P -1-log 3-1-log 4-1 thenA- 2-53- 1-5- p -2D- 2- p -2-5

The correct option is B p 3p=1log_3π+1log_4π+1 =log_π3+log_π4+1 =log_π12+1 We now that, 12 π^2 ⇒log_π12log_ππ^2 ⇒ nbsp;log_π122 Hence p3 ...

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Byju's Answer

Standard XII

Mathematics

Condition for Monotonically Decreasing Function

p =1/log 3π+1...

Question

If $p = \frac{1}{{log}_{3} π} + \frac{1}{{log}_{4} π} + 1$ then

1.5 < p < 2

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2 < p < 2.5

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2.5 < p < 3

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p > 3

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Solution

The correct option is D $p > 3$
$p = \frac{1}{{log}_{3} π} + \frac{1}{{log}_{4} π} + 1 = {log}_{π} 3 + {log}_{π} 4 + 1 = {log}_{π} 12 + 1$

We now that,
$12 > π^{2}$
$\Rightarrow {log}_{π} 12 > {log}_{π} π^{2} \Rightarrow {log}_{π} 12 > 2$

Hence $p > 3$

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If p=1log3π+1log4π+1 then

The correct option is D p>3p=1log3π+1log4π+1=logπ3+logπ4+1=logπ12+1 We now that, 12>π2 ⇒logπ12>logππ2⇒logπ12>2 Hence p>3

If $p = \frac{1}{{log}_{3} π} + \frac{1}{{log}_{4} π} + 1$ then

The correct option is D $p > 3$
$p = \frac{1}{{log}_{3} π} + \frac{1}{{log}_{4} π} + 1 = {log}_{π} 3 + {log}_{π} 4 + 1 = {log}_{π} 12 + 1$

We now that,
$12 > π^{2}$
$\Rightarrow {log}_{π} 12 > {log}_{π} π^{2} \Rightarrow {log}_{π} 12 > 2$

Hence $p > 3$