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Easy

Question

If k less or equal than sin to the power of negative 1 end exponent invisible function application x plus cos to the power of negative 1 end exponent invisible function application x plus tan to the power of negative 1 end exponent invisible function application x less or equal than K, then

  1. k equals 0 comma K equals pi
  2. k equals 0 comma K equals pi divided by 2
  3. k equals pi divided by 2 comma K equals pi
  4. None of these

hintHint:

In this question, we have to find the k and K, we have given k ≤ sin-1x + cos-1x + tan-1x ≤ K. Here, it is asking about the period of this function.

The correct answer is: k equals 0 comma K equals pi


    Here we have to find the range of function which , k and K.
    Firstly, we have k ≤ sin-1x+ cos-1x +tan-1x ≤ K
    We can write,
    sin-1x + cos-1x + tan-1x = straight pi over 2+ tan-1x [we know sin-1x + cos-1x = straight pi over 2
]
    since,
    k ≤ straight pi over 2
+ tan-1x ≤ K
    so , sin-1x and cos-1x only defined if -1 ≤ x ≤ 1
    we know that,
    (fraction numerator negative straight pi over denominator 4 end fraction) ≤ tan-1x ≤(straight pi over 4)
    (fraction numerator negative straight pi over denominator 4 end fraction plus straight pi over 2) ≤ straight pi over 2 + tan-1x ≤(straight pi over 4 plus straight pi over 2 )
    (straight pi over 4) ≤ straight pi over 2 + tan-1x ≤(fraction numerator 3 straight pi over denominator 4 end fraction)
    Therefore, k = straight pi over 4 and K = fraction numerator 3 straight pi over denominator 4 end fraction
    The correct answer is None of these.

    In this question, we have to find the range of function from k to K. Here we know that sin-1x + cos-1x = straight pi over 2 and it only defines if -1 ≤ x ≤ 1 . And tan^-1x lines in (fraction numerator negative straight pi over denominator 4 end fraction) to ( straight pi over 4) .

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