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Mathematics Question on Statistics

Let f(x)={−2sin⁡x,x≤−π/2 asin⁡x+b,\-π/2<x<π/2 cos⁡ x,x≥π/2f(x) = \begin{cases} -2 \sin x, &x \leq - \pi /2 \\\ a \sin x +b, & \- \pi /2 < x < \pi /2 \\\ \cos \, x , & x \geq \pi /2 \end{cases} then the? values of a and b so that f(x)f(x) is continuous are

A

a=1,b=1a = 1, b = 1

B

a=1,b=−1a= 1,b= -1

C

a=−1 b=la=-1 \: b=l

D

a=−1,b=−1a = - 1,b = - 1

Answer

a=−1 b=la=-1 \: b=l

Explanation

Solution

f(x) = \begin{cases} -2 \sin x, &x \leq - \pi /2 \\\ a \sin x +b, & \- \pi /2 < x < \pi /2 \\\ \cos \, x , & x \geq \pi /2 \end{cases}\(Given thatf(x)) is continuous.
∴  lim⁡x→π−2f(x)=lim⁡x→π+2f(x)=f(π2)\therefore \:\: \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)
lim⁡x→π−2asin⁡x+b=lim⁡x→π+2cos⁡x=cos⁡x=cos⁡(π2)\lim _{x\to \frac{\pi ^{-}}{2}} a \sin x + b = \lim _{x\to \frac{\pi ^{+}}{2}} \cos x = \cos x = \cos\left(\frac{\pi}{2}\right)
lim⁡h→0asin⁡(π2−h)+b=lim⁡h→0cos⁡(π2+h)=0\lim_{h \rightarrow0} a \sin \left(\frac{\pi }{2}-h\right)+b = \lim _{h \rightarrow 0} \cos\left(\frac{\pi}{2}+h\right) = 0
⇒a+b=0\Rightarrow a+b =0 ......(i)
Now for x=−π2x = -\frac{\pi}{2}
lim⁡x→π−2f(x)=lim⁡x→π+2f(x)=f(−π2)\lim_{x \rightarrow \frac{\pi^{-}}{2}} f\left(x\right) = \lim _{x \rightarrow \frac{\pi ^{+}}{2}} f\left(x\right) = f\left(-\frac{\pi }{2}\right)
⇒lim⁡x→π−2(−2sin⁡x)\Rightarrow \lim _{x \rightarrow \frac{\pi ^{-}}{2}} \left(-2 \sin x \right)
=lim⁡x→π+2(asin⁡x+b)=−2sin⁡(−π2)= \lim _{x\to \frac{\pi ^{+}}{2}} \left(a \sin x + b\right) = - 2\sin \left(- \frac{\pi }{2}\right)
⇒lim⁡h→0−2sin⁡(−π2−h)\Rightarrow \lim _{h \rightarrow 0} - 2 \sin \left(- \frac{\pi }{2} -h \right)
=lim⁡h→0asin⁡(−π2+h)+b=2= \lim _{h \rightarrow 0} a \sin \left(- \frac{\pi }{2} +h \right)+ b=2
⇒2=−a+b=2⇒b−a=2\Rightarrow 2 = -a+b =2 \Rightarrow b-a = 2 ....(ii)
Adding (i) and (ii) we get
2b=2⇒b=1⇒a=−12b = 2\Rightarrow b = 1 \Rightarrow a =-1
Hence,f(x)f(x) is continuous for a=−1,b=1a = -1 , b = 1

The ability to trace a function's graph with a pencil without taking the pencil off the paper is a feature of many functions. These are referred to as continuous functions. If a function's graph does not break at a particular point, it is said to be continuous at that location. In general, an introductory calculus course will give a precise explanation of how the limit concept applies to the continuity of a real function. First, a function f with variable x is continuous at point "a" on the real line if and only if the limit of f(x) equals the value of f(x) at "a," i.e., f(a), as x approaches "a."

Following are some mathematical definitions of continuity:

If the following three conditions are met, a function is said to be continuous at a given point.

  • f(a) is defined
  • lim x→a f(x) exists
  • limx→a f(x)=lim x→a f(x)=f(a)

When a graph can be traced without lifting the pen from the sheet, the function is said to be a continuous function. A function, on the other hand, is said to be discontinuous if it contains any gaps in between.