Question

If the function
$\tt\[ f(x)=\left\{\begin{array}{c} \tt \dfrac{(1-\cos 4 x)}{x^{2}}, \text { if } x<0 \\ \\ \tt a, \text { if } x=0 \\ \\ \tt \dfrac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, \text { if } x>0 \end{array}\right. \]$
continuous at x=0 , then the value of a is
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Answer 1

Answer :

a = 8.

To Find :

The value of a.

Solution :

If the function f( x ) is continuous at x = 0. then,

$\tt \implies LHL = RHL = f( 0) \\ \\ \tt or, \\ \\ \implies \tt \displaystyle \lim_{x \to 0^{\pm}} \tt f(x) = f(0)$

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LHL,

$\implies \: \displaystyle \lim_{x \to {0}^{ - } } \tt \dfrac{1 - cos4x}{ {x}^{2} }$

$\implies \: \displaystyle \lim_{x \to {0}^{ - } } \tt \dfrac{2 {sin}^{2}2x }{ {(2x)}^{2} } \times {(2)}^{2}$

$\implies \tt 4 \times 2 \: \displaystyle \lim_{x \to {0}^{ - } } \tt {\Bigg(\dfrac{ sin2x }{ 2x}\Bigg)}^{2}$

$\implies \tt4 \times 2 .$

$\implies \tt8 .$

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RHL,

$\implies \: \displaystyle \lim_{x \to {0}^{ + } } \tt \dfrac{ \sqrt{x} }{ { \sqrt{16 + \sqrt{x} - 4} } }$

$\implies \: \displaystyle \lim_{x \to {0}^{ + } } \tt \dfrac{ \sqrt{x} }{ { \sqrt{16 + \sqrt{x} } - 4 } } \times \tt \dfrac{ \sqrt{16 + \sqrt{x} } + 4}{ { \sqrt{16 + \sqrt{x} } + 4 } }$

$\implies \: \displaystyle \lim_{x \to {0}^{ + } } \tt \dfrac{ \cancel{\sqrt{x}} \bigg( \sqrt{16 + \sqrt{x} } + 4 \bigg)} { \cancel{16} + \cancel{\sqrt{x}} - \cancel{ 16} }$

$\implies \: \displaystyle \lim_{x \to {0}^{ + } } \tt \sqrt{16 + \sqrt{0} } + 4$

$\implies \: \tt8.$

So,

=> f ( 0 ) = a.

=> a = 8.

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More Information

Continuous in R.

• Polynomial function.
• Modulus function.
• Constant function.
• Exponential function.
• Sin x.
• Cos x.

Answer 2

SOLUTION :

please check the attached file