(i) {tex}2^{2x}-2^{x+3}+2^4=0{/tex}

{tex}=> (2^x)^{2}-2^{x}2^{3}+16=0{/tex}

Put 2^x= y

{tex}=> y^2-8y+16=0{/tex}

{tex}=> y^2-4y-4y+16=0{/tex}

{tex}=> y(y-4)-4(y-4)=0{/tex}

=> (y-4)(y-4) = 0

=> y = 4

{tex}=> 2^x= 4{/tex}

=> {tex}2^x= 2^2{/tex}

So value of x = 2

(x+4)(x+10)=x(x+10)+4(x+10)

                     =x²+10x+4x+4×10

                     =x²+14x+40

= (x+4)(x+10)

= x²+10x+4x+40

= x²+14x+40

(X+y-3)+k(2x-y-2)=0

Now, at (1,1), K=-1

Putting the value of K ,we get,

x+y-3 -2x+y+2=0

Or, - x+2y-1=0

Or, x-2y=-1

This is the equation of the bisector of the angle.

{tex}(3\sqrt 3+2\sqrt 3)(2\sqrt 3+3\sqrt 2){/tex}

= {tex}(5\sqrt 3)(2\sqrt 3+3\sqrt 2){/tex}

= {tex}10\times 3+15\sqrt 6{/tex}

= {tex}30+15\sqrt 6{/tex}

Given    

1st term a₁ = -2 ,  ..............10th term a₁₀ = 16 , .................100th term  a₁₀₀ = ?

we know that  

 an = a₁ + (n -1)d  ...........   a₁₀ =a₁ + (10 - 1) d...................... 16 = -2 + 9d ..................... d = 18/9 ....... .... d = 2

a_{100 =} a₁ + (100 - 1) 2...........a₁₀₀ = -2 + 99x2 ........................ a₁₀₀ = -2 + 198 ............... a₁₀₀ = 196      

First term of AP a = -2

Let common difference = d

As we know

{tex}a_n = a+(n-1)d{/tex}

=> {tex}a_{10} = a+(10-1)d{/tex}

=> 16 = -2+9d

=> 9d= 18

=> d= 2

{tex}a_{100} = a+99d{/tex}

=> {tex}a_{100}=-2+(99\times 2){/tex}

{tex}=>> a_{100}= -2+198=196{/tex}

{tex}9x^2+6x+1-25y^2{/tex}

{tex}= (3x)^2+2\times 3x\times 1+(1)^2-(5y)^2{/tex}

{tex}= (3x+1)^2-(5y)^2{/tex}

{tex}= (3x+1-5y)(3x+1+5y){/tex}

{tex}9x^2+6x+1-25y^2{/tex}.......................... {tex}(3x)^2 + (2x3xx1) + (1)^2 -25y^2{/tex}.................................applying Rule {tex}(a+b)^2 = a^2+b^2+2ab {/tex}

.{tex}(3x + 1)^2 - (5y)^2{/tex} ........................... . {tex}(3x+1+5y)(3x+1-5y){/tex} .................................. .... applying Rule {tex}a^2 - b^2 = (a+b) (a-b){/tex}

Ans. 

{tex}9x^2+6x+1-25y^2{/tex}

{tex}= (3x)^2+(2\times 3x\times 1)+(1)^2-25y^2{/tex} [using {tex}(a+b)^2 = a^2+2ab+b^2{/tex}]

{tex}= (3x+1)^2-(5y)^2{/tex}

{tex}= (3x+1-5y)(3x+1+5y){/tex} [using {tex}a^2-b^2 = (a+b)(a-b)]{/tex}

Let the side of equilateral triangle be a and let the perpendicular drawn to sides be OP = 10 cm, OQ = 14 cm and OR = 6 cm.

Join OA, OB and OC.

According to question,

Area of equilateral triangle ABC = Area of {tex}\Delta{/tex}OBC + Area of {tex}\Delta{/tex}OCA + Area of {tex}\Delta{/tex}OAB

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2} \times {\rm{BC}} \times {\rm{OP}} + {1 \over 2} \times {\rm{AC}} \times {\rm{OQ}} + {1 \over 2} \times {\rm{AB}} \times {\rm{OR}}{/tex}

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2} \times 10 \times a + {1 \over 2} \times 14 \times a + {1 \over 2} \times 6 \times a{/tex}

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = {1 \over 2}a\left( {10 + 14 + 6} \right){/tex}

=>     {tex}{{\sqrt 3 } \over 4}a = {1 \over 2} \times 30{/tex}

=>     {tex}a = {{15 \times 4} \over {\sqrt 3 }} = {{60} \over {\sqrt 3 }}{/tex} cm

Now, Area of equilateral triangle

{tex}{{\sqrt 3 } \over 4}{a^2} = {{\sqrt 3 } \over 4} \times {{60} \over {\sqrt 3 }} \times {{60} \over {\sqrt 3 }}{/tex}

=>     {tex}{{\sqrt 3 } \over 4}{a^2} = 300\sqrt 3 {/tex} cm²

We know that

{tex}{x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right){/tex}

Here, if {tex}x+y+z=0{/tex}, then R.H.S. = 0

Then,

{tex}{x^3} + {y^3} + {z^3} - 3xyz = 0{/tex}

=>     {tex}{x^3} + {y^3} + {z^3} = 3xyz {/tex}

So, we can say that if {tex}a+b+c=0{/tex}, then {tex}{a^3} + {b^3} + {c^3} = 3abc{/tex}

Given : x = {tex}2-\sqrt 5{/tex}

Now 

{tex}{1\over x} = {1\over 2-\sqrt 5}{/tex}

Rationalize the denominator,

We get 

{tex}{1\over x} = {1\over 2-\sqrt 5}\times {2+\sqrt 5\over 2+\sqrt 5}{/tex}

{tex}=> {1\over x} = {2+\sqrt 5\over (2)^2-(\sqrt 5)^2} {/tex}

{tex}=> {1\over x} = {2+\sqrt 5\over 4-5}= -2-\sqrt 5{/tex}

Now, 

{tex}x^2+{1\over x^2}= (2-\sqrt 5)^2+(-2-\sqrt5)^2{/tex}

= {tex}4+5-4\sqrt5+4+5+4\sqrt5{/tex}= 18

1) Yes, y+3 is a factor of f(x)= 2y³ + 3y² - 7y + 6

 2) Since (x-1) is a factor of f(x)= 4x³ + 3x² - 4x +k , x= 1 is a root of f(x). Therefore,

     f(1)= 4*(1)³ + 3*(1)² - 4(1) +k = 0

          or, 4*1 +3*1 - 4 + k = 0

          or, 4 + 3 - 4 + k = 0

          or, 3 + k = 0

          or, k = -3

     Therefore, value of k is -3

3)   x³ + y³ 

    = (x+y)³ - 3xy(x+y)                   [ since x³ + y³ = (x + y)³  - 3xy(x+y)]

    = (5)³ - 3 * (6) * (5)                    [ x+y = 5 and xy = 6 given]

    = 125 - 90

    = 35

    Therefore,  the value of x³ + y³ is 35

{tex}{3 \over 5} + {2 \over {\sqrt 3 }} = {{3\sqrt 3 + 10} \over {5\sqrt 3 }}{/tex}

= {tex}{{3\sqrt 3 + 10} \over {5\sqrt 3 }} \times {{\sqrt 3 } \over {\sqrt 3 }}{/tex}

= {tex}{{9 + 10\sqrt 3 } \over {15}}{/tex}

Given: {tex}x = {{2ab} \over {{b^2} + 1}}{/tex}

{tex}{{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} - \sqrt {a - x} }}{/tex}

= {tex}{{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} - \sqrt {a - x} }} \times {{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} + \sqrt {a - x} }}{/tex}

= {tex}{{{{\left( {\sqrt {a + x} } \right)}^2} + {{\left( {\sqrt {a - x} } \right)}^2} + 2\sqrt {a + x} .\sqrt {a - x} } \over {{{\left( {\sqrt {a + x} } \right)}^2} - {{\left( {\sqrt {a - x} } \right)}^2}}}{/tex}

= {tex}{{a + x + a - x + 2\sqrt {{a^2} - {x^2}} } \over {a + x - a + x}}{/tex}

= {tex}{{2a + 2\sqrt {{a^2} - {x^2}} } \over {2x}}{/tex}

= {tex}{{a + \sqrt {{a^2} - {x^2}} } \over x}{/tex}

= {tex}{{a + \sqrt {{a^2} - {{\left( {{{2ab} \over {{b^2} + 1}}} \right)}^2}} } \over {{{2ab} \over {{b^2} + 1}}}}{/tex}

= {tex}a + \sqrt {{a^2} - {{4{a^2}{b^2}} \over {{{\left( {{b^2} + 1} \right)}^2}}}} \times {{{b^2} + 1} \over {2ab}}{/tex}

= {tex}a + a\sqrt {{{{{\left( {{b^2} + 1} \right)}^2} - 4{b^2}} \over {{{\left( {{b^2} + 1} \right)}^2}}}} \times {{{b^2} + 1} \over {2ab}}{/tex}

= {tex}a\left[ {1 + {1 \over {{b^2} + 1}}\sqrt {{b^4} + 1 + 2{b^2} - 4{b^2}} } \right] \times {{{b^2} + 1} \over {2ab}}{/tex}

= {tex}{{{b^2} + 1 + \sqrt {{b^4} + 1 - 2{b^2}} } \over {{b^2} + 1}} \times {{{b^2} + 1} \over {2b}}{/tex}

= {tex}{{{b^2} + 1 + \sqrt {{{\left( {{b^2} - 1} \right)}^2}} } \over {2b}}{/tex}

= {tex}{{{b^2} + 1 + {b^2} - 1} \over {2b}}{/tex}

= {tex}{{2{b^2}} \over {2b}}{/tex}

= {tex}b{/tex}

{tex}{(0.04)^{2}\over 0.008\times (0.2)^6}=(0.2)^k{/tex}

{tex}=> {((0.2)^2)^{2}\over( 0.2)^3\times (0.2)^6}=(0.2)^k{/tex}

{tex}=> {(0.2)^4\over( 0.2)^9}=(0.2)^k{/tex}

{tex}=> (0.2)^{4-9}=(0.2)^k{/tex}

{tex}=> (0.2)^{-5}=(0.2)^k{/tex}

On comparing both sides, we get

k = -5

Perimeter of Isosceles triangle = 30 cm

Each of equal side of given triangle = 12 cm

Let the third side be x cm

=>          12 + 12 + x = 30

=>          x = 30 - 24 = 6 cm

Now, Semi-perimeter (s) = {tex}{{30} \over 2} = 15{/tex} cm

Using Heron's formula,

Area of given triangle = {tex}\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} {/tex}

= {tex}\sqrt {15\left( {15 - 12} \right)\left( {15 - 12} \right)\left( {15 - 6} \right)} {/tex}

= {tex}\sqrt {15 \times 3 \times 3 \times 9} {/tex}

= {tex}9\sqrt {15} {/tex} sq. cm

= 9 x 3.87

= 34.83 sq. cm

{tex}p\left( x \right) = {x^3} - 10{x^2} - 53x - 42{/tex}

Putting x = -1 in p(x),

{tex}p\left( { - 1} \right) = {\left( { - 1} \right)^3} - 10{\left( { - 1} \right)^2} - 53\left( { - 1} \right) - 42{/tex} = {tex} - 1 - 10 + 53 - 42 = 0{/tex}

Since {tex}p\left( { - 1} \right) = 0{/tex}, therefore, (x + 1) is a factor of p(x)

Now, on dividing p(x) by (x + 1), we get the quotient {tex}{x^2} - 11x - 42{/tex}

Therefore, {tex}p\left( x \right) = {x^3} - 10{x^2} - 53x - 42{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} - 11x - 42} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} - 14x + 3x - 42} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {x\left( {x - 14} \right) + 3\left( {x - 14} \right)} \right]{/tex}

{tex}\Rightarrow{/tex}         {tex}p\left( x \right) = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 14} \right){/tex}

1) x^{<font size="2">4</font>}+4x^{<font size="2">2</font>}+16

= {tex}{x^4} + 16 + 4{x^2}{/tex}

= {tex}{\left( {{x^2}} \right)^2} + {\left( 4 \right)^2} + 2 \times {x^2} \times 4 - 2 \times {x^2} \times 4 + 4{x^2}{/tex}

= {tex}{\left( {{x^2} + 4} \right)^2} - 8{x^2} + 4{x^2}{/tex}

= {tex}{\left( {{x^2} + 4} \right)^2} - 4{x^2}{/tex}

= {tex}{\left( {{x^2} + 4} \right)^2} - {\left( {2x} \right)^2}{/tex}

= {tex}\left( {{x^2} + 4 + 2x} \right)\left( {{x^2} + 4 - 2x} \right){/tex}                 [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

= {tex}\left( {{x^2} + 2x + 4} \right)\left( {{x^2} - 2x + 4} \right){/tex}

1) 2√2x^{<font size="2">3</font>}+5√2x^{<font size="2">2</font>}-7√2

{tex}p\left( x \right) = 2\surd 2{x^3} + 5\surd 2{x^2} - 7\surd 2{/tex}

Putting x = 1, in p(x),

{tex}p\left( 1 \right) = 2\surd 2{\left( 1 \right)^3} + 5\surd 2{\left( 1 \right)^2} - 7\surd 2{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( 1 \right) = 2\surd 2 + 5\surd 2 - 7\surd 2{/tex} = {tex}7\surd 2 - 7\surd 2 = 0{/tex}

Since p(1) = 0, therefore, (x - 1) is a factor of p(x)

Dividing p(x) by (x -1), we get the quotient {tex}2\sqrt 2 {x^2} + 7\sqrt 2 {/tex}

Therefore,

{tex}p\left( x \right) = 2\surd 2{x^3} + 5\surd 2{x^2} - 7\surd 2{/tex}

{tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \left( {x - 1} \right)\left[ {2\sqrt 2 {x^2} + 7\sqrt 2 } \right]{/tex}

{tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \left( {x - 1} \right)\left[ {\sqrt 2 \left( {2{x^2} + 7} \right)} \right]{/tex}

{tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \sqrt 2 \left( {x - 1} \right)\left( {2{x^2} + 7} \right){/tex}

3) {tex}{p^4} - 81{q^4}{/tex}

= {tex}{\left( {{p^2}} \right)^2} - {\left( {9{q^2}} \right)^2}{/tex}                               [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

= {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {{p^2} - 9{q^2}} \right]{/tex}

= {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {{{\left( p \right)}^2} - {{\left( {3q} \right)}^2}} \right]{/tex}

= {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {\left( {p + 3q} \right)\left( {p - 3q} \right)} \right]{/tex}        [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

= {tex}\left( {{p^2} + 9{q^2}} \right)\left( {p + 3q} \right)\left( {p - 3q} \right){/tex}

2) p(x) = x^{<font size="2">3</font>}+13x^{<font size="2">2</font>}+32x+20

putting x = -1 in given polynomial,

p(-1) = (-1)^{<font size="2">3</font>}+ 13(-1)^{<font size="2">2</font>}+ 32(-1) +20

         = -1 + 13 - 32 + 20 = 0

Since p(-1) = 0, therefore (x + 1) is a factor of p(x).

Now, on dividing p(x) by (x + 1), we get the quotient x² + 12x + 20

{tex}p\left( x \right) = {x^3} + 13{x^2} + 32x + 20{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} + 12x + 20} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} + 10x + 2x + 20} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {x\left( {x + 10} \right) + 2\left( {x + 10} \right)} \right]{/tex}

{tex}\Rightarrow{/tex}         {tex}p\left( x \right) = \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 10} \right){/tex}

Yes zero is an even number.

{tex}\sqrt 2 ,\root 3 \of 4 ,\root 4 \of 3 {/tex}

= {tex}{2^{{1 \over 2}}},{4^{{1 \over 3}}},{3^{{1 \over 4}}}{/tex}

= {tex}{2^{{6 \over {12}}}},{4^{{4 \over {12}}}},{3^{{3 \over {12}}}}{/tex}             [Making all the powers as like terms]

= {tex}\root {12} \of {{2^6}} ,\root {12} \of {{4^4}} ,\root {12} \of {{3^3}} {/tex}

= {tex}\root {12} \of {64} ,\root {12} \of {256} ,\root {12} \of {27} {/tex}

Here {tex}\root {12} \of {256} {/tex} is greater.

Therefore, {tex}\root 3 \of 4 {/tex} is greater.

p(x) = k²x³-kx²+3kx-k

As x-3 is factor of p(x) so x= 3 is zero of this polynomial.

Using remainder theorem

p(3)=0

=> k​​​​^{​​​​​​2}(3)³-k(3)²+3k(3)-k=0

=> 27k²-9k+9k-k=0

=> 27k²-k =0

=> k(27k-1)=0

=> k = 0 or 27k-1=0

=> k = 0 or {tex}1\over 27{/tex}

We know that {tex}{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca{/tex}

{tex}\Rightarrow{/tex}          {tex}{a^2} + {b^2} + {c^2} = {\left( {a + b + c} \right)^2} - 2ab - 2bc - 2ca{/tex}

{tex}\Rightarrow{/tex}          {tex}{a^2} + {b^2} + {c^2} = {\left( {a + b + c} \right)^2} - 2\left( {ab + bc + ca} \right){/tex}

Putting the given values of a + b + c and ab + bc + ca, we get

{tex}\Rightarrow{/tex}          {tex}{a^2} + {b^2} + {c^2} = {\left( 9 \right)^2} - 2\left( {26} \right){/tex}

{tex}\Rightarrow{/tex}          {tex}{a^2} + {b^2} + {c^2} = 81 - 52 = 29{/tex}

Given : {tex}x^2+{1\over x^2}= 27{/tex}

We know,

{tex}(a+b)^2=a^2+b^2+2ab{/tex}

Put a = x and b={tex}{1\over x} {/tex}

We get

{tex}=> (x+{1\over x})^2 = x^2+{1\over x^2}+2\times x\times {1\over x}{/tex}

{tex}=> (x+{1\over x})^2 = 27+2=29{/tex}

{tex}=> x+{1\over x} =\pm \sqrt{29}{/tex}

{tex}{1\over2}\sqrt{486}-{\sqrt{27}\over2}{/tex}

{tex}{1\over2}\sqrt{2\times3\times3\times3\times3\times3}{/tex}—{tex}\sqrt{3\times3\times3}\over2{/tex}

{tex}{1\over2}\times3\times3\sqrt{6}{/tex}—{tex}3\sqrt{3}\over2{/tex}

{tex}{9\sqrt{6}\over2}—{3\sqrt{3}\over2}{/tex}

{tex}9\sqrt{6}—3\sqrt{3}\over2{/tex}

{tex}3(3\sqrt{6}—\sqrt{3})\over2{/tex}

Given : {tex}x^2+{1\over x^2} = 7{/tex}

To find : {tex}2x^2-{2\over x^2}{/tex}

Solution :

We know,

{tex}(a+b)^2 = a^2+b^2+2ab{/tex} ...(1)

Put a = x and b = {tex}1\over x{/tex}
{tex}(x+{1\over x})^2= x^2+{1\over x^2}+2{/tex}

=> {tex}(x+{1\over x})^2= 7+2=9{/tex}

=> {tex}x+{1\over x}= 3{/tex}...... (2)

Also 

{tex}(x-{1\over x})^2= x^2+{1\over x^2}-2{/tex}

{tex}=> (x-{1\over x})^2= 7-2{/tex}= 5

{tex}=> x-{1\over x}= \sqrt 5{/tex} ... (3)

Multiply (2) and (3)

{tex}x^2-{1\over x^2}= 3\sqrt 5{/tex}

Multiply both side by 2

{tex}2x^2-{2\over x^2}= 6\sqrt 5{/tex}