The value of a so that the function fx-1-cos 4xx2- x0-is continuous at x-0 is

F(0)^ =lim_x → 0^ f(x)=lim_x → 0^ 1 cos 4xx^2 =lim_x → 0^ 2sin^2 2xx^2=8 f(0)^+=lim_x → 0^+ f(x) =lim_x → 0^+√(x)√(16+√(x)) 4 =lim_x → 0^+√(x)· ( √(16+√(x))+4 ...

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Standard XII

Mathematics

Properties of Determinants

The value of ...

Question

The value of $a$ so that the function $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & x < 0 a, & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{matrix}$
is continuous at $x = 0$ is

2

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4

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6

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8

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Solution

The correct option is D $8$
$f (0)^{-} = lim x \to 0^{-} f (x) = lim x \to 0^{-} \frac{1 - cos 4 x}{x^{2}} = lim x \to 0^{-} \frac{2 {sin}^{2} 2 x}{x^{2}} = 8$
$f (0)^{+} = lim x \to 0^{+} f (x)$
$= lim x \to 0^{+} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$
$= lim x \to 0^{+} \frac{\sqrt{x} \cdot (\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16}$
$= lim x \to 0^{+} (\sqrt{16 + \sqrt{x}} + 4) = 8$
For $f (x)$ to be continuous,
$f (0^{-}) = f (0^{+}) = f (0) \Rightarrow a = 8$

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Similar questions

Q. If

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & w h e n x < 0 a, & w h e n x = 0 \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}, & w h e n x > 0 \end{matrix}

is continuous at

x = 0

, then the value of a will be.

Q. If

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & i f & x < 0 a, & i f & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{4}} - 4}, & i f & x > 0 \end{matrix}

, then what value of

a, f

is continuous at

x = 0

Q. If

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, x < 0 a, x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, x > 0 \end{matrix}

, then the correct statement is-

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \begin{matrix} \frac{{cos}^{2} x - {sin}^{2} x - 1}{\sqrt{x^{2} + 4} - 2}, & x \neq 0 \end{matrix} \begin{matrix} a, & x = 0 \end{matrix} \end{matrix}

then the value of

a

in order that

f (x)

may be continuous at

x = 0

Q. lf the function

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}} & i f & x < 0 a & i f & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} & i f & x > 0 \end{matrix}

is continuous at

x = 0

then

a =

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The value of a so that the function f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1−cos4xx2,x<0a,x=0√x√16+√x−4,x>0 is continuous at x=0 is

The correct option is D 8f(0)−=limx→0−f(x)=limx→0−1−cos4xx2=limx→0−2sin22xx2=8 f(0)+=limx→0+f(x) =limx→0+√x√16+√x−4 =limx→0+√x⋅(√16+√x+4)16+√x−16 =limx→0+(√16+√x+4)=8 For f(x) to be continuous, f(0−)=f(0+)=f(0)⇒a=8

The value of $a$ so that the function $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}}, & x < 0 a, & x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{matrix}$
is continuous at $x = 0$ is