7+x^3\rightarrow8}{({7+x^3})^{1\over3}-8^{1\over3}\over 7+x^3-8} (x^2+x+1)-lim_{3+x^2\rightarrow4}{(3+x^2)^{1\over2}-4^{1\over2}\over 3+x^2-4}\times(x+1)\)
\({1\over3}(8)^{{1\over3}-1}(1^2+1+1)-{1\over2}(4^{{1\over2}-1})(1+1)\) \([\because lim_{x \rightarrow a}x^na^n=na^{n-1}]\)

\(={1\over 8^{2\over3}}-{1\over 4^{1\over2}}={1\over (2^{3})^{2\over3}}-{1\over (2^{2})^{1\over2}}\)\(={1\over2^2}-{1\over2}={1\over4}-{1\over2}\)
\(={1-2\over4}={-1\over4}\)
\(\)

Let \(\overrightarrow{a}=\) m (\(\hat{i}+\hat{j}+\hat{k}\))
\(|\overrightarrow{a}|=m\sqrt{1^2+1^2+1^2}=m\sqrt{3}\)
To make \(\overrightarrow{a}\) as a unit vector, \(|\overrightarrow{a}|=\pm1\)
\(\therefore m \sqrt{3} =\pm 1\Rightarrow m = \pm {1\over \sqrt{3}}.\)

\(LHS=\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix}^2=\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix}\times \begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix}\)
\(=\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix} \times(-1)(-1)\begin{vmatrix} 1 &x &x \\ -x & -1 &-x \\- x &-x &-1 \end{vmatrix} \)
= \(\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix} \times \begin{vmatrix} 1 &x &x \\ -x & -1 &-x \\- x &-x &-1 \end{vmatrix} \)
= \(\begin{vmatrix} 1-x^2-x^2 &x-x-x^2 &x-x^2-x \\ x-x-x^2 & x^2-1-x^2 &x^2-x-x \\ x-x^2-x &x^2-x-x &x^2-x^2-1 \end{vmatrix}\)
= \(\begin{vmatrix} 1-2x^2 &-x^2 &-x^2 \\- x^2 & -1 &x^2-2x \\ -x^2 &x^2-2x &-1 \end{vmatrix}\)
 = RHS.

Given \(A=\left[\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right]\left[\begin{array}{lll} + & - & + \\ - & + & - \\ + & - & + \end{array}\right]\)
Cofactors of elements of second row
\(|A|=-2\left|\begin{array}{ll} 3 & 8 \\ 2 & 3 \end{array}\right|+0-1\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|\)
= 2(9-16)-1(10-3)
= -2(-7) - 1(7) = 14 - 7= 7
\(|A|=7\)

Let p(n) be the statement that 3n² > (n + 1)² with n\(\ge\) 2. Therefore the first stage is n = 2.
Now, P(2) = 3 x 2² = 12 and 3² = 9. As 12> 9 we get P(2) is true.
We assume that p(n) is true for n = k.
Now,
P(k+ 1) 3(k+ 1)² = 3k² + 6k+ 3
P(k) + 6k+ 3
> (k + 1)² + 6k + 3
k² + 8k+4
k² + 4k+4 + 4k
(k+ 2)² + 4k
> (k+ 2)² since k> 0.
This is the statement P(k + 1). The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction, for all n\(\ge\) 2, 3n² > (n + 1)².

sin θ + cos θ = \(\sqrt2\)
Multiplying by\(\frac{1}{\sqrt2}\)  on both sides
\(\frac { 1 }{ \sqrt { 2 } } sin\theta +\frac { 1 }{ \sqrt { 2 } } cos\theta =\frac { 1 }{ \sqrt { 2 } } \sqrt { 2 } =1\)
\(cos\theta cos\frac { \pi }{ 4 } +sin\theta sin\frac { \pi }{ 4 } =1\)
The general soln is cos (θ - \(\pi\)/4) = 1
cos (θ - \(\pi\)/4) = cos θ
θ - \(\pi\)/4 = 2n\(\pi\) ± 0
​​​​​​​θ = 2n\(\pi\)+ \(\pi\)/4

\(x-\frac { 1 }{ 4 } >\frac { 1 }{ 2 } x-\frac { 3 }{ 4 } \) (or)  \(x-\frac { 1 }{ 4 } <\frac { 3 }{ 4 } -\frac { 1 }{ 2 } \)
\(x-\frac { 1 }{ 2 } x>-\frac { 3 }{ 4 } -\frac { 1 }{ 4 } \left( =-\frac { 2 }{ 4 } =\frac { -1 }{ 2 } \right) \)
\({1\over 2}x>{-1\over 2}\)
⇒ x >-1        ....(1)
\(x+\frac { 1 }{ 2 } x<\frac { 3 }{ 4 } +\frac { 1 }{ 4 } =1\)
\(\frac { 3 }{ 2 } x<1\)
\(x<\frac { 2 }{ 3 } \)  .....(2)
From (1) and (2)
\(-1

If equation of line is 12x² + 2kxy + 2y² + 11x - 5y + 2 = 0
Here a = 12, 2h = 2k, b = 2, 2g = 11, 2f = -5 , c = 2
\(h=k\ g=\frac { 11 }{ 2 } f=\frac { -5 }{ 2 } ,c=-5\)
The condition for pair of straight line is abc + 2fgh - af²- bg²- ch² = 0
\(\Rightarrow \quad 12(2)(2)+)-5)\left( \frac { 1 }{ 2 } \right) (k)-12\left( \frac { 25 }{ 4 } \right) -2\left( \frac { 121 }{ 4 } \right) -2\left( { k }^{ 2 } \right) =0\) 
\(\Rightarrow \quad 48-\frac { 55k }{ 2 } -\frac { 300 }{ 4 } -\frac { 242 }{ 4 } -2{ k }^{ 2 }=0\)
Multiplying by 4 we get,
192 - 110k - 300 - 242 - 8k² = 0
\(\Rightarrow\) -8k²-110k-350 = 0
\(\Rightarrow\)  4k² + 55k + 175 = 0
\(\Rightarrow\) (k-5)(4k + 35) = 0
\(\Rightarrow \ k=-5\ or\ k=\frac { -35 }{ 4 } \)

Let T_n be the nth term of the given series
Then T_n = 1 + 4 + 4² + 4³ + ...
\(=1\left(4^n-1\over 4-1\right)\)
\(={4^n-1\over 3}\)
Let S_n be the sum to n terms of the given series
Then \(S_n={\sum_{k=1}^n}T_k=\sum_{k=1}^n{4^n-3\over3}\)
\(⇒\ S_n={1\over3}\left[ \sum_{k=1}^n4^n-\sum_{k=1}^n3\right]\)
\(⇒\ S_n= {1\over3}[4^1+4^]+...+4^n-3^n\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]={1\over 3}\left[4{(4^n-1)-9n\over3}\right]\)

\({ S }_{ n }=\frac { 4 }{ 9 } \left[ \left( { 4 }^{ n }-1 \right) -n/3 \right] \)

Let y = 3x -5.
\(\Rightarrow y+5=3x\Rightarrow \frac { y+5 }{ 3 } =x\)
Let g(y) = \(\frac { y+5 }{ 3 } \)
\(gof(x)=g(f(x))=g(3x-5)=\frac { 3x-5+5 }{ 3 } =\frac { 3x }{ 3 } =y\)
Also f o g(y) = f(g(y)) = \(f\left( \frac { y+5 }{ 3 } \right) =3\left( \frac { y+5 }{ 3 } \right) -5=y+5-5=y\)
Thus g o f = Ix and fog = Iy.
This implies that f and g are bijections and inverses to each other.
Hence f is a bijection and f^-1(y) = \(\frac { y+5 }{ 3 } \)
Replacing y by x we get, f^-1(x) = \(\frac { x+5 }{ 3 } \)

\(LHS=sin\frac { \theta }{ 2 } sin\frac { 7\theta }{ 2 } +sin\frac { 3\theta }{ 2 } sin\frac { 11\theta }{ 2 } \)
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) -cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) -cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ cos(-3\theta )-cos(4\theta )+cos(-4\theta )-cos7\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos4\theta +cos4\theta -cos7\theta \right] \quad \quad \left[ \because cos(-\theta )=cos\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos7\theta \right] =\frac { 1 }{ 2 } \left[ 2sin\left( \frac { 3\theta +7\theta }{ 2 } \right) .sin\left( \frac { 7\theta -3\theta }{ 2 } \right) \right] \)
\(=sin5\theta .sin2\theta =RHS.\)

\(\text { Let } \sqrt{x}=t\)
\(\text { Then } x=t^{2}\)
\(\therefore \frac{d x}{d t}=2 t\)
\(\therefore d x=2 t d t\)
\(\therefore \int \sin \sqrt{x} d x=\int \sin t \times 2 t d t\)
\(=2 \int t \sin t d t\)
\(\text { Applying Bernoulli's formula }\)
\(=2[t(-\cos t)-1(-\sin t)]+c\)
\(\text { Putting } t=\sqrt{x}\)
\(=2[-\sqrt{x} \cos \sqrt{x}+\sin \sqrt{x}]+c\)

\(f^{\prime}\left(a^{-}\right) =\lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a}=\lim _{x \rightarrow a^{-}} \frac{(2 a-x)-(2 a-a)}{x-a} \)
\(=\lim _{x \rightarrow a^{-}} \frac{-x+a}{x-a}=-1 \)
\(f^{\prime}\left(a^{+}\right) =\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} \)
\(=\lim _{x \rightarrow a^{-}} \frac{(3 x-2 a)-(3 a-2 a)}{x-a} \)
\(=\lim _{x \rightarrow a^{-}} \frac{3 x-2 a-a}{x-a}=\lim _{x \rightarrow a^{-}} \frac{3 x-3 a}{x-a}=3 \)
\(f^{\prime}\left(a^{-}\right) \neq f^{\prime}\left(a^{+}\right) \)
\(\therefore \text { It is not differentiable at } x=a\)
\(\therefore f^{\prime}(x) \text { does not exist }\)

\(\lfloor x\rfloor= \begin{cases}2, & 2 \leq x<3 \\ 3, & 3 \leq x<4\end{cases} \)
\(\therefore \lim _{x \rightarrow 3^{-}}\lfloor x\rfloor=2 \text { and } \lim _{x \rightarrow 3^{+}}\lfloor x\rfloor=3 \)
\(\therefore \text { Limit does not exist }\)

\(\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] A=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right]\)
\(\text { Let } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\)
\(\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right]\)
\(\left[\begin{array}{cc} a+3 c & b+3 d \\ 0+c & 0+d \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right]\)
\(c=0,d=-l\)
\(a+3 c=1 \quad b+3 d=1\)
\(a+0=1 \quad b-3=1\)
\(a=1 \quad b=4\)
\(\therefore A=\left[\begin{array}{cc} 1 & 4 \\ 0 & -1 \end{array}\right]\)

Number of rectangles in the chessboard is
\({ }^{9} C_{2} \times{ }^{9} C_{2} =\frac{9 \times 8}{1 \times 2} \times \frac{9 \times 8}{1 \times 2} \)
\(=36 \times 36=1296 \)

\(x^{2}-\mathrm{k} x+\mathrm{c}=0\)
a and b are the roots
\(\therefore a+b=k, a b=c\)
To find
\(\sqrt{(a-b)^{2}+0^{2}}=a-b=\sqrt{(a+b)^{2}-4 a b}=\sqrt{k^{2}-4 c}\)
 

\(\left(2 \cos ^{2} \frac{\pi}{16}\right)\left(2 \cos ^{2} \frac{3 \pi}{16}\right)\left(2 \cos ^{2} \frac{5 \pi}{16}\right)\left(2 \cos ^{2} \frac{7 \pi}{16}\right) \)
\(=2^{4}\left[\cos \frac{\pi}{16} \cos \frac{3 \pi}{16} \cos \frac{5 \pi}{16} \cos \frac{7 \pi}{16}\right]^{2} \)
\(=\left(\cos \frac{8 \pi}{16}+\cos \frac{6 \pi}{16}\right)^{2}\left(\cos \frac{8 \pi}{16}+\cos \frac{2 \pi}{16}\right)^{2} \)
\(=\left(\cos \frac{6 \pi}{16}\right)^{2}\left(\cos \frac{2 \pi}{16}\right)^{2} \)
\(=\left\{\frac{1}{2}\left[\cos \frac{8 \pi}{16}+\cos \frac{4 \pi}{16}\right]\right\}^{2} \)
\(=\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8} \)

\(n[(A \times B) \cup(A \times C)]=n(A) \times n(B \cup C)\)
= 2 \(\times\) 3 = 6

\({ x }^{ 2 }+2xy+{ y }^{ 3 }=42\)
Differentiating both sides with respect to 'x' we get,
\(2x+2\left[ x.\frac { dy }{ dx } +y(1) \right] +3{ y }^{ 2 }\frac { dy }{ dx } =0 \Rightarrow 2x+2x\frac { dy }{ dx } +2y+3{ y }^{ 2 }\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } (2x+3{ y }^{ 2 })=-2x-2y \Rightarrow \frac { dy }{ dx } =\frac { -2\left( x+y \right) }{ 2x+3{ y }^{ 2 } } \)

\(=x^2+4 x+2 \)
\(=\int \frac{1}{\sqrt{x^2+4 x+2}} d x \)
\(=\left(x^2+4 x\right)+2 \)
Completing the square on x
\( =\left(x^2+4 x+2^2\right)-2^2+2\)
\(=(x+2)^2-2 \)
\(=(x+2)^2-(\sqrt{2})^2 \)
\(\therefore \int \frac{1}{\sqrt{x^2+4 x+2}} d x =\int \frac{1}{\sqrt{(x+2)^2-(\sqrt{2})^2}} d x \)
\(=\int \frac{1}{\sqrt{u^2-a^2}} d u, u=x+2, a=\sqrt{2}\)
\(=\frac{1}{2 a} \log \left|u+\sqrt{u^2-a^2}\right|+c\)
\(=\frac{1}{2 \sqrt{2}} \log \left|x+2+\sqrt{x^2+4 x+2}\right|+c\)

\( \int \frac{1}{1+x^2} d x =\tan ^{-1} x+c\)
\( \therefore \int \frac{1}{1+36 x^2} d x =\int \frac{1}{1+(6 x)^2} d x \)
\(=\frac{\tan ^{-1} 6 x}{6}+c \)

Let the H.P. be \(\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},...\)
\(\therefore { T }_{ m }=\frac { 1 }{ a+\left( m-1 \right) d } =n\)   and \(\therefore { T }_{ n }=\frac { 1 }{ a+\left( n-1 \right) d } =m\)
a + (m - 1)d = \(\frac{1}{n}(1)\) and a + (n - 1)d = \(\frac{1}{m}\)
(1) - (2) \(\Rightarrow\) (m - 1 - n + 1)d = \(\frac{1}{n}-\frac{1}{m}\) \(\Rightarrow\) (m - n) d = \(\frac{m-n}{mn}\Rightarrow d=\frac{1}{mn}\)
\({ T }_{ p }=\frac { 1 }{ a+\left( p-1 \right) d } =\frac { 1 }{ \frac { 1 }{ mn } +\left( b-1 \right) \frac { 1 }{ mn } } =\frac { mn }{ 1+p-1 } \)
\({ T }_{ p }=\frac { mn }{ p } \)

Let OL be the perpendicular from the origin to the required line.
Given OL = p = \(3\sqrt { 2 } \) and \(\angle XOL\) = 75^o
p = \(3\sqrt { 2 } \)  and \(\alpha \) = 75^o
Equation of the line is normal form is \(x\cos { \alpha } +y\sin { \alpha } =p\)
\(\Rightarrow x\cos { { 75 }^{ o } } +y\sin { { 75 }^{ o } } =3\sqrt { 2 }......(1)\)
Now cos 75^o = cos (45^o + 30^o)
= cos 45^o cos 30^o - sin 45^o sin 30^o
= \(\frac { \sqrt { 3 } }{ 2\sqrt { 2 } } -\frac { 1 }{ 2\sqrt { 2 } } =\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } }.....(2)\)
sin 75^o = sin (45^o + 30^o) = sin 45^o cos 30^o + cos 45^o sin 30^o
= \(\frac { \sqrt { 3 } }{ 2\sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } .......(3)\)
Substituting (2) and (3) in (1) we get,
\(x\left( \frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } \right) +y\left( \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) =3\sqrt { 2 } \)
\(\Rightarrow x\left( \sqrt { 3 } -1 \right) +y\left( \sqrt { 3 } +1 \right) =3\sqrt { 2 } \times 2\sqrt { 2 } =12\)             
\(\Rightarrow\) \(x(\sqrt{3}-1)+y(\sqrt{3}+1)\)=12 which is the required equation

Given equation of line is x + y - 2 = 0  .......(1)
Any line perpendicular to x + y - 2 = 0 will be of the form x - y + k = 0
This line passes through (-10, -2)
\(\therefore\)  - 10 + 2 + k = 0
\(\Rightarrow\)  k = 8
\(\therefore\) Equation of AB is x - y + 8 = 0....(2)
The foot of the perpendicular is the point of intersection of (1) and (2).
 
(1)  \(\Rightarrow \)   x + y - 2 = 0
(2) \(\Rightarrow \)   x - y + 8 = 0
________________
              2x + 6 = 0
 \(\Rightarrow \)           x = -3
Substituting x = -3 in (1) we get,
-3 + y - 2 = 0
\(\Rightarrow \)  y = 5
\(\therefore\)  Co-ordinate of the foot of the perpendicular B is (-3, 5). ...(2)
Now length of the perpendicular from (-10, -2) to the line x + y - 2 = 0 is
\(=\pm \left( \frac { -10-2-2 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } \right) =\pm \left( \frac { -14 }{ \sqrt { 2 } } \right) =\frac { 14 }{ \sqrt { 2 } } \)
\(\frac { 14 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 14\sqrt { 2 } }{ 2 } =7\sqrt { 2 } \)
[ Since length of the perpendicular cannot be negative]
\(=\frac { 14 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 14\sqrt { 2 } }{ 2 } =7\sqrt { 2 } \)

Given log 2 = 0.3010
log (0.5)¹⁰ = \(log\left({{1}\over{2}} \right)^{10}=log\ 2^{-10}\)
= -10 log 2 = - 10 (0.3010) = - 3.010
= -3-0.0010- (-3-1) + (-0.010)
= -4 + 0.990 = 4.990.
\(\therefore\) Characteristic of log (0.5)¹⁰ = 4 ie -4
\(\therefore\) Number of Zeroes immediately after the decimal part = 4 - 1 = 3.
\(\therefore\) First significant digit is at 4th place after decimal.