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Sia ? 6 years, 6 months ago
{tex}f(x) = \sqrt {9 - {x^2}} {/tex}
The function is not defined when 9 - x2 < 0
{tex}\therefore {/tex} Domain of function {tex}=\{ x:9 - {x^2} \geqslant 0\} = \{ x:{x^2} - 9 \leqslant 0\} {/tex}
{tex} = \{ x:(x + 3)(x - 3) \leqslant 0\} {/tex}
= [-3, 3]
Now {tex}f(x) = \sqrt {9 - {x^2}} \geqslant 0{/tex} for {tex}x \in [ - 3,3]{/tex}
{tex}\therefore {/tex} Range of function = {tex}\left[ {0,\infty } \right){/tex}
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Since {tex}(3k-2), (4k-6)\ and \ (k+2){/tex} are in AP, we have
{tex}(4k-6)-(3k-2)=(k+2)-(4k-6){/tex}
{tex}\Rightarrow{/tex} {tex}4k - 6 - 3k + 2 = k + 2 - 4k + 6{/tex}
{tex}\Rightarrow{/tex} {tex}k - 4 = -3k + 8{/tex}
{tex}\Rightarrow{/tex} 4k = 12
{tex}\Rightarrow{/tex} k = 3.
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