Let M and m respectively be the maximum and minimum values of the function f(x) = tan^-1 (sin x + cos x) in $[0, \frac{π}{2}] .$ Then the value of tan(M – m) is equal to

Option 1 -

3 - $2 \sqrt[]{2}$

Option 2 -

$2 + \sqrt[]{3}$

Option 3 -

3 + $2 \sqrt[]{2}$

Option 4 -

$2 - \sqrt[]{3}$

5 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 1

Detailed Solution:

f(x) = tan^-1 (sin x + cos x)

$a s x \in [0, \frac{π}{2}]$ so 1 $\leq s i n x + c o s x \leq \sqrt[]{2}$

so $f (x) \in [t a n^{- 1} 1, t a n^{- 1} \sqrt[]{2}]$

->M = tan^-1 $\sqrt[]{2} & m = t a n^{- 1} 1 = \frac{π}{4}$

Now, tan (M – m) = tan $(t a n^{- 1} \sqrt[]{2} - \frac{π}{4})$

$= \frac{\sqrt[]{2} - 1}{1 + \sqrt[]{2} . 1} = \frac{\sqrt[]{2} - 1}{\sqrt[]{2} + 1} \times \frac{\sqrt[]{2} - 1}{\sqrt[]{2} - 1} = 3 - 2 \sqrt[]{2}$

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Let M and m respectively be the maximum and minimum values of the function f(x) = tan^-1 (sin x + cos x) in $[0, \frac{π}{2}] .$ Then the value of tan(M – m) is equal to

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Let M and m respectively be the maximum and minimum values of the function f(x) = tan-1 (sin x + cos x) in [ 0 , π 2 ] . Then the value of tan(M – m) is equal to

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Let M and m respectively be the maximum and minimum values of the function f(x) = tan^-1 (sin x + cos x) in $[0, \frac{π}{2}] .$ Then the value of tan(M – m) is equal to