Y={10}

More than 230 litres but less than 920 litres]

(a) Let E₁ denotes the event that exactly one child is a girl.
Then, E₁={BBG,BGB,GBB}
\(\Rightarrow \)  \(P({ E }_{ 1 })=\frac { 3 }{ 8 } \)
(b) Let E₂ denotes the event that atleast two children are girls.
Then, E_{2 =} {BGG,GBG,GGB,GGG}
\(\Rightarrow \) \(P({ E }_{ 2 })=\frac { 4 }{ 8 } =\frac { 1 }{ 2 } \)
(c) Let E₃ denotes the event that no child is a girl.
then, E₃ = {BBB}
\(\Rightarrow \)  \(P({ E }_{ 3 })=\frac { 1 }{ 8 } \)

If the triangle is not equilateral, then all three sides of the triangle are not equal.

Let p : x < y and q : x + 5 < y + 5.
Then, converse of p \(\Rightarrow \) q is q \(\Rightarrow \) p i.e. 'If x + 5 < y + 5, then x < y'.

\(\frac { dy }{ dx } =\frac { 1 }{ \sqrt [ 2 ]{ x } } -\frac { 1 }{ 2x^{ 3/2 } } \)
Ans: 0

We have, \(f\left( x \right) =\tan { \left( ax+b \right) } \)
By first principle of derivative, we have
\(f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( x+h \right) -f\left( x \right) }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { \tan { \left[ a\left( x+h \right) +b \right] -\tan { \left( ax+b \right) } } }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { \frac { \sin { \left( ax+ah+b \right) } }{ \cos { ax+ah+b } } -\frac { \sin { \left( ax+b \right) } }{ \cos { ax+b } } }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { \left[ \sin { \left( ax+ah+b \right) \cos { ax+b } - } \sin { \left( ax+b \right) \cos { ax+ah+b } } \right] }{ h\cos { \left( ax+b \right) \cos { \left( ax+ah+b \right) } } } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { a\sin { \left( ah \right) } }{ a.h\cos { \left( ax+b \right) \cos { \left( ax+ah+b \right) } } } } \)
\(\left[ \therefore \sin { A\cos { B-\cos { A\sin { B=\sin { \left( A-B \right) } } } } } \right] \)
\(=\lim _{ h\rightarrow 0 }{ \frac { a }{ \cos { \left( ax+b \right) \cos { \left( ax+ah+b \right) } } } \lim _{ h\rightarrow 0 }{ \frac { \sin { ah } }{ ah } } } \)
\(=\frac { a }{ \cos { ^{ 2 }\left( ax+b \right) } } \times 1\)
\(=a\sec { ^{ 2 }\left( ax+b \right) } \)

\(\lim_ { x\rightarrow 0 }{ lim } \frac { sin\left( 2+x \right) -sin\left( 2-x \right) }{ x } \)
\(=\lim_ { x\rightarrow 0 }{ lim } x\frac { 2cos\quad \left( \frac { 2+x+2-x }{ 2 } \right) sin\left( \frac { 2+x-2+x }{ 2 } \right) }{ x } \)
\(\left[ \because sin \ C- sin\ D=2cos\left( \frac { C+D }{ 2 } \right) .sin\left( \frac { C-D }{ 2 } \right) \right] \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 2cos\quad 2\quad sin\quad x }{ x } \)
\(=2cos\quad 2\underset { x\rightarrow 0 }{ lim } \frac { sin\quad x }{ x } =2cos\quad 2\quad \left[ \because \quad \underset { \theta \rightarrow 0 }{ lim } \frac { sin\theta }{ \theta } =1 \right] \)

\(\lim_ { y\rightarrow 0 }{ lim } \frac { \left( x+y \right) sec\left( x+y \right) -x\quad sec\quad x }{ y } \)
\(\lim_ { y\rightarrow 0 }{ lim } x\frac { x\left( sec\left( x+y \right) -sec\quad x \right) +\quad y\quad sec(x+y) }{ y } \)
\( \Rightarrow\lim_ { y\rightarrow 0 }{ lim } \quad x\left\{ \frac { sec(x+y)-sec\quad x }{ y } \right\} +\underset { y\rightarrow 0 }{ lim } \frac { ysec(x+y) }{ y } \)
\(=\lim_ { y\rightarrow 0 }{ lim } \frac { \frac { x }{ cos(x+y) } -\frac { x }{ cos\quad x } }{ y } +\lim_ { y\rightarrow 0 }{ lim } sec(x+y)\)
\(=\lim_{ y\rightarrow 0 }{ lim } x\left\{ \frac { cosx\quad -\quad cos(x+y) }{ ycos\quad x\quad cos\quad (x+y) } \right\} +\lim_ { y\rightarrow 0 }{ lim } sec(x+y)\)
\(=\lim_ { y\rightarrow 0 }{ lim } \left\{ \frac { cos\quad x-cos\quad (x+y) }{ y } \times \frac { x }{ cos\quad x\quad cos(x+y) } \right\} +\lim_ { y\rightarrow 0 }{ lim } \)
\(=\lim_ { y\rightarrow 0 }{ lim } \left\{ \frac { 2sin\left( x+\frac { y }{ 2 } \right) sin\left( \frac { y }{ 2 } \right) }{ 2\left( \frac { y }{ 2 } \right) } \times \frac { x }{ cos\quad x\quad cos\left( x+y \right) } \right\} +\lim_ { y\rightarrow 0 }{ lim } sec(x+y)\)
\(=\lim_ { y\rightarrow 0 }{ lim } sin\left( x+\frac { y }{ 2 } \right) \times \lim_ { y\rightarrow 0 }{ lim } \frac { sin\left( \frac { y }{ 2 } \right) }{ \frac { y }{ 2 } } \times \lim_ { y\rightarrow 0 }{ lim } \frac { x }{ cos\quad x\quad cos(x+y) } +\lim_{ y\rightarrow 0 }{ lim } sec(x+y)\)
\(=sin\quad x\times 1\times \frac { x }{ { cos }^{ 2 }x } +sec\quad x= x\quad tan\quad x\quad sec\quad x+sec\ x\)

The centroid of a triangle is equal to the centroid of the triangle formed by mid-points of its sides.

(1, 1, -2)

L(3,0,0), M(0,4,0) and N(0,0,5)

\(e=\frac { 5 }{ 8 } ,2ae=10\Rightarrow a=\frac { 5 }{ 5/8 } =8\quad \therefore \quad e=\sqrt { 1-\frac { { b }^{ 2 } }{ 64 } } \)
\(\frac { 25 }{ 64 } =\frac { 64-{ b }^{ 2 } }{ 64 } \Rightarrow { b }^{ 2 }=39.\quad Latusrectum=\frac { 2{ b }^{ 2 } }{ a } \)
Ans. \(\frac { 39 }{ 4 } \)

Let equation of line            
Since, the coordinates of the middle point are P(3, 2)
\(\therefore \)   \(3=\frac { 0+a }{ 3 } \Rightarrow 3=\frac { a }{ 2 } \Rightarrow a=6\)            
Similarly, b = 4
Ans. 2x + 3y = 12

Given equation of line is
\(\frac { x }{ a } +\frac { y }{ b } =1\) ....(i)
Perpendicular length from the origin to the line (i) is
\(p=\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } =\frac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \Rightarrow { p }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } \)
Since, a² , p² and b² are in AP.
\(\therefore { 2p }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\Rightarrow \frac { 2{ a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } ={ a }^{ 2 }{ +b }^{ 2 }\)
\(\Rightarrow \) 2a²b² = (a² + b²)² \(\Rightarrow \)2a²b² = a⁴+b⁴+2a²b²
\(\Rightarrow \) a⁴ + b⁴ = 0

(i) Given a, b, c, d are in AP and we know that AM > GM, then for the first three terms.
b > \(\sqrt { ac } \)  [here, \(\frac { a+c }{ 2 } \) = b]
On squaring both sides, we get
b² > ac.....(i)
Similarly, for the last three terms
C > \(\sqrt { bd } \)   [here, \(\frac { b+d }{ 2 } \) = c]
\(\Rightarrow \)  c² > bd...(ii)
On multiplying Eqs. (i) and (ii), we get 
b² c² > (ac) (bd)
\(\Rightarrow \)  bc  > ad

Let A be the first term and D be the common difference of the given AP. Then,
\({ T }_{ p }=a\Rightarrow A+(p-1)D=a\quad \quad \quad \quad \quad ...(i)\)
\(and\quad { T }_{ q }=b\Rightarrow A+(q-1)D=b\quad ...(ii)\)
On subtracting Eq.(ii) from Eq.(i), we get
\((p-q)D=a-b\Rightarrow D=\frac { a-b }{ p-q } \quad \quad \quad ...(ii)\)
On adding Eqs.(i) and (ii), we get
\(2A+(p+q-2)D=(a+b)\)
\( \Rightarrow 2A+pD+qD-2D=a+b\)
\(\Rightarrow A+pD+qD-D=a+b+D\)
\(\Rightarrow 2A+(p+q-1)D=a+b+D\)
\(\Rightarrow (p+q-1)D=a+b+\left( \frac { a-b }{ p-q } \right) \) [from Eq.(iii)]  ....(iv)
Now,  \({ S }_{ p+q }=\frac { p+q }{ 2 } [2A+(p-q-1)D]\)
\(\frac { p+q }{ 2 } \left[ a+b+\frac { a-b }{ p-q } \right] \) [from Eq.(iv)
Hence proved.

Let us first seat  the 5 girls (or 5 boys). This can be done in 5! ways.
Now, for each such arrangement, the 5 boys can be seated only at the cross marked the place as shown below:
(I)  \(\times G_{ 1\times }G_{ 2 }\times { G }_{ 3 }\times { G }_{ 4 }\times { G }_{ 5 }\times \)
(II) \(G_{ 1\times }G_{ 2 }\times { G }_{ 3 }\times { G }_{ 4 }\times { G }_{ 5 }\times \)
In case I, 5 boys can be seated in P = 5! ways
Thus, number of ways of seating = 5! \(\times \) 5! = \({ (5! })^{ 2 }\)
Similarly, in case II, number of ways of seating = \({ (5! })^{ 2 }\) 
Hence, required number of ways = \({ (5! })^{ 2 }\) + \({ (5! })^{ 2 }\) \({ =2.(5! })^{ 2 }=28800\)
For solving this type of problem, we used the following steps
Step I Firstly, decide that from how many digits the required number will be formed.
Step II Fill up the places on which restrictions are present and let the number of ways of filling up these places be k.
Step III Find the number of ways of filling the remaining places with remaining digits by using the formula ⁿP _r  .
Step IV Required number of numbers is k. P . ⁿP _r .

We have 5 boys and 5 girls,
Since no two girls sit together, therefore the possible choices for girls are the places marked as '\(\times\)'.
B1  \(\times\)B2   \(\times\)B3 \(\times\)B4 \(\times\)B5 \(\times\)
Clearly, the girls can be arranged in ⁶P5 ways and the boys can be arranged in 5! ways.
Hence, by fundamental principle multiplication, required number of ways = ⁶P_{5 \(\times \)5!}
 \(=\frac { 6! }{ (6-5)! } = 5!=6!\times 5!=86400\)
For solving this type of problem, we used the following steps
Step I Firstly, decide that from how many digits the required number will be formed.
Step II Fill up the places on which restrictions are present and let the number of ways of filling up these places be k.
Step III Find the number of ways of filling the remaining places with remaining digits by using the formula  ⁿP_r
Step IV Required number of numbers is k. ⁿ P _r .

There will be as many ways as there are ways of filling 2 vacant places  in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.

Use \(|x|\le a\Longrightarrow -a\le x\le a\).
[-1,3]

\(\frac { { \left( 1-i \right) }^{ 3 } }{ 1-{ i }^{ 3 } } =\frac { { 1 }^{ 3 }-{ i }^{ 3 }-3i+3{ i }^{ 2 } }{ 1+i } =\frac { 1+i-3i-3 }{ 1+i } =\frac { -2-2i }{ 1+i } \)
Ans.-2

\(z=\frac { 1+icos\theta }{ 1-2icos\theta } \times \frac { 1+2icos\theta }{ 1+2icos\theta } =\frac { 1-2{ cos }^{ 2 }\theta +3icos\theta }{ 1+4{ cos }^{ 2 }\theta } \)
\(For\quad Re(z),put\quad 3cos\theta =0\Rightarrow cos\theta =0\)
Ans. \(2n\pi \pm \frac { \pi }{ 2 } \)

Given,
\(f\left( x \right) =2{ x }^{ 2 }-1\)
\(g\left( x \right) =1-3x\)
\(Sice,\ f\left( x \right) =g\left( x \right) \)
\(\therefore 2{ x }^{ 2 }-1=1-3x\)
\(\Rightarrow \ 2{ x }^{ 2 }+3x-2=0\)
\(\Rightarrow \ 2{ x }^{ 2 }+4x-x-2=0\)
\(\Rightarrow \ 2x(x+2)-1(x+2)=0\)
\(\Rightarrow \ (2x-1)(x+2)=0\)
\(\Rightarrow \ 2x-1=0\ or\ x+2=0\)
\(\Rightarrow \ x=\frac { 1 }{ 2 } orx=-2\)
Thus, domain for which the function \(f\left( x \right) =g\left( x \right) \ is \ \left\{ \frac { 1 }{ 2 } ,-2 \right\} \)

Let the two sets be A and B such that n(A) =m and n(B) =n.
Then, number of subsets of set A=2^m
and the number of the subset of set B=2ⁿ
According to given condition, we have
2^m=112+2ⁿ\(\Rightarrow\)2^m-2ⁿ=2⁷-2⁴
On comparing both sides, we get
2^m=2⁷ and 2ⁿ = 2⁴ \(\Rightarrow\)m=7 and n= 4

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)