Question

Please Don't Spam
and Options (D) is correct.
I want Full Explanation ​

Answer 1

Topic :-

Inverse Trigonometric Function

Given :-

$(\cot^{-1}x)^2-7(\cot^{-1}x)+10>0$

To Find :-

Interval in which all 'x' satisfying the given inequality lies.

Solution :-

$\sf {Consider\:\cot^{-1}x=t,}$

So, given inequality changes to,

$\sf{t^2-7t+10>0}$

Solving it,

$\sf {t^2-5t-2t+10>0}$

$\sf {t(t-5)-2(t-5)>0}$

$\sf {(t-5)(t-2)>0}$

$\sf {It\:is\:true\:when\:\:t<2\:and\:t>5.}$

So,

$\sf{\cot^{-1}x>5}$

We will reject this case as range of cot inverse is (0, π).

( 5 is exceeding the range so this case get rejected. )

$\sf{\cot^{-1}x<2}$ or

$\sf{x>\cot2}$

$So,\:x \:\in \:(\cot2,\infty).$

Answer :-

Interval in which all 'x' satisfying the given inequality lies is $(\cot2,\infty)$ which is option D.

Answer 2

Step-by-step explanation:

Topic :-

Inverse Trigonometric Function

Given :-

$(\cot^{-1}x)^2-7(\cot^{-1}x)$

To Find :-

Interval in which all 'x' satisfying the given inequality lies.

Solution :-

$\sf {Consider\:\cot^{-1}x=t,}$

So, given inequality changes to,

$\sf{t^2-7t+10 > 0}$

Solving it,

$\sf {t^2-5t-2t+10 > 0}$

$\sf {t(t-5)-2(t-5) > 0}t(t−5)−2(t−5)>0$

$\sf {(t-5)(t-2) > 0}(t−5)(t−2)>0$

$\sf {It\:is\:true\:when\:\:t < 2\:and\:t > 5.}$

So,

$\sf{\cot^{-1}x > 5}cot$

We will reject this case as range of cot inverse is (0, π).

( 5 is exceeding the range so this case get rejected. )

$\sf{\cot^{-1}x < 2}cot$

$\sf{x > \cot2}x>cot2$

$So,\:x \:\in \:(\cot2,\infty).So,x∈(cot2,∞)$

Answer :-

Interval in which all 'x' satisfying the given inequality lies is (\cot2,\infty)(cot2,∞) which is option D.