Total number of bulbs = 10
Number of defective bulbs = 2 .
ஃ Number of good bulbs = 10 - 2 = 8
Now selecting 5 from the 10 bulbs can be done in ¹⁰C₅ ways.
(i.e.,) \(n(S)=^{ 10 }{ C }_{ 5 }=\cfrac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } =252\) 
Let A be the event of selecting 5 good bulbs (from 8 good bulbs)
\(\therefore n(A)={ 8C }_{ 5 }=8{ C }_{ 3 }=\cfrac { 8\times 7\times 6 }{ 3\times 2\times 1 } =56\)
\(\therefore P(A)=\cfrac { n(A) }{ n(S) } =\cfrac { 56 }{ 252 } =\cfrac { 2 }{ 9 } \)

We have,\(\underset { x\rightarrow 1 }{ lim } f(x)=f(1)\) 
\(\Leftrightarrow \underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=\underset { x\rightarrow ^{ + } }{ lim } f(x)=f(1)\) 
\(\Leftrightarrow \underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=f(1)and,\underset { x\rightarrow ^{ + } }{ lim } f(x)=f(1)\) 
\(\Leftrightarrow \underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=4and,\underset { x\rightarrow 0 }{ lim } f(1+h)=4\) 
\(\Leftrightarrow \underset { x\rightarrow 0 }{ lim } \left\{ a+b(1-h) \right\} =4and,\underset { x\rightarrow 0 }{ lim } \left\{ b-a\left( 1+h \right) \right\} =4\) 
\(\Leftrightarrow a+b=4and,b-a=4\) 
\(\Leftrightarrow a=0,b=4\)

Let I = \(\int { \frac { { e }^{ -x } }{ 16+9{ e }^{ -2x } } } \) = \(\int { \frac { { e }^{ -x }dx }{ { 4 }^{ 2 }+(3e^{ -x })^{ 2 } } } \)
Let  3e^-x = t \(\Rightarrow\) -3e^x dx = dt \(\Rightarrow\) e^-x dx = -\(\frac { dt }{ 3 } \)
\(\therefore\) I = \(\int { \frac { \frac { -dt }{ 3 } }{ { 4 }^{ 2 }+{ t }^{ 2 } } } =-\frac { 1 }{ 3 } \int { \frac { dt }{ { 4 }^{ 2 }+{ t }^{ 2 } } =-\frac { 1 }{ 3 } \times \frac { 1 }{ 4 } { tan }^{ -1 }\left( \frac { t }{ 4 } \right) } +c\)
= \(\frac { -1 }{ 12 } { tan }^{ -1 }\left( \frac { 3e^{ -x } }{ 4 } \right) +c\)

Differentiating the function implicitly with respect to x, we get
\({dy\over dx}={{dy\over dt}\over {dx\over dt}}={a \ cos \ t\over -a \ sin \ t}=-{cos \ t\over sin \ t}\)
\({d^2y\over dx^2}={d\over dx}({dy\over dx})\)
\(={d\over dx}({-cos \ t \over sin \ t})\)
\(=\frac{d}{d t}\left(\frac{-\cos t}{\sin t}\right) \frac{d t}{d x}=-\left[-\operatorname{cosec}^2 t\right] \times \frac{1}{x^{\prime}(t)}\)
\(=cosec^2t\times {1\over -a \ sin \ t}\)
\(=-\frac{\operatorname{cosec} e^3 t}{a}\)

We know that, y = e^x is the inverse function of y = log_ex and hence y = e^x is the reflection of y -= log_ex about y = x.

We have \({\left( 2x-{1\over 2x} \right)}^{4}\) = ⁴C₀(2x)⁴ \({\left(-{1\over 2x} \right)}^{0}\) +⁴C₁(2x)³\({\left(-{1\over 2x} \right)}^{1}\) + ⁴C₂(2x)²\({\left(-{1\over 2x} \right)}^{2}\)+⁴C₃(2x)¹\({\left(-{2\over x} \right)}^{3}\) + ⁴C₄(2x)⁰\({\left(-{1\over 2x} \right)}^{4}\)
= (2x)⁴ - 4(2x)³\({\left({1\over 2x} \right)}\) + 6(2x)²\({\left({1\over 2x} \right)}^{2}\)- 6(2x)\({\left({1\over 2x} \right)}^{2}+{\left({1\over 2x} \right)}^{4}\) 
\(=16x^4-16x^2+6-{3\over2x^2}+{1\over16x^4}\)

Observe that \(\sqrt{x^2-10x+25}=\sqrt{(x-5)^2}=|x-5|\)

Note that the degree of f(x) is 3 and the leading coefficient is 1. Since g(x) divides f(x), we have f(x) = (x + b) g(x), for some b ∈ R. Thus, x³ - 3px + 2q = (x + b) (x² + 2ax + a² ).
Equating like coefficients on both sides, we have 2a + b = 0, a² + 2ab = -3p and 2q = ba², Thus b = -2a, p = a² and q = -a³
Now, q = -a³ = -a(a²) = -ap, which gives ap + q = 0.

There are 10 questions in Part-A, out of which 8 questions can be chosen in 10C₈ ways.
Similarly from Part-B, contains using 10 questions, 5 questions can be chosen in 10C₅ ways.
Hence, the total number of ways of selecting 8 questions from Part-A and 5 from Part-B.
= 10C₈ \(\times\)10C₅ = \(\frac { 10! }{ 8!2! } \times \frac { 10! }{ 5!\times 5! } \)

= 11340.

Given \(log_2\left({{\sqrt [ 3 ]{4 } }\over{4^2\sqrt{8}}} \right)\)
= \({log}_{2}\sqrt [ 3 ]{4 }-{log}_{2}4^2(\sqrt{8})\)
= \(log_24^{1/3}-[log_24^2+log_2\sqrt{8}]\)
= log²(2²)^1/3- log²(2²)²- log²(2³)^1/2
= log²2^1/3- log²2⁴- log²2^3/2
\(={{2}\over{3}}(1)-4(1)-{{3}\over{2}}(1)\)   \([\because {log}^{2}_{2}=1]\)
\(={{4-24-9}\over{6}}={{-29}\over{6}}\)

\(LHS=1+cos2x+cos4x+cos6x\quad \left[ \because 1+cos2x=2{ cos }^{ 2 }x \right] \)
\(=2{ cos }^{ 2 }x+cos4x+cos6x\)
\(=2{ cos }^{ 2 }x+2cos\left( \frac { 4x+6x }{ 2 } \right) .cos\left( \frac { 4x-6x }{ 2 } \right) \quad \left[ \because cosA+cosB=2cos\left( \frac { A+B }{ 2 } \right) .cos\left( \frac { A-B }{ 2 } \right) \right] \)
= 2cos²x + 2cos 5x.cos(-x)
= 2cos²x + 2cos 5x cos x
\(=2cosx\left( 2cos\frac { x+5x }{ 2 } .cos\left( \frac { x-5x }{ 2 } \right) \right) \)
= 2cos x.2cos 3x.cos(-2x)
= 4cos x cos 3x.cos 2x
= RHS   \(\quad \left[ \because cos(-\theta =cos\theta \right] \)

\(A=\left[ \begin{matrix} 1 & -1 \\ 2 & -1 \end{matrix} \right] ;B=\left[ \begin{matrix} x & 1 \\ y & -1 \end{matrix} \right] \) 
\(A+B=\left[ \begin{matrix} 1 & -1 \\ 2 & -1 \end{matrix} \right] +\left[ \begin{matrix} x & 1 \\ y & -1 \end{matrix} \right] =\left[ \begin{matrix} 1+x & 0 \\ 2+y & -2 \end{matrix} \right] \) 

= \(\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right] \) 
\({ B }^{ 2 }=\left[ \begin{matrix} x & 1 \\ y & -1 \end{matrix} \right] \left[ \begin{matrix} { x }^{ 2 } & 1 \\ y & -1 \end{matrix} \right] =\left[ \begin{matrix} { x }^{ 2 }+y-1 & y-1 \\ xy-y & y+1 \end{matrix} \right] \) 
So, \({ A }^{ 2 }+{ B }^{ 2 }=\left[ \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right] +\left[ \begin{matrix} { x }^{ 20 }+y & x-1 \\ xy-y & y+1 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} { x }^{ 2 }+y-1 & x-1 \\ xy-y & y+1-1 \end{matrix} \right] =\left[ \begin{matrix} { x }^{ 2 }+y-1 & x-1 \\ xy-y & y \end{matrix} \right] \)    ..(2)
Given \((1)=(2)\Rightarrow \left[ \begin{matrix} (1+x)^{ 2 } & 0 \\ (2x+y)(1+x)-2(2+y) & 4 \end{matrix} \right] =\left[ \begin{matrix} { x }^{ 2 }+y-1 & x-1 \\ xy-y & y \end{matrix} \right] \) 
(i,e) x - 1 = 0 ⇒ x = 1
y = 4 So, x = 1, y = 4

Given, one of the root is x = 1
Using synthetic division we get,
(x -1)(x - 2)(x - 3) = 0
Other roots 2, 3
\(\therefore\) x = 1,2,3

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,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 6 } \times \frac { 1 }{ 100 } }{ \frac { 1 }{ 6 } \times \frac { 1 }{ 100 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 100 } +\frac { 1 }{ 2 } \times \frac { 15 }{ 100 } } =\frac { \frac { 1 }{ 600 } }{ \frac { 1 }{ 600 } +\frac { 1 }{ 100 } +\frac { 15 }{ 200 } } =\frac { \frac { 1 }{ 600 } }{ \frac { 52 }{ 600 } } \)
\(=\frac { 1 }{ 600 } \times \frac { 600 }{ 52 } =\frac { 1 }{ 52 } \)

\(\lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \left( \frac { \pi }{ 2 } -x \right) \tan { x } } \)
\(Put\quad \frac { \pi }{ 2 } -x=\theta ;\quad \Rightarrow x=\frac { \pi }{ 2 } -\theta \\ Also,\quad \theta =\quad \frac { \pi }{ 2 } -x\quad \rightarrow 0\quad as\quad x\rightarrow \frac { \pi }{ 2 } \)
\(\therefore \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \left( \frac { \pi }{ 2 } -x \right) \tan { x } } =\lim _{ \theta \rightarrow 0 }{ \theta .\tan { \left( \frac { \pi }{ 2 } -\theta \right) = } } \lim _{ \theta \rightarrow 0 }{ \theta } \cot { \theta } =\lim _{ \theta \rightarrow 0 }{ \frac { \theta }{ \tan { \theta } } } =\frac { 1 }{ \lim _{ \theta \rightarrow 0 }{ \theta } \left( \frac { \tan { \theta } }{ \theta } \right) } =\frac { 1 }{ 1 } =1\)

5x - 7 = A\({d\over dx}(3x-x^2-2)+B\)
5x - 7 = \(A(3-2x)+B\)
Comparing the coefficients of like terms, we get
\(-2A=5 \Rightarrow A=-{5\over2};3A+B=-7 \Rightarrow B={1\over2}\)
\(I=\int {{-5\over 2}(3-2x)+{1\over2}\over \sqrt{3x-x^2+2}}dx\)
\(I=-{5\over2}\int {3-2x\over \sqrt{3x-x^2+2}}+{1\over2}\int {1\over( \sqrt{17\over 2})^2-(x-{3\over2})^2}dx\)
\(=-5\sqrt{3x-x^2+2}+{1\over 2}sin^{-1}({x-{3\over2}\over{\sqrt{17}\over 2}})+c\)
Thus, I = \(-5\sqrt{3x-x^2+2}+{1\over2}sin^{-1}({2x-3\over \sqrt{17}})+c\)​​​​

Given \(\vec{a}, \vec{b}\) and \(\vec{c}\)are unit vectors.
\(\Rightarrow \left| \vec { a } \right| =\left| \vec { b } \right| =\left| \vec { c } \right| =1\)
\(\vec{a}.\vec{b}=\vec{a}.\vec {c}\)= 0, and angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{3}\)
\(\vec{a}.\vec{b}=\vec{a}.\vec {c}\)= 0 \(\Rightarrow \vec{a}\) is \({ \bot }^{ r }\) to both\(\vec{b}\) and \(\vec{c}\)
\(\vec{a}\) is \({ \bot }^{ r }\) to \(\vec { b } \times \vec { c } \Rightarrow \vec { a } =\lambda (\vec { b } \times \vec { c } )\) for some scalar \(\lambda\).
\(\therefore { \left| \vec { a } \right| }^{ 2 }={ \lambda }^{ 2 }{ \left| \vec { b } \times \vec { c } \right| }^{ 2 }\)
\(\Rightarrow 1={ \lambda }^{ 2 }[{ \left| \vec { b } \right| }^{ 2 }{ \left| \vec { c } \right| }^{ 2 }-{ (\vec { b } \times \vec { c } ) }^{ 2 }]\left[ \therefore \left| \vec { a } \right| =1{ and\left| \vec { a } \times \vec { b } \right| }^{ 2 } \right] ={ \left| \vec { a } \times \vec { b } \right| }^{ 2 }={ \left| \vec { a } \right| }^{ 2 }{ \left| \vec { b } \right| }^{ 2 }-{ (\vec { a } .\vec { b } ) }^{ 2 }]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[(1)(1)-{ \left| \vec { b } \right| }^{ 2 }{ \left| \vec { c } \right| }^{ 2 }{ cos }^{ 2 }\frac { \pi }{ 3 } ]\) [\(\therefore\)angle between \(\vec{a}\)and\(\vec{b}\) is\(\frac{\pi}{3}\) ]
\(\Rightarrow 1={ \lambda }^{ 2 }[1-{ cos }^{ 2 }\frac { \pi }{ 3 } ][\because \left| \vec { b } \right| \left| \vec { c } \right| =1]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1-\frac { 1 }{ 4 } ]\Rightarrow 1={ \lambda }^{ 2 }\left( \frac { 3 }{ 4 } \right) \)
\(\Rightarrow 1={ \lambda }^{ 2 }\frac { 4 }{ 3 } \Rightarrow 1={ \lambda }\pm \frac { 2 }{ \sqrt { 3 } } \)
Substituting \(\)\({ \lambda }\pm \frac { 2 }{ \sqrt { 3 } } \) in (1) we get,
\(\vec { a } =\pm \frac { 2 }{ \sqrt { 3 } } (\vec { b } \times \vec { c } )\)

Let \(\overrightarrow{a}=5\hat{i}+6\hat{j}+7\hat{k}\)
\(\overrightarrow{b}=\)7 \(\hat{i}\) -8\(\hat{j}\) +9 \(\hat{k}\)
\(\overrightarrow{c}=\)3\(\hat{i}\)+20\(\hat{j}\) +5\(\hat{k}\)
Let \(\overrightarrow{a}=s\overrightarrow{b}+t \overrightarrow{c}\)
\(\Rightarrow 5\hat { i } +6\hat { j } +7\hat { k } =s(7\hat { i } -8\hat { j } +9\hat { k } )+t(3\hat { i } +20\hat { j } +5\hat { k } )\)
\(\Rightarrow 5\hat { i } +6\hat { j } +7\hat { k } =(7s+3t)\hat { i } +(-8s+20t)\hat { j } +(9s+5t)\hat { k } \)
Equating the like components, both sides we get.
                   5 = 7s + 3t  .....(1)
                  -8s + 20t = 6 ....(2)
                   9s + 5t = 7 ......(3)

                  164s  = 82    \(\Rightarrow \quad s=\frac { 82 }{ 164 } =\frac { 1 }{ 2 } \)
Substituting  \(\\ s=\frac { 1 }{ 2 } \) in (1) we get,
\(7\left( \frac { 1 }{ 2 } \right) +3t=5\quad \Rightarrow 3t=5-\frac { 7 }{ 2 } =\frac { 10-7 }{ 2 } =\frac { 3 }{ 2 } \)
\(\Rightarrow t=\frac { 3 }{ 2\times 3 } =\frac { 1 }{ 2 } \)
Substituting \(s=\frac { 1 }{ 2 } ,t=\frac { 1 }{ 2 } \) in (3) we get,
\(9\left( \frac { 1 }{ 2 } \right) +5\left( \frac { 1 }{ 2 } \right) =7\)
\(\Rightarrow \frac { 9 }{ 2 } +\frac { 5 }{ 2 } =7\)
\(\Rightarrow \frac { 14 }{ 2 } =7\)
\(\Rightarrow\) 7 = 7 which satisfies the (3) equation.
Thus, one vector is a linear combination of other two vectors.
Hence, the given vectors are co-planar.                                 

Any number divisible by 5, its unit place must have 0 or 5. We have to find 4-digit number greater than 6000 and less than 7000.
So, the unit place can be filled with 2 ways (0 or 5) since, repetition is not allowed.
\(\therefore\) Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.
But the required number is greater than 6000 and less than 7000. So, thousand place can be filled with I digits (i.e) 6.

Th	H	T	O
1	8	7	2

So, the total number of integers = 1 \(\times\) 8 \(\times\) 7 \(\times\) 2 = 112
Hence, the required number of integers = 112

Let P(n) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ n+1 } \) 
Substituting the value of n = 1, in the statement we get 
P(1) = \(\frac { 1 }{ 1.2 } =\frac{1}{2}\)
Hence, P(1) is true. Let us assume that the statement is true for n = k. Then
P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ k+1 } \)
We need to show that P(k + 1) is true. Consider,
\(P(k+1) =\underbrace{\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } ...+\frac { 1 }{ k(k+1) }} +\frac { 1 }{ (k+1)(k+2) } \)

= \(P(k)+\frac{1}{(k+1)(k+2)}\)
= \(\frac{ k} {(k + 1)} + \frac{1}{ (k + 1)(k + 2)}\)
= \(\frac{1} {k + 1} (\frac{ k}{ 1} + \frac{1} {k + 2})\)
= \(\frac{1} {k + 1} (\frac{ k ^2 + 2k + 1}{ k + 2 })\)
= \(\frac{1}{(k+1) }\frac{(k+1)^2}{k+2}=\frac{k+1}{k+2}\)
This implies, P(k + 1) is true
The validity of P(k + 1) follows from that of P(k)
Therefore by the principle of mathematical induction, for any natural number n,
\(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ n+1 } \).

Given equation of pair of lines is 6x² + 5xy - py² + 7x + qy - 5 = 0.
6x²+ 5xy - py²+ 7x + qy - 5 = 0
Here a = 6, b = -p, 2h = 5, 2g = 7, 2f = q
h = \(\frac { 5 }{ 2 } \) g = \(\frac { 7 }{ 2 } \) f = \(\frac { q }{ 2 } ,\) c = -5
The condition to represent perpendicular lines is a + b = 0 \(\Rightarrow\) 6 - p = 0 \(\Rightarrow\) p = 6
The condition to represent pair of lines is abc + 2fgh - af²- bg²- ch² = 0
\(\Rightarrow \ 6-(-6)(-5)+\left( \frac { 7 }{ 2 } \right) \left( \frac { 5 }{ 2 } \right) -6\left( \frac { { q }^{ 2 } }{ 4 } \right) +6\left( \frac { 4q }{ 4 } \right) +5\left( \frac { 25 }{ 4 } \right) =0\)
Multiplying by 4 throughout we get,
720 + 35q - 6q²+294 + 125 = 0
\(\Rightarrow \quad 35q-6q^{ 2 }+1139=0\)
\(\Rightarrow \quad q=\frac { 35\pm \sqrt { { (-35) }^{ 2 }-4(6)(1139) } }{ 2\times 6 } \)
\(\Rightarrow \quad q=\frac { 35\pm \sqrt { 1225+27336 } }{ 12 } \)
\(\Rightarrow \quad q=\frac { 35\pm \sqrt { 28561 } }{ 12 } =\frac { 35\pm 169 }{ 12 } \)
\(\Rightarrow \quad q=\frac { 204 }{ 12 } or-\frac { 134 }{ 12 } \)
\(\Rightarrow \quad q=17\quad or\quad -\frac { 67 }{ 6 } \)
Hence P = 6, q = 17 or \(-\frac{67}{6}\)

We know (1 +x)n = nC₀ +nC₁ x +nC₂ x² +...+nC_n-1 x^n-1 +nC_nXⁿ 
Putting x = 8
(1 +8)ⁿ = nC₀ + nC₁ (8) + nC₂ (64) +..+ nC_n-1 8^n-1 + nC_n . 8N
9ⁿ = 1 + 8n + nC² (64) + nC₃ + ..+ nC_n 8^n-2
9ⁿ - 8n -1 is divisible by 64 for all positive integer n
putting N = n +1 we get
9ⁿ⁺¹ -8(n+1) -1 is divisible by 64 for all positive integer n
(9^n-1 -8n -8 -1 ) is divisible by 64
9^n-1 - 8n -9 is always divisible by 4

(i) y = sin (-x)

Let y = sin x.
Then sin(-x) is the reflection of the graph of sin x, about y-axis.
(ii) y = -sin(-x)

-sin(-x) is the reflection of the graph of sin(-x) about the x-axis.
(iii) \(y=sin\left( {\pi\over 2}+x\right)\)
Let y = sinx.
Then \(sin\left( {\pi\over 2}+n\right)\)causes the shift to the left for \(\pi\over 2\) unit to the sin x curve.
(iv) \(y=sin\left({\pi\over 2}-x \right)\)

Let y = sin x. Then \(sin\left( {\pi\over 2}-n\right)\) causes the shift to the left for \(\pi\over 2\) unit to the sin (-x) curve.

\(5 R, 5 B \text { one ball is drawn }\)
\(\mathrm{P}(\text { getting red ball })=\frac{5}{10}=\frac{1}{2}\)
\(\text { Now } 2 \text { red balls are added }=7 \mathrm{R}, 5 \mathrm{~B}\)
\(P(\text {getting red ball })=\frac{7}{12}\)
\(\text {Supdose one black ball is obtained in first draw }\)
\(\mathrm{P}(\text {Black ball })=\frac{5}{10}=\frac{1}{2}\)
\(\text {Now } 2 \text { black balls are added }=5 R, 7 B\)
\(P(\text {getting red ball })=\frac{5}{12}\)
\(\text {Required Probality }=\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}\)
\(=\frac{7}{24}+\frac{5}{24}=\frac{12}{24}=\frac{1}{2}\)

Given \(\frac{d y}{d x} =\frac{x^{2}-4}{x^{2}} \)
\(=1-\frac{4}{x^{2}} \)
\(d y =\left(1-\frac{4}{x^{2}}\right) d x \)
\(\int d y =\int\left(1-\frac{4}{x^{2}}\right) d x \)
\(y =x-4\left(\frac{-1}{x}\right)+c \)
\(y =x+\frac{4}{x}+c \)
\(y=7 \text { when } x=2\)
\(7=2+\frac{4}{2}+c \)
\(7=4+c \Rightarrow c=3 \)
\(\therefore \text { The equation of the required curve }\)
\(y=x+\frac{4}{x}+3\)

\(y =\cos \left(\sin x^{2}\right) \)
\(\frac{d y}{d x} =-\sin \left(\sin x^{2}\right) \cos \left(x^{2}\right)(2 x) \)
\(\text { At } x =\sqrt{\pi / 2}, \frac{d y}{d x}=-\sin \sin \left(\frac{\pi}{z}\right) \cos (\pi / 2) 2(\sqrt{\pi / 2}) \)
\(=(\sin 1)(0) 2\left(\frac{\sqrt{\pi}}{2}\right)=0 \quad[\because \cos \pi / 2=0] \)

\(\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{x} \quad(\text { replacing } \sin x \text { by } x)\)
= 1

\(\text {Given} \ \overrightarrow{O A}=\hat{i}+2 \hat{j}+4 \hat{k}, \overrightarrow{O B}=2 \hat{i}-3 \lambda \hat{j}-3 \hat{k} \)
\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=2 \hat{i}-3 \lambda \hat{j}-3 \hat{k}-\hat{i}-2 \hat{j}-4 \hat{k}\)
\(\hat{i}+5 \hat{j}-7 \hat{k}=\hat{i}+(-3 \lambda-2) \hat{j}-7 \hat{k}\)
\(-3 \lambda-2=+5\) \((\because \text { compare coefft of } \hat{j}\text { on both sides) }\)
\(-3 \lambda=7 \)
\(\lambda=-\frac{7}{3} \)

\(A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \)
\(a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=\frac{-1}{2} \)
\(a_{21}=\frac{1}{2}(3(2)-2)=\frac{4}{2}=2 ; a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \)
\(\therefore A=\left[\begin{array}{ll} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{array}\right] \)

\(\mathrm{e}^{-2 x}=1-\frac{2 x}{1 !}+\frac{(2 x)^{2}}{2 !}-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{4}}{4 !}-\frac{(2 x)^{5}}{5 !}+\ldots\)
\(\text { Coefficient of } x^{5} \text { is } \frac{-2^{5}}{5 !}=\frac{-32}{120}=\frac{-4}{15}\)

No. of ways of answering = 5³ = 125
Correct answer - 1
.'. Number of incorrect answer = 125 - 1 = 124

Given P(A) = \(\frac { 1 }{ 2 } \), P(B) = \(\frac { 7 }{ 12 } \) and P(\(\bar { A } \cup \bar { B } \)) = \(\frac { 1 }{ 4 } \).
Now, \(P(\bar { A } \cup \bar { B } )=P(\overline { A\cap B } )=1-P(A\cap B)\)
\(\Rightarrow \frac { 1 }{ 4 } =1-P(A\cap B)\quad \Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
Now P(A) \(\times\) P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A)\times P(B)\)
Thus, A and B are not independent.

\(\lim _{ x\rightarrow a }{ \frac { \sqrt { x } +\sqrt { a } }{ x+a } } =\frac { \sqrt { a } +\sqrt { a } }{ a+a } =\frac { 2\sqrt { a } }{ 2a } =\frac { 1 }{ \sqrt { a } } \)

Let y = \(\sin { (\sqrt { 3 } \sin { x } +\cos { x } ) } \)
\(\frac { dy }{ dx } =\cos { (\sqrt { 3 } \sin { x } +\cos { x } ) } \frac { d }{ dx } (\sqrt { 3 } \sin { x } +\cos { x } )=\cos { (\sqrt { 3 } \sin { x } +\cos { x } )\left[ \sqrt { 3 } \cos { x } -\sin { x } \right] } \)

Applying C₁ ⟶ C₁ - C₂ we get
\(\triangle =\left| \begin{matrix} { cosec }^{ 2 }\theta & -{ cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta & -cose{ c }^{ 2 }\theta & -1 \\ 42 & 40 & 2 \end{matrix} \right| =\left| \begin{matrix} 1 & { cot }^{ 2 }\theta & 1 \\ -1 & cose{ c }^{ 2 }\theta & -1 \\ 2 & 40 & 2 \end{matrix} \right| \)
= 0 \([\therefore { C }_{ 1 }\equiv { C }_{ 2 }]\)

\(\int x^{11} d x=\frac{x^{11+1}}{11+1}+c=\frac{x^{12}}{12}+c\)

⁸P₄ = 8 \(\times\)7 \(\times\) 6 \(\times\) 5 = 1680

Given equation is 9x² + 12xy + 4y² + 6x + 4y - 3 = 0
\(\Rightarrow (3x+2y)^2+2(3x+2y)-3=0\)
\(\Rightarrow\) y²+ 2y - 3 = 0 where y = 3x + 2y
\(\Rightarrow\) (y+3) (y-1) = 0
\(\Rightarrow\) (3x + 2y + 3) (3x + 2y - 1) = 0[y = 3x + 2y]
Hence the separate equation are 3x + 2y + 3 = 0 and 3x + 2y - 1 = 0
\(\Rightarrow\) a = 3, b = 2, c₁ = 3 and c₂ = -1
Now, Distance between parallel lines \(=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|\)
\(=\left|\frac{3-(-1)}{\sqrt{3^2+2^2}}\right|=\left|\frac{4}{\sqrt{9+4}}\right|=\left|\frac{4}{\sqrt{13}}\right|\)
\(=\frac{4}{\sqrt{13}}\)

\((1+x)^{ m }=1+mx+\frac { m(m-1) }{ 2! } { x }^{ 2 }+..\)    [Binomial theorem for rational index]
∴ Co-efficient of x² = \(\frac { m(m-1) }{ 2 } \)
Given \(\frac { m(m-1) }{ 2 } \) = 6 ⇒ m²- m = 12
⇒ m²-m -12 = 0 ⇒ (m - 4) (m + 3) = 0
⇒ m = 4 or -3
∴ Negative value of m is -3

Given (n+1)! = 12\(\times\)(n-1)!
\(\Rightarrow\) (n+1)(n)(n-1)! = 12(n-1)!
\(\Rightarrow\) (n+1)n = 12
\(\Rightarrow\) n²+n-12 = 0
\(\Rightarrow\) (n+4)(n-3) = 0
\(\Rightarrow\) n = -4 or 3
\(\Rightarrow\) n = 3