मराठी

If F ( X ) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 1 − Sin 2 X 3 Cos 2 X , X < π 2 a , X = π 2 B ( 1 − Sin X ) ( π − 2 X ) 2 , X > π 2 . Then, F (X) is Continuous at X = π 2 (A) a = 1 3 , (B) a = 1 3 , B = 8 3 - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

पर्याय

  • \[a = \frac{1}{3},\] b = 2

  • \[a = \frac{1}{3}, b = \frac{8}{3}\]

  • \[a = \frac{2}{3}, b = \frac{8}{3}\]
  • none of these

MCQ
Advertisements

उत्तर

 \[a = \frac{1}{3} , b = \frac{8}{3}\]

Given:  

\[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x}, \text{ if }x < \frac{\pi}{2} \\ a, \text{ if }x = \frac{\pi}{2} \\ \frac{b\left( 1 - \ sinx \right)}{\left( \pi - 2x \right)^2}, \text{ if } x > \frac{\pi}{2}\end{cases}\]

We have
(LHL at x = \[\frac{\pi}{2}\] =  \[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right)\]

\[= \lim_{h \to 0} \left( \frac{1 - \sin^2 \left( \frac{\pi}{2} - h \right)}{3 \cos^2 \left( \frac{\pi}{2} - h \right)} \right)\]
\[ = \lim_{h \to 0} \left( \frac{1 - \cos^2 h}{3 \sin^2 h} \right)\]
\[ = \frac{1}{3} \lim_{h \to 0} \left( \frac{\sin^2 h}{\sin^2 h} \right)\]
\[ = \frac{1}{3}\]

(RHL at x = \[\frac{\pi}{2}\] = \[\lim_{x \to \frac{\pi}{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} + h \right)\]

\[= \lim_{h \to 0} \left( \frac{b\left[ 1 - \sin \left( \frac{\pi}{2} + h \right) \right]}{\left[ \pi - 2\left( \frac{\pi}{2} + h \right) \right]^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{b\left( 1 - \cos h \right)}{\left[ - 2h \right]^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{2b \sin^2 \frac{h}{2}}{4 h^2} \right)\]
\[ = \lim_{h \to 0} \left( \frac{2b \sin^2 \frac{h}{2}}{16\frac{h^2}{4}} \right)\]
\[ = \frac{b}{8} \lim_{h \to 0} \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^2 \]
\[ = \frac{b}{8} \times 1\]
\[ = \frac{b}{8}\]

Also,

\[f\left( \frac{\pi}{2} \right) = a\]

If f(x) is continuous at x = \[\frac{\pi}{2}\], then 

\[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{x \to \frac{\pi}{2}^+} f\left( x \right) = f\left( \frac{\pi}{2} \right)\]
\[\Rightarrow \frac{1}{3} = \frac{b}{8} = a\]
\[\Rightarrow a = \frac{1}{3} \text{ and } b = \frac{8}{3}\]
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Continuity - Exercise 9.4 [पृष्ठ ४७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 9 Continuity
Exercise 9.4 | Q 39 | पृष्ठ ४७

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Examine the following function for continuity:

f(x) = x – 5


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.

 

 


Discuss the continuity of the following functions at the indicated point(s): 

(ii) \[f\left( x \right) = \left\{ \begin{array}{l}x^2 \sin\left( \frac{1}{x} \right), & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]


Show that 

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


Find the value of 'a' for which the function f defined by

\[f\left( x \right) = \begin{cases}a\sin\frac{\pi}{2}(x + 1), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0\end{cases}\]  is continuous at x = 0.
 

 


Determine the value of the constant k so that the function 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{x^2 - 3x + 2}{x - 1}, if & x \neq 1 \\ k , if & x = 1\end{array}\text{is continuous at x} = 1 \right.\] 


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.


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:   \[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if }  0 \leq x \leq 1\end{cases}\]


Prove that
\[f\left( x \right) = \begin{cases}\frac{\sin x}{x} , & x < 0 \\ x + 1 , & x \geq 0\end{cases}\] is everywhere continuous.

 


If \[f\left( x \right) = \begin{cases}\frac{x^2 - 16}{x - 4}, & \text{ if }  x \neq 4 \\ k , & \text{ if }  x = 4\end{cases}\]  is continuous at x = 4, find k.


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


The value of f (0), so that the function 

\[f\left( x \right) = \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]   becomes continuous for all x, given by

The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


Show that \[f\left( x \right) =\]`{(12x, -,13, if , x≤3),(2x^2, +,5, if x,>3):}` is differentiable at x = 3. Also, find f'(3).


Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: 

\[f\left( x \right) = \begin{cases}3x - 2, & 0 < x \leq 1 \\ 2 x^2 - x, & 1 < x \leq 2 \\ 5x - 4, & x > 2\end{cases}\]

If f is defined by f (x) = x2, find f'(2).


Discuss the continuity and differentiability of f (x) = |log |x||.


Discuss the continuity and differentiability of f (x) = e|x| .


Discuss the continuity and differentiability of 

\[f\left( x \right) = \begin{cases}\left( x - c \right) \cos \left( \frac{1}{x - c} \right), & x \neq c \\ 0 , & x = c\end{cases}\]

Let f (x) = |x| and g (x) = |x3|, then


The function f (x) = e|x| is


If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]


The function f (x) =  |cos x| is


If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is


The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is


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

                     = k                    for x = 0

is continuous at x = 0. 


Discuss continuity of f(x) =`(x^3-64)/(sqrt(x^2+9)-5)` For x ≠ 4 

= 10 for x = 4  at x = 4


Discuss the continuity of the function f at x = 0

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

         = 1,   for x = 0


If f(x) = `(e^(2x) - 1)/(ax)` .                for x < 0 , a ≠ 0
         = 1.                             for x = 0
         = `(log(1 + 7x))/(bx)`.        for x > 0 , b ≠ 0
is continuous at x = 0 . then find a and b


If f (x) = `(1 - "sin x")/(pi - "2x")^2` , for x ≠ `pi/2` is continuous at x = `pi/4` , then find `"f"(pi/2) .`


If the function f is continuous at x = 2, then find 'k' where

f(x) = `(x^2 + 5)/(x - 1),` for  1< x ≤ 2 
      = kx + 1 , for x > 2


If Y = tan-1 `[(cos 2x - sin 2x)/(sin2x + cos 2x)]` then find `(dy)/(dx)`


If the function
f(x) = x2 + ax + b,         x < 2

      = 3x + 2,                 2≤ x ≤ 4

      = 2ax + 5b,             4 < x

is continuous at x = 2 and x = 4, then find the values of a and b


Show that the function f defined by f(x) = `{{:(x sin  1/x",", x ≠ 0),(0",", x = 0):}` is continuous 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?


f(x) = |x| + |x − 1| at x = 1


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

f(x) = `{((1-cos"kx")/("x"sin"x")","  "x" ≠ 0),(1/2","  "x" = 0):}`

is continuous at x = 0 is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×