Maths

23. Given f (x) = $\{\begin{array}{ll}{x}^{2}sin\frac{1}{x}, if x\ne 0.\hfill & 0 if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f (c) = $c^{2}sin\frac{1}{c}$

$\underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}{x}^{2}sin\frac{1}{x}=c^{2}sin\frac{1}{c}.$

So, f is continuous for $x\ne 0.$

For x = 0,

f (0) = 0

$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\left(x^{2}sin\frac{1}{x}\right)$

As we have sin $\frac{1}{x}\in [-1,1]$

$\underset{x\to 0}{lim}$ f (x) = 02 a where $a\in [-1,1]$

= 0 = f (0).

∴ f is also continuous at x = 0.

13. sin 765°

We know that value of sun x repeats after an interval of 2π or 360°.

Sin (765°) = sin (2*360°+45°)

= sin 45°

= 1/√2.

12. Kindly go through the solution

22. Given f(x) = $\{\begin{array}{ll}\frac{sinx}{x},\hfill & if x<0.\hfill \\ x+1,\hfill & if x\ge 0.\hfill \end{array}$

For x = c < 0,

f(c) = $\frac{sinc}{c}$

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ $\frac{sinx}{x}=\frac{sinc}{c}=f(c)$

So, f is continuous for x < 0

For x = c > 0

f(c) = c + 1

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ x + 1 = c + 1 = f(c)

So, f is continuous for x > 0.

For x = 0.

L.H.L. = $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{sinx}{x}=1.$

R.H.S. = $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}x+1=0+1=1$

And f(0) = 0 + 1 = 1

L.H.L = R.H.L. = f(0)

So, f is continuous at x = 1.

Hence, discontinuous point of x does not exit.

11. Kindly go through the solution

10. Kindly go through the solution

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