The value of k if the equation kx - sin-1x2-8x-17 - cos-1x2-8x-17 - 9-2 has atleast one solution is

The correct option is B π x^2 8x+17=(x 4)^2+1∈ [ 1,1] (x 4)^2∈ [ 2,0] x 4=0 x=4 kx+sin^ 1(x^2 8x+17)+cos^ 1(x^2 8x+17)=9π2 4k+π2=9π2 4 ...

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Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

The value of ...

Question

The value of $k$ if the equation $k x + {sin}^{- 1} (x^{2} - 8 x + 17) + {cos}^{- 1} (x^{2} - 8 x + 17) = \frac{9 π}{2}$ has atleast one solution is

2 π

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π

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1

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\frac{π}{2}

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Solution

The correct option is B $π$
$⟹ x^{2} - 8 x + 17 = (x - 4)^{2} + 1 \in [- 1, 1]$
$⟹ (x - 4)^{2} \in [- 2, 0] ⟹ x - 4 = 0 ⟹ x = 4$
$k x + {sin}^{- 1} (x^{2} - 8 x + 17) + {cos}^{- 1} (x^{2} - 8 x + 17) = \frac{9 π}{2}$
$4 k + \frac{π}{2} = \frac{9 π}{2} ⟹ 4 k = 4 π ⟹ k = π$

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The value of k if the equation kx+ sin−1(x2−8x+17)+ cos−1(x2−8x+17)=9π2 has atleast one solution is

The correct option is B π⟹x2−8x+17=(x−4)2+1∈[−1,1]⟹(x−4)2∈[−2,0]⟹x−4=0⟹x=4kx+sin−1(x2−8x+17)+cos−1(x2−8x+17)=9π24k+π2=9π2⟹4k=4π⟹k=π

The value of $k$ if the equation $k x + {sin}^{- 1} (x^{2} - 8 x + 17) + {cos}^{- 1} (x^{2} - 8 x + 17) = \frac{9 π}{2}$ has atleast one solution is

The correct option is B $π$
$⟹ x^{2} - 8 x + 17 = (x - 4)^{2} + 1 \in [- 1, 1]$
$⟹ (x - 4)^{2} \in [- 2, 0] ⟹ x - 4 = 0 ⟹ x = 4$
$k x + {sin}^{- 1} (x^{2} - 8 x + 17) + {cos}^{- 1} (x^{2} - 8 x + 17) = \frac{9 π}{2}$
$4 k + \frac{π}{2} = \frac{9 π}{2} ⟹ 4 k = 4 π ⟹ k = π$