TSA of the article = 2{tex}\pi{/tex}rh + 2(2{tex}\pi{/tex}r²)
= 2{tex}\pi{/tex} (3.5)(10) + 2[2{tex}\pi{/tex} (3.5)²]
= 70{tex}\pi{/tex} + 49{tex}\pi{/tex}
= 119 {tex}\pi{/tex}
= 119 {tex}\times{/tex} {tex}\frac {22}7{/tex}
= 374 cm ²

S = 3 + 6 + 9 + 12 + .... + 24
= 3(1 + 2 + 3 + ... + 8)
= 3 {tex}\times{/tex} {tex}\frac{8 \times 9}{2}{/tex} = 108

{tex}\sqrt {{x^2} + {y^2}} {/tex}

It is given that x = 2 and x = 3 are roots of the equation 3x² - 2kx + 2m = 0.

{tex}\therefore{/tex}  3 {tex}\times{/tex} 2² - 2k {tex}\times{/tex} 2 + 2m = 0 and 3 {tex}\times{/tex} 3² - 2k {tex}\times{/tex} 3 + 2m = 0
{tex}\Rightarrow{/tex} 12 - 4k + 2m = 0 and 27 - 6k + 2m = 0
{tex}\Rightarrow{/tex} 12 = 4k - 2m...(i) and 27 = 6k - 2m...(ii)
Solving i and ii equation, we get k = {tex}\frac { 15 } { 2 }{/tex} and m = 9

LHS
{tex} = {\sin ^6}\theta + {\cos ^6}\theta {/tex}
{tex} = {\left( {{{\sin }^2}\theta } \right)^3} + {\left( {{{\cos }^2}\theta } \right)^3}{/tex}
{tex} = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^3} - 3{\sin ^2}\theta {\cos ^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right){/tex}
{tex}\left[ {\because {a^3} + {b^3} = {{(a + b)}^3} - 3ab(a + b)} \right]{/tex}
{tex} = {(1)^3} - 3{\sin ^2}\theta {\cos ^2}\theta \times 1{/tex} {tex}\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]{/tex}
{tex} = 1 - 3{\sin ^2}\theta {\cos ^2}\theta {/tex}
RHS {tex} = \frac{{4 - 3{{({x^2} - 1)}^2}}}{4}{/tex}
{tex} = \frac{{4 - 3{{\left\{ {{{\left( {\sin \theta + \cos \theta } \right)}^2} - 1} \right\}}^2}}}{4}{/tex} {tex}\left[ {given\;x = \sin \theta + \cos \theta } \right]{/tex}
{tex} = \frac{{4 - 3{{\left\{ {{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta - 1} \right\}}^2}}}{4}{/tex}
{tex} = \frac{{4 - 3{{\left\{ {1 + 2\sin \theta \cos \theta - 1} \right\}}^2}}}{4}{/tex} {tex}\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]{/tex}
{tex} = \frac{{4 - 3{{\left( {2\sin \theta \cos \theta } \right)}^2}}}{4}{/tex}
{tex} = \frac{{4 - 3 \times 4{{\sin }^2}\theta {{\cos }^2}\theta }}{4}{/tex}
{tex} = \frac{{4\left( {1 - 3{{\sin }^2}\theta {{\cos }^2}\theta } \right)}}{4}{/tex}
LHS = RHS
Hence proved.

we have, 2x² + x - 4 = 0

Dividing both sides by 2, we have

{tex}x^2 +{1 \over 2}x - 2 =0{/tex}

{tex}\implies x^2 + {1\over2} x +{1 \over 16} = 2 + {1 \over 16}{/tex}

{tex}\implies (x)^2 + 2 (x) {1 \over 4} + ({1 \over 4})^2 = {32+1 \over16}{/tex}

{tex}\implies (x + {1 \over 4})^2 = {33 \over 16}{/tex}

{tex}\implies (x + {1 \over 4}) = \pm {\sqrt 33 \over 4}{/tex}

{tex}\implies x = {\sqrt33 \over 4} - {1 \over4}, \, -{\sqrt33 \over 4} - {1 \over4}{/tex}

{tex}\implies x = {\sqrt 33 - 1 \over 4},\, {-\sqrt33 - 1 \over 4}{/tex}

{tex}\therefore x = {\sqrt 33 - 1 \over 4},\, {-\sqrt33 - 1 \over 4}{/tex} are the required roots.

The zeros of f(x) are given by f(x) = 0
{tex} \Rightarrow {/tex}x² + 7x + 10 = 0
{tex}\Rightarrow {/tex}(x + 5)(x + 2) = 0
{tex}\Rightarrow {/tex} x + 5 = 0 or, x + 2 = 0
{tex}\Rightarrow {/tex} x = -5 or, -2
Thus, the zeroes of f(x) = x² +7x + 12 are {tex}\alpha = - 5 \text { and } \beta = - 2{/tex}

 {tex}\alpha + \beta{/tex}= (-5) + (-2) = -7.......(1)

and  {tex}\alpha\beta=(-5)(-2)=10.….....(2){/tex}

In the eqn x^{​​​​​2}​​​ +7x+10=0

a=1,b=7,c=10

So  {tex}\alpha +\beta =\frac{-b}{a}=\frac{-7}{1}=-7..........(3){/tex}

and {tex}\alpha\beta=\frac ca=\frac{10}{1}=10........(4)..{/tex}

Hence from (1) and (3) sum of zeros and from (2) ,and (4) product of zeros is verified.

L.H.S. {tex}= (\sin \theta + \cos \theta )(\tan \theta + \cot \theta ){/tex}
{tex}= (\sin \theta + \cos \theta )\left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right){/tex}
{tex}= (\sin \theta + \cos \theta )\left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}} \right){/tex}
{tex} = (\sin \theta + \cos \theta )\left( {\frac{1}{{\sin \theta \cos \theta }}} \right){/tex}
{tex} = \frac{{\sin \theta + \cos \theta }}{{\sin \theta \cos \theta }}{/tex}
{tex} = \frac{{\sin \theta }}{{\sin \theta \cos \theta }} + \frac{{\cos \theta }}{{\sin \theta \cos \theta }}{/tex}
{tex}= \frac{1}{{\cos \theta }} + \frac{1}{{\sin \theta }}{/tex}
{tex}= \sec \theta + \cos ec\theta {/tex}
= R.H.S.

{tex}\text {Series of odd is }1, 3, 5, ..., 49.{/tex}

{tex}\therefore \text { first number } a = 1 \text { last number } l = 49{/tex}

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

{tex}\Rightarrow 49=1+(n-1)(2){/tex}

{tex}\Rightarrow n=25{/tex}

{tex}\text {Total odd numbers from 0 to 50 are}{/tex}

{tex}\text {Now, sum of odd numbers from 1 to 50 } S_n \text { is }{/tex}

{tex}S_n = \cfrac {n}{2}(a+l){/tex}

{tex}\Rightarrow S_n = \cfrac {25}{2} (1+49){/tex}

{tex}\Rightarrow S_n = 25 \times 25=625{/tex}

{tex}\text {Thus, required sum is 625.}{/tex}

{tex}\text {Given two equations are } x+y=0 \text { and } x-y=0{/tex}

{tex}\text {On solving these two equations, algebrically with any of methods knows, } i.e., \text { substitution, elimination, etc., we get,}{/tex}

{tex}x=0 \text { and } y = 0{/tex}

H.C.F = H.C.F of numerators / L.C.M of denominators

L.C.M = L.C.M of numerators / H.C.F of denominators