Let α and β be the roots of the equation x2−5x+5=0 . If a=αβ+βα and t=x2−4x+3a−15+1x2−4x+9,x∈R then

α+β=5, αβ=5 ∴a=α2+β2αβ=25−105=3 ∴t=x2−4x+9−15+1x2−4x+9,x∈R= (x−2)2+5−15+1(x−2)2+5=(x−2)2+5+1(x−2)2+5−15tmin when x=2⇒tmin=5+15−15=5 ∴ minimum of a + t is 8Option–C t≥5⇒log15t≤log155 log15t≤−1Option–D t≥5⇒log5t≥1⇒cot−1(log5t)≤cot−1⇒cot−1(log5t)≤π4

Let α and β be the roots of the equation x2−5x+5=0 . If a=αβ+βα and t=x2−4x+3a−15+1x2−4x+9,x∈R then

α+β=5, αβ=5 ∴a=α2+β2αβ=25−105=3 ∴t=x2−4x+9−15+1x2−4x+9,x∈R= (x−2)2+5−15+1(x−2)2+5=(x−2)2+5+1(x−2)2+5−15tmin when x=2⇒tmin=5+15−15=5 ∴ minimum of a + t is 8Option–C t≥5⇒log15t≤log155 log15t≤−1Option–D t≥5⇒log5t≥1⇒cot−1(log5t)≤cot−1⇒cot−1(log5t)≤π4

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Let $α$ and $β$ be the roots of the equation $x^{2} - 5 x + 5 = 0$ . If $a = \frac{α}{β} + \frac{β}{α}$ and $t = x^{2} - 4 x + 3 a - \frac{1}{5} + \frac{1}{x^{2} - 4 x + 9}, x \in R$ then

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Minimum value of a + t is $\frac{24}{5}$

Minimum value of a + t is 8

The maximum value of $l o g_{\frac{1}{5}} t$ is –1

The range of $y = \cot^{- 1} (l o g_{5} t)$ is $(0, \frac{π}{4}]$

answer is B, C, D.

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Detailed Solution

$α + β = 5, α β = 5 ∴ a = \frac{α^{2} + β^{2}}{α β} = \frac{25 - 10}{5} = 3 ∴ t = x^{2} - 4 x + 9 - \frac{1}{5} + \frac{1}{x^{2} - 4 x + 9}, x \in R$
= ${(x - 2)}^{2} + 5 - \frac{1}{5} + \frac{1}{{(x - 2)}^{2} + 5} = {(x - 2)}^{2} + 5 + \frac{1}{{(x - 2)}^{2} + 5} - \frac{1}{5}$
$t_{\min}$ when $x = 2 \Rightarrow t_{\min} = 5 + \frac{1}{5} - \frac{1}{5} = 5$ $∴$ minimum of a + t is 8
Option–C $t \geq 5 \Rightarrow l o g_{\frac{1}{5}} t \leq l o g_{\frac{1}{5} 5} l o g_{\frac{1}{5}} t \leq - 1$
Option–D $t \geq 5 \Rightarrow l o g_{5} t \geq 1 \Rightarrow \cot^{- 1} (l o g_{5} t) \leq \cot^{- 1} \Rightarrow \cot^{- 1} (l o g_{5} t) \leq \frac{π}{4}$