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प्रश्न
The value of f (0), so that the function
पर्याय
a3/2
a1/2
−a1/2
−a3/2
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उत्तर
\[- a^\frac{1}{2}\]
Given:
\[\Rightarrow f\left( x \right) = \frac{\left( \sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( a^2 - ax + x^2 - \left( a^2 + ax + x^2 \right) \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( - 2ax \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( - 2ax \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( a + x - a + x \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{\left( - 2ax \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( 2x \right)\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \Rightarrow f\left( x \right) = \frac{- a\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
If \[f\left( x \right)\] is continuous for all x, then it will be continuous at x = 0 as well.
So, if \[f\left( x \right)\] is continuous at x = 0, then
\[ \Rightarrow \left[ \frac{- 2a\left( \sqrt{a} \right)}{\left( \sqrt{a^2} + \sqrt{a^2} \right)} \right] = f\left( 0 \right)\]
\[ \Rightarrow \left[ \frac{- 2a\left( \sqrt{a} \right)}{\left( a + a \right)} \right] = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = - \sqrt{a}\]
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