
Let f(x) $f:R\to R$be a function defined as
$f(x)=\left\{ \begin{align}
& 5,\text{ if x}\le \text{1} \\
& a+bx\text{ if 1x3} \\
& b+5x\text{ if 3}\le \text{x5} \\
& 30\text{ if x}\ge \text{5} \\
\end{align} \right\}$
Then the function is
$\begin{align}
& \text{a) Continuous if a = 5 and b = 5} \\
& \text{b) Continuous if a = -5 and b = -10} \\
& \text{c) Continuous if a = 0 and b = 5} \\
& \text{d) Not continuous for any values of a and b} \\
\end{align}$
Answer
508.8k+ views
Hint: we know that a function f (x) is continuous at a, if $\displaystyle \lim_{x \to a}$ from left is equal to $\displaystyle \lim_{x \to a}$ from right is equal to f (a). Hence we write it as $\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Complete step-by-step answer:
Now consider the function f (x) $f:R\to R$
$f(x)=\left\{ \begin{align}
& 5,\text{ if x}\le \text{1} \\
& a+bx\text{ if 1x3} \\
& b+5x\text{ if 3}\le \text{x5} \\
& 30\text{ if x}\ge \text{5} \\
\end{align} \right\}$
We know that the function is continuous if it is continuous on every point.
Let us say it is continuous at x = 1
Now we have $f(1)=5$ , $\displaystyle \lim_{x \to {{1}^{+}}}f(x)=a+b(1)$ and $\displaystyle \lim_{x \to {{1}^{-}}}f(x)=5$
Now we know that for a function to be continuous
$\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Hence we get $a+b=5.................(1)$
Now let us say that the function is also continuous at x = 3
Now we have $f(3)=b+5(3)=b+15$ , $\displaystyle \lim_{x \to {{3}^{+}}}f(x)=b+3(5)=b+15$ and \[\displaystyle \lim_{x \to {{3}^{-}}}f(x)=a+3b\]
Now we know that for a function to be continuous
$\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Hence we get
\[\begin{align}
& a+3b=b+15 \\
& a+2b=15.....................(2) \\
\end{align}\]
Now let us say the function is also continuous at x = 5
Now we have \[f(5)=30\] , \[\displaystyle \lim_{x \to {{5}^{+}}}f(x)=30\] and $\displaystyle \lim_{x \to {{5}^{-}}}f(x)=b+5(5)=b+25$
Now we know that for a function to be continuous
$\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Hence we get $b+25=30..................(3)$
Now consider equation (3)
If we take 25 to RHS we get $b=30-25=5$
Hence b = 5.
Now let us substitute b = 5 in equation (1) we get $a+5=5$
Hence taking 5 to RHS we get $a=5-5=0$
Hence the value of a = 0
Now we have a = 0 and b = 5. Now we will substitute the values a = 0 and b = 5 in equation (2)
We get $0+2(5)=15$this is a contradiction as 5 (2) = 10 and is not equal to 15.
Hence no value of a, b, c is true for all three equations.
Hence we get the function is not continuous.
So, the correct answer is “Option d”.
Note: Since here we have three equations all three must be true at once as the function is said to be continuous if it is continuous at all points. Now once we find the value of a and b substitute it in the equation we have not used to find the values of a and b and check if the equation holds.
Complete step-by-step answer:
Now consider the function f (x) $f:R\to R$
$f(x)=\left\{ \begin{align}
& 5,\text{ if x}\le \text{1} \\
& a+bx\text{ if 1x3} \\
& b+5x\text{ if 3}\le \text{x5} \\
& 30\text{ if x}\ge \text{5} \\
\end{align} \right\}$
We know that the function is continuous if it is continuous on every point.
Let us say it is continuous at x = 1
Now we have $f(1)=5$ , $\displaystyle \lim_{x \to {{1}^{+}}}f(x)=a+b(1)$ and $\displaystyle \lim_{x \to {{1}^{-}}}f(x)=5$
Now we know that for a function to be continuous
$\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Hence we get $a+b=5.................(1)$
Now let us say that the function is also continuous at x = 3
Now we have $f(3)=b+5(3)=b+15$ , $\displaystyle \lim_{x \to {{3}^{+}}}f(x)=b+3(5)=b+15$ and \[\displaystyle \lim_{x \to {{3}^{-}}}f(x)=a+3b\]
Now we know that for a function to be continuous
$\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Hence we get
\[\begin{align}
& a+3b=b+15 \\
& a+2b=15.....................(2) \\
\end{align}\]
Now let us say the function is also continuous at x = 5
Now we have \[f(5)=30\] , \[\displaystyle \lim_{x \to {{5}^{+}}}f(x)=30\] and $\displaystyle \lim_{x \to {{5}^{-}}}f(x)=b+5(5)=b+25$
Now we know that for a function to be continuous
$\displaystyle \lim_{x \to {{a}^{+}}}=\displaystyle \lim_{x \to {{a}^{-}}}=f(a)$
Hence we get $b+25=30..................(3)$
Now consider equation (3)
If we take 25 to RHS we get $b=30-25=5$
Hence b = 5.
Now let us substitute b = 5 in equation (1) we get $a+5=5$
Hence taking 5 to RHS we get $a=5-5=0$
Hence the value of a = 0
Now we have a = 0 and b = 5. Now we will substitute the values a = 0 and b = 5 in equation (2)
We get $0+2(5)=15$this is a contradiction as 5 (2) = 10 and is not equal to 15.
Hence no value of a, b, c is true for all three equations.
Hence we get the function is not continuous.
So, the correct answer is “Option d”.
Note: Since here we have three equations all three must be true at once as the function is said to be continuous if it is continuous at all points. Now once we find the value of a and b substitute it in the equation we have not used to find the values of a and b and check if the equation holds.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

The non protein part of an enzyme is a A Prosthetic class 11 biology CBSE

Which of the following blood vessels in the circulatory class 11 biology CBSE

What is a zygomorphic flower Give example class 11 biology CBSE

The deoxygenated blood from the hind limbs of the frog class 11 biology CBSE

What is the function of copulatory pads in the forelimbs class 11 biology CBSE
