AP ICET 2013

we know that the expansion of sum of even term and sum of odd term are respective given below 

$$(a + x)^{n}$$  =  n$$C_{0}$$ $$a^{n}$$ +  n$$C_{1}$$ $$a^{n-1}$$x +  n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$  + ............. + n$$C_{n}$$ $$x^{n}$$

$$(a - x)^{n}$$   =   n$$C_{0}$$ $$a^{n}$$ - n$$C_{1}$$ $$a^{n-1}$$x + n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ - ............. + $$(-1)^n$$ n$$C_{n}$$ $$x^{n}$$

adding both we get 

$$(a + x)^{n}$$ + $$(a - x)^{n}$$ = 2( n$$C_{0}$$ $$a^{n}$$ +  n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ + ............. +  n$$C_{n}$$ $$x^{n}$$)

hence  

P = ($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2        equation 1  

Q = ($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2        equation 2 

there fore PQ we get 

PQ =   ($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2 $$\times$$ ($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2  = ($$(a + x)^{2n}$$ + $$(a - x)^{2n}$$)\4

or 4PQ = ($$(a + x)^{2n}$$ + $$(a - x)^{2n}$$)

$$P^{2}$$ - $$Q^{2}$$ =  (($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2)^2  -  (($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2)^2

$$P^{2}$$ - $$Q^{2}$$ =  $$(a + x)^{n}$$ $$(a - x)^{n}$$

$$P^{2}$$ - $$Q^{2}$$ =  ($$(a)^{2}$$ - $$(x)^{2}$$)^n  Answer

given that A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$

means that

A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$ = |A|I = 4 $$\begin{bmatrix}1 & 0 & 0\\0 & 1&0 \\ 0 & 0 & 1\end{bmatrix}$$

 ∣A∣=4

det (2Adj A)  =  ∣A∣^(n−1) , where n is number of rows in a matrix

                     =  2$$\times$$4^(3-1)
                     =  2$$\times$$ 16   

                     = 32  Answer 

given that 

$$\lim_{x \rightarrow 5} \frac{x^{2} - 3x - 10}{x^{2} - 7x + 10} $$

we can write this as 

$$\lim_{x \rightarrow 5} \frac{x^{2} - 5x + 2x - 10}{x^{2} - 5x -2x + 10} =$$  $$\lim_{x \rightarrow 5} \frac{x(x-5) + 2(x - 5)}{x(x-5) - 2(x- 5)} =$$

$$\lim_{x \rightarrow 5} \frac{(x - 5)(x +2)}{(x - 5)(x -2)} =$$  $$\lim_{x \rightarrow 5} \frac{(x + 2)}{(x - 2)} =$$

put the value of x we get 

7\3 Answer 

given that 

$$x = t - \frac{1}{t}, y = t + \frac{1}{t}$$ 

then

$$\frac{dy}{dx}$$ = $$\frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}}$$  

then $$\frac{\text{d}y}{\text{d}t}$$ = 1\y and $${\text{d}x}{\text{d}t}$$ = 1\x  

$$\frac{dy}{dx}$$ = 1\y\1\x = x\y Answer 

we know that the 

$$A. A^{T}$$ = $$A^{T}.A$$ 

so when we solve this always get symmetric matrix we found  

we know that the area of the eqitriangle = $$a^{2}$$ $$\times$$ $$\sqrt{3}$$ \4 = $$121 \sqrt{3}$$

a = 22

then the length of the wire = perimeter of the triangle 

                                         = 3a = 3*22 = 66

then the radius of the circle = 2$$\pi$$r = 2*$$\frac{22}{7})$$*r = 66 

r = 21\2  

then area of the circle =  $$\pi$$ $$\times$$ $$r^{2}$$ =  $$\frac{22}{7})$$  $$\times$$ $$\frac{21}{2})$$  $$\frac{21}{2})$$

then area of the circle = 346.5 cm square  Answer 

we know that the  sum of all the internal angles of a regular hexagon, in radians, is = 720 = 4$$\pi$$       Answer 

If circles are touching internally the distance between their centers will be = radius of bigger circle - radius of smaller circle

    = 10 - 3

    =  3  cm   Answer

note - 

If circles are touching externally the distance between their centers will be = = radius of bigger circle + radius of smaller circle

let the mid points of the slides AB and AC of the triangle is D(4,-2) and F(-8,14) 

which is parallel to the BC and BC = 2DF 

then DF Length  is  DF = (-8 +4) + (!4 + 2) = 20 

then BC = 2* 20 = 40 Answer 

we know the fourmula of distance between two line 

$$(A_{1}x +B_{1}y +C)\div\\sqrt({A_{1}^{2}+B_{1}^{2}})$$ = $$\pm$$ $$(A_{2}x +B_{2}y +C)\div\\sqrt({A_{2}^{2}+B_{2}^{2}})$$

then we have here  $$(A_{1}$$ = 1\3  $$(B_{1}$$ = 1\4 and $$(c_{1}$$ = -1 and 

                             $$(A_{2}$$ = 8 $$(B_{2}$$ = 8  and $$(c_{2}$$ = -5 

then 

|$$(A_{1}x +B_{1}y +C)\div\\sqrt({A_{1}^{2}+B_{1}^{2}})$$| = $$\pm$$ |$$(A_{2}x +B_{2}y +C)\div\\sqrt({A_{2}^{2}+B_{2}^{2}})$$|

|$$\frac{1\3 + 1\4 - 1}\sqrt{{1\3}^{2}+{1\4}^{2}}$$| = $$\pm$$ |$$\frac{8 +6 -5}\sqrt{{8}^{2}+{6}^{2}}$$|

1 + 9\10 = 10\19  Answer