Question

hello dear friends
solve it Now.
Q . 18/19/20/21
it is very urgent​

Answer 1

$\bf \:\large \red{AηsωeR : 18} ✍$

$\tt \:  f(x) = \dfrac{ {x}^{2} - x}{x + 3}$

$\tt \:  ⟼f(1) = \dfrac{1 - 1}{1 + 3} = 0$

$\tt \:  ⟼ \: f(2) = \dfrac{4 - 2}{2 + 3} = \dfrac{2}{5}$

$\tt \:  ⟼ \: f( - 2) = \dfrac{4 + 2}{ - 2 + 3} = 6$

$\tt \:  ⟼ \: f(0) = \dfrac{0 - 0}{0 + 3} = 0$

$\tt \:  ⟼ \: Consider \: \dfrac{f(1) + f(2)}{f( - 2) + f(0)}$

$\tt \:  = \: \dfrac{0 + \dfrac{2}{5} }{6 + 0}$

$\tt \:  = \: \dfrac{2}{30}$

$\tt \:  = \dfrac{1}{15}$

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$\bf \:\large \red{AηsωeR : 19} ✍$

$\tt \:  ⟼ \: f(x) = {x}^{2} {(x - 1)}^{2}$

$\tt \:  Hence, f(x + 1) = {(x + 1)}^{2} {(x + 1 - 1)}^{2}$

$\tt \:  ⟼ \: f(x + 1) = {(x + 1)}^{2} {x}^{2}$

$\tt \:  Consider \: f(x + 1) \: - \: f(x)$

$\tt \:  = {(x + 1)}^{2} {x}^{2} - {x}^{2} {(x - 1)}^{2}$

$\tt \:  = {x}^{2} \bigg( {(x + 1)}^{2} - {(x - 1)}^{2} \bigg)$

$\tt \:  = {x}^{2} (4x)$

$\tt \:  = 4 {x}^{3}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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$\bf \:\large \red{AηsωeR : 20} ✍$

$\tt \:  ⟼f(n) = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4}$

$\tt \:  Hence, f(n - 1) = \dfrac{ {(n - 1)}^{2} {n}^{2} }{4}$

$\tt \:  ⟼ \: Consider \: f(n) - f(n - 1)$

$\tt \:  = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} - \dfrac{ {(n - 1) }^{2} {n}^{2} }{4}$

$\tt \:  = \dfrac{ {n}^{2} }{4} \bigg( {(n + 1)}^{2} - {(n - 1)}^{2} \bigg)$

$\tt \:  = \dfrac{ {n}^{2} }{4} \times 4n$

$\tt \:  = {n}^{3}$

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$\bf \:\large \red{AηsωeR : 21} ✍$

$\tt \:  ⟼f(x) = 5 {x}^{2} + 7x + 2$

$\bf\implies \: {5x}^{2} + 7x + 2 = 110$

$\tt \:  ⟼ {5x}^{2} + 7x - 108 = 0$

$\tt \:  ⟼ {5x}^{2} + 27x - 20x - 108 = 0$

$\tt \:  ⟼x(5x + 27) - 4(5x + 27) = 0$

$\tt \:  ⟼(5x + 27)(x - 4) = 0$

$\bf\implies \:x = - \dfrac{27}{5} \: or \: x \: = 4$

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