मराठी

Determine the values of a, b, c for which the function f(x) = forfor/for{sin(a+1)x+sinxxfor x<0xfor x=0x+bx2-xbx3/2for x>0 is continuous at x = 0. - Mathematics

Advertisements
Advertisements

प्रश्न

Determine the values of a, b, c for which the function f(x) = `{((sin(a + 1)x + sin x)/x, "for"   x < 0),(x, "for"  x = 0),((sqrt(x + bx^2) - sqrtx)/(bx^(3"/"2)), "for"  x > 0):}` is continuous at x = 0.

बेरीज

उत्तर

Given:

f(x) is continuous at x = 0

For f(x) to be continuous at x = 0, f(0)- = f(0)+ = f(0)

LHL = f(0)- = `lim_(x->0) (sin (a + 1)x + sinx)/x`

`=> lim_(h->0)(sin (a + h)h + sinh)/h`

`=> lim_(h->0)(sin (a + 1)h)/h + lim_(h->0)sinh/h`

`=> lim_(h->0)(sin(a + 1)h)/h xx ((a + 1))/((a + 1)) + lim_(h->0)sinh/h`

`=> lim_(h->0)(sin (a + 1)h)/((a + 1)) xx ((a + 1))/1 + lim_(h->0)sinh/h`

`lim_(h->0)(sin(a + 1)h)/((a + 1)h) = 1`

`lim_(h->0)sinh/h = 1`

⇒ 1 × (a + 1) + 1

⇒ (a + 1) + 1

f(0)- ⇒ a + 2       ...(1)

RHL = f(0+) = `lim_(x->0)(sqrt(x + bx^2) - sqrtx)/(bx^(3/2))`

`=> lim_(x->0)(sqrt(x + bx^2)- sqrtx)/(bx^(3/2))`

`=> lim_(h->0)(sqrt(h + bh^2) - sqrth)/(bh^(3/2))`

`=> lim_(h->0)(sqrt(h + bh^2) - sqrth)/(b xx h xx h^(1/2))`

`=> lim_(h->0) (sqrt(h + bh^2)-sqrth)/(b xx h xx sqrth)`

`=> lim_(h ->0)(sqrt(h(1 + bh))- sqrth)/(b xx h xx sqrth)`

`=> lim_(h ->0)(sqrth(sqrt(1 + bh))- sqrt1)/(bh xx sqrth)`

`=> lim_(h->0)((sqrt(1 + bh))- sqrt1)/(bh)`

Take the complex conjugate of 

`(sqrt(1 + bh)- sqrt 1)`,

i.e, `(sqrt(1 + bh)- sqrt 1)` and multiply it with numerator and denominator

`=> lim_(h->0)((sqrt(1 + bh))- sqrt1)/(bh) xx ((sqrt(1 + bh)) + sqrt1)/((sqrt(1 + bh)) + sqrt1)`

`lim_(h->0) ((sqrt(1 + bh))^2 - (sqrt1)^2)/(bh)`

∴ (a + b)(a − b) = a2 − b2

`=> lim_(h->0)((1 + bh - 1))/(bh(sqrt(1 + bh))+ sqrt1)`

`=> lim_(h->0)((bh))/((sqrt(1 + bh)) sqrt1)`

`=> 1/((sqrt(1 + b xx 0)) + sqrt1)`

`=> 1/(1 + 1)`

f(0)+ = `1/2`   ...(2)

since, f(x) is continuous at x = 0, From (1) & (2), we get,

⇒ a + 2 = `1/2`

⇒ a = `1/2 - 2`

⇒ a = `(-3)/2`

Also, 

f(0)- = f(0)+ = f(0)

⇒ f(0) = c

⇒ c = a + 2 = `1/2`

⇒ c = `1/2`

So the values of a = `(-3)/2,` c = `1/2` and b = R-{0}(any real number except 0)

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Continuity - Exercise 9.1 [पृष्ठ १९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 9 Continuity
Exercise 9.1 | Q 26 | पृष्ठ १९

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Discuss the continuity of the function f, where f is defined by `f(x) = {(-2,"," if x <= -1),(2x, "," if -1 < x <= 1),(2, "," if x > 1):}`


Let \[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x^2}, when & x \neq 0 \\ 1 , when & x = 0\end{cases}\] Show that f(x) is discontinuous at x = 0.

 

 


Show that 

\[f\left( x \right) = \begin{cases}1 + x^2 , if & 0 \leq x \leq 1 \\ 2 - x , if & x > 1\end{cases}\]


Discuss the continuity of the function f(x) at the point x = 1/2, where \[f\left( x \right) = \begin{cases}x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \\ 1 - x, \frac{1}{2} < x \leq 1\end{cases}\] 


If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\]  is continuous at x = 4, find ab.

 


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; 

\[f\left( x \right) = \begin{cases}kx + 1, \text{ if }  & x \leq \pi \\ \cos x, \text{ if }  & x > \pi\end{cases}\] at x = π

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \binom{\frac{x^3 + x^2 - 16x + 20}{\left( x - 2 \right)^2}, x \neq 2}{k, x = 2}\] 

 


For what value of k is the following function continuous at x = 2? 

\[f\left( x \right) = \begin{cases}2x + 1 ; & \text{ if } x < 2 \\ k ; & x = 2 \\ 3x - 1 ; & x > 2\end{cases}\]

Let\[f\left( x \right) = \left\{ \begin{array}\frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{ if }  x < \frac{\pi}{2} \\ a , & \text{ if }  x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x )^2}, & \text{ if }  x > \frac{\pi}{2}\end{array} . \right.\] ]If f(x) is continuous at x = \[\frac{\pi}{2}\] , find a and b.

 

In the following, determine the value of constant involved in the definition so that the given function is continuou:   \[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if }  0 \leq x \leq 1\end{cases}\]


The function  \[f\left( x \right) = \begin{cases}\frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ 0 , & x = 0\end{cases}\]

 


Let f (x) = | x | + | x − 1|, then


\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & 0 \leq x \leq 1\end{cases}\]is continuous in the interval [−1, 1], then p is equal to

 


The value of b for which the function 

\[f\left( x \right) = \begin{cases}5x - 4 , & 0 < x \leq 1 \\ 4 x^2 + 3bx , & 1 < x < 2\end{cases}\] is continuous at every point of its domain, is 

The points of discontinuity of the function 

\[f\left( x \right) = \begin{cases}2\sqrt{x} , & 0 \leq x \leq 1 \\ 4 - 2x , & 1 < x < \frac{5}{2} \\ 2x - 7 , & \frac{5}{2} \leq x \leq 4\end{cases}\text{ is } \left( \text{ are }\right)\] 


Find whether the function is differentiable at x = 1 and x = 2 

\[f\left( x \right) = \begin{cases}x & x \leq 1 \\ \begin{array} 22 - x  \\ - 2 + 3x - x^2\end{array} & \begin{array}11 \leq x \leq 2 \\ x > 2\end{array}\end{cases}\]

Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.


If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text {  is }\] 


The function f (x) =  |cos x| is


If \[f\left( x \right) = \begin{cases}\frac{1}{1 + e^{1/x}} & , x \neq 0 \\ 0 & , x = 0\end{cases}\]  then f (x) is 


Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is 


Find k, if f(x) =`log (1+3x)/(5x)` for x ≠ 0

                     = k                    for x = 0

is continuous at x = 0. 


`f(x)=(x^2-9)/(x - 3)` is not defined at x = 3. what value should be assigned to f(3) for continuity of f(x) at = 3?


Discuss the continuity of the function f at x = 0

If f(x) = `(2^(3x) - 1)/tanx`, for x ≠ 0

         = 1,   for x = 0


Find `dy/dx if y = tan^-1 ((6x)/[ 1 - 5x^2])`


If f(x) = `{{:((x^3 + x^2 - 16x + 20)/(x - 2)^2",", x ≠ 2),("k"",", x = 2):}` is continuous at x = 2, find the value of k.


If f(x) = `(sqrt(2) cos x - 1)/(cot x - 1), x ≠ pi/4` find the value of `"f"(pi/4)`  so that f (x) becomes continuous at x = `pi/4`


Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if"  x ≠ 0),(0",",  "if"  x = 0):}` is discontinuous at x = 0.


Let f(x) = `{{:((1 - cos 4x)/x^2",",  "if"  x < 0),("a"",",  "if"  x = 0),(sqrt(x)/(sqrt(16) + sqrt(x) - 4)",", "if"  x > 0):}`. For what value of a, f is continuous at x = 0?


The value of k which makes the function defined by f(x) = `{{:(sin  1/x",",  "if"  x ≠ 0),("k"",",  "if"  x = 0):}`, continuous at x = 0 is ______.


The number of points at which the function f(x) = `1/(log|x|)` is discontinuous is ______.


f(x) = `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0


f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if"  x ≠ 2),(5",", "if"  x = 2):}` at x = 2


f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0 


f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1


Prove that the function f defined by 
f(x) = `{{:(x/(|x| + 2x^2)",",  x ≠ 0),("k",  x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.


Show that f(x) = |x – 5| is continuous but not differentiable at x = 5.


The set of points where the function f given by f(x) = |2x − 1| sinx is differentiable is ______.


If the following function is continuous at x = 2 then the value of k will be ______.

f(x) = `{{:(2x + 1",", if x < 2),(                 k",", if x = 2),(3x - 1",", if x > 2):}`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×