Ans. y^x = x^y

Take log both side, we get 

log y^x = log x^y

=> x log y = y log x

differentiate w.r.t x, we get

=> \(x {1\over y} {dy \over dx} + log y = y {1\over x} + log x {dy \over dx}\)

=> \( {x\over y} {dy \over dx} - log x {dy \over dx} = {y\over x} - log y\)

=> \(( {{x\over y} - log x} ) {dy \over dx} = {y\over x} - log y\)

=> \( {dy \over dx} = { ( { {y\over x} - log y } ) \over ( {{x\over y} - log x} )}\)

Ans. S_n = 8( 1 + 11 + 111 + 1111 + ..... Upto n terms  )

=> 8/9 (9 + 99 + 999 + 9999 + ……… upto n terms)

=> 8/9 [ (10-1) + (100-1) + (1000-1) + upto n terms]

=> 8/9 [ (10 + 100 + 1000 +  ...... n terms ) -(1+ 1 + 1 + ………… n terms)]

=> 8/9 [ 10(10ⁿ -1)/ (10-1) - n ]

=> 8/9 [ 10(10ⁿ-1)/9 - n]

=> 80(10ⁿ -1) /81 - 8n/9

Ans. f(x) = (x²+2x+1)/(x²-8x+12)

To find domain as the function is rational so its denominator must not be equal to zero(0).

x²-8x+12 = 0

=> x²-6x-2x+12=0

=> x(x-6)-2(x-6)=0

=> (x-6)(x-2)=0

=> x = 2,6

For these two values this function 'll become undefined. So  domain = R - {2,6}

To Find Range:

Let f(x)=y

=> (x²+2x+1)/(x²-8x+12) = y

=> x²+2x+1 = yx²+

-8xy+12y

=> yx²-8xy +12y -x²-2x-1 =0 

=> (y-1)x² -(8y+2)x +(12y-1) = 0

Now D ≥ 0

=> b² -4ac ≥ 0

=> (8y+2)² - 4(y-1)(12y-1) ≥ 0 

= (8y+2)² ≥ 4(12y² -13y +1)

=> 64y² + 4 + 32y ≥ 48y² -52y +4

=> 16y² + 84y ≥  0

divide by 4

=> 4y² + 21y ≥ 0

=> y(4y+21) ≥ 0

=> y ≥ 0 and 4y +21 ≥ 0

y ≥ 0  and y ≥ -21/4

So domain = R - (-21/4, 0)

2tan x - cot x + 1 = 0

2tan x - 1/tan x + 1 = 0

2tan²x - 1 + tan x = 0

2tan²x + tan x - 1 = 0

2tan²x + 2tan x - tan x - 1 = 0

2tan x(tan x + 1) - 1 (tan x + 1) = 0

(tan x + 1)(2tan x - 1) = 0

tan x + 1 = 0 or 2tan x - 1 = 0

tan x = -1 or tan x = 1/2

x = 3π/4, 7π/4 or x = 3.6052, 0.4636

Since the family of lines passing through the intersection of given lines is

(2x + 3y - 4) k(x - 5y + 7) = 0

This line meets x-axis i.e. y = 0, then

2x - 4 + k(x - 5y + 9) = 0

x = (4 - 7k)/(2 + k), which is the x-intercept.

Therefore, -4 = (4 - 7k)/(2 + k)

k = 4

Putting the value of k in the family equation

(2x + 3y - 4) + 4(x - 5y + 7) = 0

6x - 17y + 24 = 0, which is the required equation

Ans. $2x + 3y - 4 = 0 => 2x +3y = 4 ........\left(1\right)$

$x - 5y = 7 ........\left(2\right)$

First we need to find intersection of the lines, For that solve these equation for x and y 
multiply equation (2) by 2, We get 

$2x - 10y = 14 ......\left(3\right)$

Subtract (1) from (3), we get 
$-13y = 10 => y = \frac{-10}{13}$

Put value of y in (1), we get  $x = \frac{41}{13}$
 
Let the Equation of line in intercept form is 
$\frac{x}{a} + \frac{y}{b} = 1$

as x intecept is -4  and point $\left(\left(\frac{41}{13}\right),\left(\frac{-10}{13}\right)\right)$ satisfies this equation.

putting all Values, we get

$\frac{41}{13\times -4} + \frac{-10}{13b} = 1$

=> $-\frac{41}{52} - 1 = \frac{10}{13b}$

=> $\frac{10}{13b} = \frac{-93}{52}$

=> $b = -\frac{40}{93}$

So the Equation of line ll be 

$\frac{x}{-4} - \frac{93y}{40} = 1$

In that case, post the question along with your answer and we will give the correct explanation for your wrong answers.! 

Ans. Infinity multiplied by zero is UNDEFINED. It is one of the 7 indeterminate forms in limits.

Domain: If f : X $\to$Y is a function, then the set X is called the Domain and Y is called the co-domain of the function f.

               Example: Find domain of the function $f\left(x\right) = \frac{1}{x-5}$

                               The given function is not defined at x = 5.

                               Therefore, Domain of the given function is R - {5}, where R is set of all real numbers.

Range: The set of second elements of the ordered pairs defining a function is called the Range of the function.

            Example: Find the range of function $f\left(x\right) = \frac{1}{x-5}$

                            Now, $y= \frac{1}{x-5}$

                            $\Rightarrow \frac{1}{y}=x-5$

                             $\Rightarrow x = \frac{5y+1}{y}$

                 Clearly, x is not defined when y = 0.

                 Therefore, Range of given funciton is R - {0}.

$Domain=R$

$Range=R$

$\frac{2x-3}{3x-7}>0$

$Case-1:$

$2x-3>0 and 3x-7>0$

$\Rightarrow x>\frac{3}{2} and x>\frac{7}{3}$

$\Rightarrow x>\frac{7}{3}$

$Case-2$

$2x-3<0 and 3x-7<0$

$\Rightarrow x<\frac{3}{2} and x<\frac{7}{3}$

$\Rightarrow x<\frac{3}{2}$

$\left(-\infty ,\frac{3}{2}\right)\cup \left(\frac{7}{3},\infty \right) is the solution set$

Proper Subset is a subset of the set which is having the elemnts from the Main set but not all. It is 2ⁿ - 1 in numbers.

Non Empty Proper subsets are all the subsets excluding the "Empty Subset" i.e. { }. It is 2ⁿ - 2 in numbers.

For e.g. If A = {1,2} then the number of subsets it has is 2ⁿ where n represents the no. of elements in that set means A will have 2² = 4 subsets which are {1}, {2}, {1,2} & { } sets. 

Now, here {1}, {2} & { } are the proper subsets & {1}, {2} are the non empty proper subsets.

Hope it clears better.

If A and B are two sets, then A is called the proper subset of B if A ⊆ B but B ⊇ A i.e., A ≠ B. The symbol ‘⊂’ is used to denote proper subset. Symbolically, we write A ⊂ B. 

E.g

A = {1, 2, 3, 4}

Here n(A) = 4

B = {1, 2, 3, 4, 5} 

Here n(B) = 5 

We observe that, all the elements of A are present in B but the element ‘5’ of B is not present in A. 

So, we say that A is a proper subset of B.
Symbolically, we write it as A ⊂ B

cos x is the derivative of sin x

The derivative of sin(x) from first principles.

Setting aside the limit for now, our first step is to evaluate the fraction with f(x) = sin x.

On the right hand side we have a difference of 2 sines, so we apply the formula in (A2) above:

Simplifying the right hand side gives:

Now to put it all together and consider the limit:

We make use of (3), fraction on a fraction, to bring that 2 out front down to the bottom:

Now, the limit of a product is the product of the limits, so we can write this as:

Now, the first limit is in the form of Limit of sin θ/θ that we met in (A1) above.

We know it has value 1.

For the right hand limit, we simply obtain cos x.

So we can conclude that

The sum of cubes of even numbers is equal to 2 n^2 (n+1)^2

So 2^3+ 4^3+ 6^3 +........+2n^2 = 2n^2(n+1)^2 

2 n^2 (n+1)^2  = k n^2 (n+1) ^2

Cancelling the common terms, we get 

k=2

It is given that sum of n terms of an A.P. is 3n² + 5n

=>  S₁ = 3(1)² + 5(1) = 3 + 5 = 8 which will also be 1st term as "a₁".

Further, S₂  = 3 (2)² + 5 (2) = 12 + 10 = 22

And, we know, a₂ = S₂ - S₁ = 22 - 8 = 14.... So, d = a₂ - a₁ = 14 - 8 = 6

Now, it's given that a_m = 164 => a₁ + (m - 1)d = 164

8 + (m - 1)$\times$6  = 164  => (m - 1)$\times$6 = 156 => (m - 1) = 26 => m = 27 (Answer)

According to question,

S₁ = a = 3(1)² + 5 x 1 = 8

S_{<font size="2">2</font>} = 3(2)^{<font size="2">2</font>} + 5 x 2 = 22                                Then,         a₂ = S₂ - S₁ = 22 - 8 = 14

S_{<font size="2">3</font>} = 3(3)^{<font size="2">2</font>} + 5 x 3 = 42                                Then,         a_{<font size="2">3</font>} = S_{<font size="2">3</font>} - S_{<font size="2">2</font>} = 42 - 22 = 22

Therefore, d = 14 - 8 = 6

Now, a_m = a + (m - 1)d = 164

=>          164 = 8 + (m - 1)6

=>          156 = (m - 1)6

=>          26 = m - 1

=>          m = 27

In the expension of (1+X)ⁿ, T_(r+1) = ⁿC_r X^r

Cofficient of (r+1)th term =  ⁿC_r

_{So Cofficient of 5th, 6th and 7th term ll be} ⁿC₄ ⁿC₅ ⁿC_{6 respectivily} 

As it is given these terms are in AP

So,  2 * ( ⁿC₅ ) = ⁿC₄ + ⁿC₆

(2 * n! ) / { (n-5)! 5!} =  (n! ) / { (n-4)! 4!}  + (n! ) / { (n-6)! 6!}

i think you can proceed from here .

Given: b = (a + b)/2.            ........(i)

And.    c² = bd                      ..........(ii)

And.    d = 2ce/(c + e).        ..........(iii)

Substituting values of b and d from eq.(i) and (iii), in eq.(ii),

c² = {(a + c)/2} × {2ce/(c + e)}

c² = (ace + c²e)/(c + e)

c³ = ace

c² = ae

L.H.S. = [1/(ab)^1/2+a] + [1/(ab)^{<font size="2">1/2</font>}+b] = [1/a^1/2(a^1/2 + b^1/2)] + [1/b^{<font size="2">1/2</font>}(a^{<font size="2">1/2</font>} + b^{<font size="2">1/2</font>})]

           = (a^1/2 + b^1/2)/(ab)^1/2(a^{<font size="2">1/2</font>} + b^{<font size="2">1/2</font>})

           = 1/(ab)^1/2

           = 1/g

           = R.H.S.

L.H.S. = Cos(3π/4+x)-cos(3π/4-x)

= cos 3π/4 cos x - sin 3π/4 sin x - [cos 3π/4 cos x + sin 3π/4 sin x]

= cos 3π/4 cos x - sin 3π/4 sin x - cos 3π/4 cos x - sin 3π/4 sin x]

= -2sin 3π/4 sin x

= -2 x (1/√2) sin x

= -√2sin x

Given: ( x + iy ) ( 3 – 4i ) = (5 + 12i )

Solution:

( x + iy ) = (5 + 12i ) $/$ ( 3 – 4i )

By Rationalization, ( x + iy ) = (5 + 12i ) $/$ ( 3 – 4i ) $\times$ ( 3 + 4i )$/$ ( 3 + 4i )

= (5 + 12i ) ( 3 + 4i ) $/$ ( 3 + 4i ) ( 3 - 4i )

= (15 + 36i + 20i + 48i²) $/$ 3^2 - (4i)²                       {( 3 + 4i ) ( 3 - 4i ) = 3² - (4i)² because (a+b)(a-b)=a² - b²}

= (15 - 48 + 56i) $/$ 9 - 16i²                                        (48i^2 = -48 because i² = -1)

= ( -33 + 56i) $/$ 9 +16

= ( -33 + 56i) $/$ 25

So, ( x + iy ) = -33$/$25 + 56i$/$25 

And, ( x - iy ) = -33$/$25 - 56i$/$25 

Now, ( x + iy )( x - iy ) = (-33$/$25 + 56i$/$25 ) (-33$/$25 - 56i$/$25 )

x² + y² = (-33$/$25)² -  (56i$/$25)²

= 1089$/$625 - 3136 i²$/$ 625

= 1089$/$625 + 3136 $/$ 625

= 4225 $/$ 625 = 6.76

Now Root of x² + y² means root of $\sqrt{}$6.76 = 2.6 (Answer)

The Binomial Theorem is a quick way (okay, it's a less slow way) of expanding (or multiplying out) a binomial expression that has been raised to some (generally inconveniently large) power.

total number of terms in expansion =  (power of expression +1)

so in this expension total number of terms ll be 13 and middle term ll be 7th term

T₇ = ¹²C₆ ({x)²}⁶ (yx)⁶

Given:

n (U) = 100, n (H - G - S) = 15, n (S - G - H) = 12, n (G - H - S) = 8, n (G $\cap$ H) = 40, n (G $\cap$ S) = 20, n (H $\cap$ S) = 10, n (G) = 65

Solution:

n(G) = n (G - H - S) + n (G $\cap$ H) + n (G $\cap$ S) - n (G $\cap$ S $\cap$ H)

65 = 8 + 40 + 20 - n (G $\cap$ S $\cap$ H)

n (G $\cap$ S $\cap$ H) = 68 - 65 = 3

Similarly, n (H) = n (H - G - S) + n (H $\cap$ S) + n (G $\cap$ H) - n (G $\cap$ S $\cap$ H)

n (H) = 15 + 10 + 40 - 3 = 65 - 3 = 62 (Answer to part a)

Similarly, n (S) = n (S - G - H) + n (H $\cap$ S) + n (G $\cap$ S) - n (G $\cap$ S $\cap$ H)

n (S) = 12 + 10 + 20 - 3 = 39 (Answer to part b)

Also, we know n (G $\cup$ S $\cup$ H) = n (G) + n (S) + n (H) - n (G $\cap$ S) - n (H $\cap$ S) - n (G $\cap$ H) + n (G $\cap$ S $\cap$ H)

n (G $\cup$ S $\cup$ H) = 65 + 39 + 62 - 20 - 10 - 40 + 3 = 169 - 70 = 99

Also, n (G $\cup$ S $\cup$ H)$\text{'}$ = n (U) - (G $\cup$ S $\cup$ H) = 100 - 99 = 1 (Answer to part c)

dx = sin x / 2cos²x = 1/2[sin x - sec²x]               [By Product rule]

dx = 1/2[sin x.2sec x.sec x.tan x + sec²x.cos x]

dx = 1/2[sin x.2sec²x.tan x + sec x]

dx = (1/2) sec x[2sin²x.sec²x + 1] = (1/2) sec x[2tan²x + 1] = 1/(2cos x) [(2sin²x/cos²x) + 1]

dx = 1/(2cos³ x) [2sin²x + 1 - sin²x ] = 1/(2cos³ x) [sin²x + 1]