Question

PAGENA DATE 2012 The value of k (K<0) for which the function f defined as f(x) = {1-coskx/xsinx, x is not = 0 is continuous at x = 0 is.

(a) +- 1
(b) -1
(c) +- 1/2
(d) 1/2​

Answer 1

Answer:

(c) +- 1/2

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Step-by-step explanation:

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Answer 2

The value of k in the given function is ± 1 (A).

Given:

A function $f(x)=\frac{1-coskx}{x.sinx}$

To find:

The value of $k$.

Solution:

We have been given that the given function is continuous at $x=0$. Hence, solving the function further, we get

$f(x)=\frac{1-coskx}{x.sinx}$

Multiplying and dividing the equation by $k$, we get

$f(x)=\frac{1-coskx}{k.x} .\frac{x}{sinx}$

We know, when limit $x$ tends to $0$,

$\frac{sinx}{x}=0$ and $\frac{1-cosx}{x} =0$

Hence, we get

$f(x)=(0)(1)=0$

But, we have been given that the value of function is not equal to zero.

Thus,

From the given data,

$f(x)=\frac{1}{2}$

$\frac{1-coskx}{x.sinx} =\frac{1}{2}$

We know, $1-cos\alpha =2sin^{2}\frac{\alpha }{2}$

$\frac{2.sin^{2}\frac{kx}{2} }{x.sinx} =\frac{1}{2}$

Multiplying and dividing the equation by $(\frac{kx}{2} )^{2}$, we get

$\frac{2.sin^{2}\frac{kx}{2} (4) }{k^{2} x^{2} }.\frac{k^{2} x^{2} }{4} .(\frac{1}{x.sinx} )=\frac{1}{2}$

We know, when limit $x$ tends to $0$, $\frac{sinx}{x} =1$. Hence, we get

$\frac{k^{2} x^{2} }{2}.(\frac{1}{x.sinx} )=\frac{1}{2}$

$\frac{k^{2} }{2} (\frac{x}{sinx}) =\frac{1}{2}$

We also know that, when $x$ tends to $0$, $\frac{sinx}{x} =1$ . Hence, we get

$\frac{k^{2} }{2} =\frac{1}{2}$

$k^{2} =1$

$k=$ ± $1$

Final answer:

Hence, the value of k in the given function is ± 1 (A).

Although your question is incomplete, you might be referring to the question below.

The value of k (k < 0) for which the function f is defined as

$\left \{ {{f(x)=\frac{1-coskx}{x.sinx} } \atop {f(x)=\frac{1}{2} }} \right.$

and $x\neq 0$ , the function is continuous at $x=0$

1. ± 1
2. - 1
3. ± 0.5
4. 0.5