Let - and - be roots of x2- 6t2 - 2t - 2x - 2 - 0with - - - If an - -n - -n for n - then find the minimum value of a100 - 2a98-a99 where t - R

x ^ 2 6( t ^ 2 2t+2)x 2=0 roots are α ,β ∴α #160; ^ 2 6( t ^ 2 2t+2)α 2 α #160; ^ 2 2=6( t ^ 2 2t+2)α ∴α #160; ^ 100 =6( t ...

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Byju's Answer

Standard XII

Mathematics

Determinant

Let α and ...

Question

Let $α$ and $β$ be roots of $x^{2} - 6 (t^{2} - 2 t + 2) x - 2 = 0$
with $α > β$ . If $a_{n} = α^{n} - β^{n}$ for n $\geq$ , then find the minimum value of $\frac{a_{100} - 2 a_{98}}{a_{99}}$ (where t $\in$ R)

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Solution

$x^{2} - 6 (t^{2} - 2 t + 2) x - 2 = 0 r o o t s a r e α, β ∴ α^{2} - 6 (t^{2} - 2 t + 2) α - 2 α^{2} - 2 = 6 (t^{2} - 2 t + 2) α ∴ α^{100} = 6 (t^{2} - 2 t + 2) α^{98} α^{100} - 2 α^{98} = 6 (t^{2} - 2 t + 2) α^{99} ∴ β^{100} - 2 β^{98} = 6 (t^{2} - 2 t + 2) β^{99} a_{x} = α^{x} - β^{x} \frac{a_{100} - {2 a}_{98}}{a_{99}} = \frac{α^{100} - 2 α^{98} - β^{100} + 2 β^{98}}{α^{99} - β^{99}} = \frac{6 (t^{2} - 2 t + 2) (α^{99} - β^{99})}{(α^{99} - β^{99})} m i n i m u m v a l u e o f (t^{2} - 2 t + 2) = 1 ∴ s o m i n i m u m v a l u e = 6$

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Similar questions

Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ with $α > β$ if $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ then the value of $\frac{a_{10} - 2 a_{8}}{3 a_{9}} =$

Q. Let

α

and

β

be the roots of

x^{2} - 6 x - 2 = 0

, with

α > β

. If

a_{n} = α^{n} - β^{n} f o r n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{2 a_{9}}

Q. Let

α

and

β

be the roots of

x^{2} - 6 x - 2 = 0

. If

a_{n} = α^{n} - β^{n}

for

n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{3 a_{9}}

is:

Q. Let

α

and

β

be the roots of equation

x^{2} - 6 x - 2 = 0

. If

a_{n} = α^{n} - β^{n}

, for

n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{2 a_{9}}

Q. Let

α

and

β

be the roots of

x^{2} - x - 1 = 0,

with

α

β

, For all positive integers n, define,

a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1

b_{1} = 1

and

b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2

.
Then which of the following option is/are correct?

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Let α and β be roots of x2−6(t2−2t+2)x−2=0with α>β. If an=αn−βn for n ≥, then find the minimum value of a100−2a98a99 (where t ∈ R)

Let $α$ and $β$ be roots of $x^{2} - 6 (t^{2} - 2 t + 2) x - 2 = 0$
with $α > β$ . If $a_{n} = α^{n} - β^{n}$ for n $\geq$ , then find the minimum value of $\frac{a_{100} - 2 a_{98}}{a_{99}}$ (where t $\in$ R)