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Preeti Dabral 2 years, 7 months ago
Let {tex}\alpha{/tex} and 2{tex}\alpha{/tex} are the zeroes of the polynomial 2x2 - 5x - (2k + 1).
Then, 2{tex}\alpha{/tex}2 - 5{tex}\alpha{/tex} - (2k + 1) = 0
and 2 (2{tex}\alpha{/tex})2 - 5(2{tex}\alpha{/tex}) - (2k + 1) = 0
{tex}\Rightarrow{/tex} 2{tex}\alpha{/tex}2 - 5{tex}\alpha{/tex} = 2k + 1 ...(i)
and 8{tex}\alpha^2{/tex} - 10{tex}\alpha{/tex} = 2k + 1 ...(ii)
From Eqs. (i) and (ii), we get
2{tex}\alpha^2{/tex} - 5a = 8{tex}\alpha^2{/tex} - 10{tex}\alpha{/tex} {tex}\Rightarrow{/tex} 6{tex}\alpha^2{/tex} = 5{tex}\alpha{/tex} {tex}\Rightarrow{/tex} {tex}\alpha=\frac{5}{6}{/tex} {tex}[\because \alpha \neq 0]{/tex}
{tex}\therefore{/tex} 2{tex}\alpha{/tex} = {tex}\frac{5}{6} \times 2{/tex} {tex}=\frac{5}{3}{/tex}
Thus, the zeroes of the polynomial are {tex}\frac{5}{6}{/tex} and {tex} \frac{5}{3}{/tex}
Now, substituting {tex}\alpha=\frac{5}{6}{/tex} in Eq. (i), we get
2 {tex}\times{/tex} {tex} \frac{25}{36}-\frac{25}{6}{/tex} = 2k + 1
{tex}\Rightarrow{/tex} 2k + 1 = {tex}\frac{50-150}{36} {/tex} {tex}\Rightarrow 2 k+1{/tex} = {tex}-\frac{100}{36}{/tex}
{tex}\Rightarrow{/tex} 2k = {tex}-\frac{100}{36}-1{/tex} {tex} \Rightarrow{/tex} {tex} 2 k={/tex} {tex}-\frac{136}{36}{/tex}
{tex}\Rightarrow{/tex} k = - {tex}\frac{68}{36}{/tex} = {tex}\frac{-17}{9}{/tex}
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Preeti Dabral 2 years, 8 months ago
f(x) = 4x2 + 4x - 3
{tex}\Rightarrow \quad f \left( \frac { 1 } { 2 } \right) = 4 \left( \frac { 1 } { 4 } \right) + 4 \left( \frac { 1 } { 2 } \right) - 3{/tex}
= 1 + 2 - 3 = 0
and {tex}f \left( - \frac { 3 } { 2 } \right) = 4 \left( \frac { 9 } { 4 } \right) + 4 \left( - \frac { 3 } { 2 } \right) - 3{/tex}
= 9 - 6 - 3 = 0
{tex}\therefore \frac { 1 } { 2 } , - \frac { 3 } { 2 }{/tex}are zeroes of polynomial 4x2 + 4x - 3.
Sum of zeroes = {tex}\frac { 1 } { 2 } - \frac { 3 } { 2 } = - 1 = \frac { - 4 } { 4 }{/tex}
{tex}= - \frac { \text { Coefficient of } x } { \text { Coefficient of } x ^ { 2 } }{/tex}
Product of zeroes = {tex}\left( \frac { 1 } { 2 } \right) \left( - \frac { 3 } { 2 } \right) = \frac { - 3 } { 4 }{/tex}
{tex}= \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } }{/tex}
Verified.
Posted by Ali Ahamad Khan 2 years, 8 months ago
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Avinash Tripathi 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
17×5×11×3×2+2×11=17×5×3×22+22
=22(17×5×3+1)
=22(255+1)=2×11×256 -----(i)
Equation (i) is divisible by 2, 11 and 256, so it has more than 2 prime factors.
∴ (17×5×11×3×2+2×11) is a composite number.
Payal Kumari Kushwaha 2 years, 8 months ago
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Posted by Tharun Nayak 2 years, 8 months ago
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Preeti Dabral 2 years, 8 months ago
Height of the cylinder (h) = 14 cm,
Base diameter = 7 cm
{tex}\Rightarrow{/tex} Radius of the base of the cylinder (r) = 3.5 cm
Volume of the cylinder = {tex}\pi r ^ { 2 } h{/tex}
= {tex}\frac { 22 } { 7 } \times 3.5 \times 3.5 \times 14{/tex}
= 22 {tex}\times{/tex}3.5 {tex}\times{/tex}3.5{tex}\times{/tex}14
= 539 cm3
Radius of the conical holes (r1) = 2.1 cm,
Height of the conical holes (h1) = 4 cm,
volume of the conical hole {tex}= \frac { 1 } { 3 } \pi r _ { 1 } ^ { 2 } h _ { 1 }{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 2.1 \times 2.1 \times 4{/tex}
= 18.48 cm3
Volume of the two conical hole = 2 {tex}\times{/tex}18.48
= 36.96 cm3
Volume of the remaining solid = Volume of the cylinder - Volume of two conical hole
= 539 - 36.96
= 502.04 cm3
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Anshika Maurya 2 years, 7 months ago
1Thank You