(2x + 3y + 4z)²

Substituting x=-1 and y=-1 in 9kx+12ky=63, we get
\(\Rightarrow\) 9k(-1)+12k(-1)=63
\(\Rightarrow\) -9k-12k=63
\(\Rightarrow\) -21k=63 k=-3

\(\frac { { 2 }^{ x-1 } }{ { 2 }^{ x-2 } } -\frac { 2^{ -x } }{ 2^{ x+1 } } =0\)
\(\Rightarrow \ { 2 }^{ x-1-x+2 }-2^{ -x-x-1 }=0\)
\(\Rightarrow \ { 2 }^{ 1 }-{ -2 }^{ -2x-1 }=0\)
\(\Rightarrow \ { 2 }^{ -2x-1 }={ 2 }^{ 1 }\)
\(\Rightarrow -2x-1=1\Rightarrow \ x=-1\)

Given,   AB = BC
and   BX = BY
On subtracting Eq. (ii) from Eq. (i), we get
AB - BX = BC - BY
⇒ AX = CY

Now,length AB = 3+ 3= 6 units

10

34

(a + b + c)² = (a² + b² + c²) + 2 (ab + bc + ca)
⇒ [9]² = (a² + b² + c²) + 2 (26) = (a² + b² + c²) + 52
⇒ a² + b² + c² = 92 - 52 = 81- 52 = 29

\(\frac{1}{a}.\frac{ab}{b+a}+\frac{1}{a}\frac{ab}{b-a}=\frac{b^2-ab+b^2+ab}{b^2-a^2}\)
=\(\frac{-2b^{2}}{a^2-b^2}\)

Marks the distance 9.5 units from a fixed point A on a given line to obtain a point B such that AB = 9.5 units from B, marks a distance of 1 unit and mark the new point as C. Find the mid point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and
intersecting the semi-circle at D.
then BD = √9.5
To represent √9.5 on the number line, let us treat the line BC as the number line, with B as zero (c) as 1 and so on.
Draw an arc with centre B and radius BD which intersect the number line at E.
ஃ E represents √9.5

(i) Given 3x+2y=k
Put x=2, y=1 then
3(2)+2(1)=k \(\Rightarrow\) k=8
(ii) Given, 2x-ky=6
Put x=2, y=1, then
2(2)-k(1)=6
\(\Rightarrow\) 4-k=6\(\Rightarrow\) k=4-6=-2
(iii) Given, \(\frac{x}{4}+\frac{y}{3}=5k\)
Put x=2, y=1, then
\(\frac{2}{4}+\frac{1}{3}=5k\)
\(\Rightarrow 5k=\frac{10}{12}=\frac{5}{6}\)
\(\Rightarrow k=\frac{1}{6}\) 

(i) Let us draw mutually perpendicular axes XOX' and YOY' and choose a suitable units of distance on the axes.
Let 1 crn = 1 unit. In point P(1, 0), y-coordinate is zero, so it lies on X-axis at a distance of 1 unit from Y-axis.
In point Q (4, O),y-coordinate is zero, so it lies on X-axis at a distance of 4 units from Y-axis. The point S(1, 3) is at a distance of 1 unit from Y-axis and 3 units from X-axis.
On plotting these points, we get the below graph

Now, we need to take a point R on the graph such that PQRS is a square. For this, draw a line passing through Q and parallel to PS and draw a line passing through S and parallel to PQ. Both lines intersect each other at a point, say R. Thus, we get a square PQRS. Clearly, abscissa of R will be equal to abscissa of Q, i.e. 4 and ordinate of R will be equal to ordinate of S, i.e. 3. Hence, the coordinates of point R
are (4, 3).
(ii) We have, S (1, 3) and R (4,3)
Here, we see that y-coordinate of both points S and R are same.
\(\therefore\)  Length of the line segment SR
= Difference of x-coordinate of both points S and R
= 4 - 1 = 3 units

 We know that, \({ \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2\times x\times \frac { 1 }{ x } \ \left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)\(\therefore { \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2\)
\(\Rightarrow { \left( x+\frac { 1 }{ x } \right) }^{ 2 }=14+2=16\ \left[ \because \ { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =14,given \right] \)
\(\Rightarrow x+\frac { 1 }{ x } =\pm \sqrt { 16 } \)
\(\therefore \quad x+\frac { 1 }{ x } =\pm 4\)
Case I When \(x+\frac { 1 }{ x } =4,\) then
\({ \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3\times x\times \frac { 1 }{ x } { \left( x+\frac { 1 }{ x } \right) }\ \left[ \because { (a+b) }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }+3ab(a+b) \right] \)\(\Rightarrow { \left( x+\frac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3{ \left( x+\frac { 1 }{ x } \right) }\)
\(\Rightarrow { (4) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3\times 4\)
\(\Rightarrow 64={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +12\)
\(\therefore { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =64-12=52\)
Case II When \(x+\frac { 1 }{ x } =-4,\) then
\(\Rightarrow { \left( x+\frac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3{ \left( x+\frac { 1 }{ x } \right) }\)
\(\Rightarrow { (-4) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3\times (-4)\)
\(\therefore { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =-64+12=-52\) 

x³-y³=(x-y)(x²+y²+xy)
(x-y)[(x-y)²+8-2xy+xy]  [\(\because\)(x-y)²=x²+y²-2xy] [\(\because\)(x-y)²+2xy=x²+y²]
=(x-y)[(x-y)²+3xy]
=5[(5)²+3 x84]  [\(\because\)x-y=4 and xy=84]
=5(25+525)=5 x 277=1385

\(x^3+13x^2+32x+20\)
Let \(p(x)={ x }^{ 3 }+13{ x }^{ 2 }+32x+20\)
By trail, we find that
\(p(-1)={ (-1) }^{ 3 }+13{ (-1) }^{ 2 }+32(-1)+20\)
\(=-1+13-32+20=0\)
\(\therefore\) By Factor Theorem, x-(-1), i.e., (x+1) is a factor of p(x).
Now,
\({ x }^{ 3 }+13{ x }^{ 2 }+32x+20={ x }^{ 2 }(x+1)+12x(x+1)+20(x+1)\)
\(=(x+1)({ x }^{ 2 }+12x+20)\)
\(=(x+1)({ x }^{ 2 }+2x+10x+20)\)
\(=(x+1)\{ x(x+2)+10(x+2)\} \)
\(=(x+1)(x+2)(x+10).\)

8

(a, 3a) satisfies y=3x

\((x-\frac{1}{x})(x+\frac{1}{x})(x^2+\frac{1}{x^2})\)
\((x^2-\frac{1}{x^2})(x^2+\frac{1}{x^2})=x^4-\frac{1}{x^4}\)

Coefficient of \(x^2=3+1=4\)

\(p(x)=(p-x)^3+14=(17-x)^3+14\)
\(=(17)^3-x^3-3(17)^2(x)+3.17.x^2+14\)
\(\because\)  Degree=3

x² + 5xy + y² = (x + y)²+ 3xy
\(={ \left[ \frac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } +\frac { \sqrt { 2 } +1 }{ \sqrt { 2 } -1 } \right] }^{ 2 }+3\frac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } \times \frac { \sqrt { 2 } +1 }{ \sqrt { 2 } -1 } \)
\(={ \left( \frac { 2+1-2\sqrt { 2 } +2+1+2\sqrt { 2 } }{ 2-1 } \right) }^{ 2 }+3\)
= (6)² + 3
= 36 + 3
= 39

We have, p(x) = ax³ - 3x² + 4
Since, when p(x) is divided by x-2
leaves remainder p
\(\therefore\)  p(2) =p \(\Rightarrow\) a(2)³ = 3(2)² + 4 = p
\(\Rightarrow\)   8a - 12 + 4 = p
\(\Rightarrow\)  8a - 8 = p    ....(i)
and we have, g(x) = 2x³ - 5x +a
Since, when g(x) is divided by x - 3, leaves remainder q.
\(\therefore\)  g(3) = q
\(\Rightarrow\)  2(3)³ - 5(3) +a = q
\(\Rightarrow\)  54 - 15 + a = q
\(\Rightarrow\)  39 + a = q      .....(ii)
We have,     p - 2q = 4
\(\Rightarrow\) 8a - 8 - 2(39+a) = 4 [from Eqs. (i) and (ii)]
\(\Rightarrow\) 8a - 8 - 78 - 2a = 4
\(\Rightarrow\) 6a - 86 - 4\(\Rightarrow\) 6a = 4 + 86 = 90
\(\Rightarrow \quad a=\frac { 90 }{ 6 } =15\)

4 sq units

\(\triangle PQR\) is an equilateral triangle
PQ = PR = QR
\(\Rightarrow \) PQ = QR
\(\Rightarrow \) PQ = 4
\(\Rightarrow \) OQ = 2
In right triangle POQ
OP² + OQ² = PQ²
\(\Rightarrow \) OP + (2)² = (4)² \(\Rightarrow \) OP = \(2\sqrt { 3 } \)
\(P\rightarrow \left( 0,2\sqrt { 3 } \right) \)

We know that the if the point lies on y-axis, its x-coordinate is 0.
So, Reason is correct.
The x co-ordinate of the point (0, 4) is zero.
So, Point (0, 4) lies on y -axis.
So, Assertion is also correct
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

(a) (iii) 2x + 3y = 60
Let the number of children be x and the number of adults be y then the linear equation in two variable for the given situation is 
2x + 3y = 60.
(b) (i) 10
2x + 3y =60 ⇒ 2(15) + 3y = 60
⇒ 3y = 60 - 30 = 30 
⇒ y = 10
(c) (i) 12
2x + 3y = 60 ⇒ 2x + 3(12) = 60
⇒ 2x 60 - 36 = 24
⇒ x = 12
(d) (iii) 15
On putting x = 5 and y = 0 in the equation 3x + 5y = b, we have 
3 x 5 + 5 x 0 = b
⇒ 15 + 0 = b
⇒ b = 15
(e) (ii) 1.x - 1.y + 5 = 0
y - x = 5 ⇒ y = x + 5
⇒ x - y + 5 = 0
⇒ 1.x - 1.y + 5 = 0