\displaystyle \text{Question 1: } \text{Show that } f(x)=|x-3| \text{ is continuous but not differentiable at } x=3.\ \hspace{10cm}\text{[CBSE 2012, 2013]}

\displaystyle \text{Answer:}

\displaystyle \text{Given:} f(x)=|x-3|= \begin{cases} x-3, & x \ge 3 \\ -x+3, & x<3 \end{cases}

\displaystyle \text{Continuity at } x=3:

\displaystyle \text{(LHL at } x=3\text{)}

\displaystyle \lim_{x\to 3^-} f(x) =\lim_{h\to 0} f(3-h) =\lim_{h\to 0} [-(3-h)+3] =\lim_{h\to 0} h =0

\displaystyle \text{(RHL at } x=3\text{)}

\displaystyle \lim_{x\to 3^+} f(x) =\lim_{h\to 0} f(3+h) =\lim_{h\to 0} [(3+h)-3] =\lim_{h\to 0} h =0

\displaystyle \text{Also, } f(3)=|3-3|=0

\displaystyle \therefore \lim_{x\to 3^-} f(x)=\lim_{x\to 3^+} f(x)=f(3)

\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=3.

\displaystyle \text{Differentiability at } x=3:

\displaystyle \text{(LHD at } x=3\text{)}

\displaystyle \lim_{x\to 3^-} \frac{f(x)-f(3)}{x-3} =\lim_{x\to 3^-} \frac{(-x+3)-0}{x-3} =\lim_{x\to 3^-} \frac{-(x-3)}{x-3} =\lim_{x\to 3^-} (-1) =-1

\displaystyle \text{(RHD at } x=3\text{)}

\displaystyle \lim_{x\to 3^+} \frac{f(x)-f(3)}{x-3} =\lim_{x\to 3^+} \frac{(x-3)-0}{x-3} =\lim_{x\to 3^+} 1 =1

\displaystyle \text{Since LHD } \neq \text{ RHD,}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=3.

\displaystyle \text{Question 2: } \text{Show that } f(x)=x^{1/3} \text{ is differentiable at } x=0.

\displaystyle \text{Answer:}

\displaystyle \text{Given:} f(x)=x^{\frac{1}{3}}

\displaystyle \text{We have,}

\displaystyle \text{(LHD at } x=0\text{)}

\displaystyle \lim_{x\to 0^-} \frac{f(x)-f(0)}{x-0} =\lim_{h\to 0} \frac{f(0-h)-f(0)}{0-h-0} =\lim_{h\to 0} \frac{(0-h)^{\frac{1}{3}}-0}{-h} =\lim_{h\to 0} \frac{(-h)^{\frac{1}{3}}}{-h} =\lim_{h\to 0} (-h)^{-\frac{2}{3}} =0

\displaystyle \text{(RHD at } x=0\text{)}

\displaystyle \lim_{x\to 0^+} \frac{f(x)-f(0)}{x-0} =\lim_{h\to 0} \frac{f(0+h)-f(0)}{0+h-0} =\lim_{h\to 0} \frac{(0+h)^{\frac{1}{3}}-0}{h} =\lim_{h\to 0} \frac{h^{\frac{1}{3}}}{h} =\lim_{h\to 0} h^{-\frac{2}{3}} =0

\displaystyle \text{LHD at } (x=0)=\text{RHD at } (x=0)

\displaystyle \text{Hence, } f(x)=x^{\frac{1}{3}} \text{ is differentiable at } x=0.

\displaystyle \text{Question 3: } \text{Show that } f(x)= \begin{cases} 12x-13, & \text{if } x\le 3,\\ 2x^2+5, & \text{if } x>3 \end{cases} \text{ is differentiable at } x=3.\ \text{Also, find } f'(3)

\displaystyle \text{Answer:}

\displaystyle \text{Given: }\text{Show that } f(x)= \begin{cases} 12x-13, & x<3 \\ 2x^2+5, & x>3 \end{cases} \text{ is differentiable at } x=3.

\displaystyle \text{We have,}

\displaystyle \text{(LHD at } x=3\text{)}

\displaystyle \lim_{x\to 3^-} \frac{f(x)-f(3)}{x-3} =\lim_{x\to 3^-} \frac{(12x-13)-23}{x-3} =\lim_{x\to 3^-} \frac{12x-36}{x-3} =\lim_{x\to 3^-} \frac{12(x-3)}{x-3} =\lim_{x\to 3^-} 12 =12

\displaystyle \text{(RHD at } x=3\text{)}

\displaystyle \lim_{x\to 3^+} \frac{f(x)-f(3)}{x-3} =\lim_{x\to 3^+} \frac{(2x^2+5)-23}{x-3} =\lim_{x\to 3^+} \frac{2x^2-18}{x-3} =\lim_{x\to 3^+} \frac{2(x^2-9)}{x-3} =\lim_{x\to 3^+} 2(x+3) =2\times 6 =12

\displaystyle \text{Thus, LHD at } x=3=\text{RHD at } x=3=12.

\displaystyle \text{Hence, } f(x) \text{ is differentiable at } x=3 \text{ and } f'(3)=12.

\displaystyle \text{Question 4: } \text{Show that the function } f \text{ defined as follows } f(x)= \begin{cases} 3x-2, & 0<x\le 1,\\ 2x^2-x, & 1<x\le 2,\\ 5x-4, & x>2 \end{cases} \\ \text{is continuous at } x=2,\ \text{but not differentiable thereat.}\ \hspace{3cm}\text{[CBSE 2010]}

\displaystyle \text{Answer:}

\displaystyle \text{Given:} f(x)= \begin{cases} 3x-2, & 0<x<1 \\ 2x^2-x, & 1<x\le 2 \\ 5x-4, & x>2 \end{cases}

\displaystyle \text{First, we show that } f(x) \text{ is continuous at } x=2.

\displaystyle \text{We have,}

\displaystyle \text{(LHL at } x=2\text{)}

\displaystyle \lim_{x\to 2^-} f(x) =\lim_{h\to 0} f(2-h) =\lim_{h\to 0} \big[2(2-h)^2-(2-h)\big] =\lim_{h\to 0} (8-8h+2h^2-2+h) =\lim_{h\to 0} (6-7h+2h^2) =6

\displaystyle \text{(RHL at } x=2\text{)}

\displaystyle \lim_{x\to 2^+} f(x) =\lim_{h\to 0} f(2+h) =\lim_{h\to 0} \big[5(2+h)-4\big] =\lim_{h\to 0} (10+5h-4) =6

\displaystyle \text{Also, } f(2)=2(2)^2-2=6

\displaystyle \therefore \lim_{x\to 2^-} f(x)=\lim_{x\to 2^+} f(x)=f(2)

\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=2.

\displaystyle \text{Differentiability at } x=2:

\displaystyle \text{(LHD at } x=2\text{)}

\displaystyle \lim_{x\to 2^-} \frac{f(x)-f(2)}{x-2} =\lim_{h\to 0} \frac{f(2-h)-f(2)}{-h} =\lim_{h\to 0} \frac{(6-7h+2h^2)-6}{-h} =\lim_{h\to 0} \frac{-7h+2h^2}{-h} =\lim_{h\to 0} (7-2h) =7

\displaystyle \text{(RHD at } x=2\text{)}

\displaystyle \lim_{x\to 2^+} \frac{f(x)-f(2)}{x-2} =\lim_{h\to 0} \frac{f(2+h)-f(2)}{h} =\lim_{h\to 0} \frac{(10+5h-4)-6}{h} =\lim_{h\to 0} 5 =5

\displaystyle \text{Thus, LHD at } x=2 \neq \text{ RHD at } x=2.

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=2.

\displaystyle \text{Question 5: } \text{Discuss the continuity and differentiability of the function } f(x)=|x|+|x-1| \text{ in the interval } (-1,2).\ \hspace{7cm}\text{[CBSE 2015]}

\displaystyle \text{Answer:}

\displaystyle \text{Given: }f(x)=|x|+|x-1|

\displaystyle |x|=-x \text{ for } x<0

\displaystyle |x|=x \text{ for } x>0

\displaystyle |x-1|=-(x-1)=-x+1 \text{ for } x<1

\displaystyle |x-1|=x-1 \text{ for } x>1

\displaystyle \text{Now,}

\displaystyle f(x)=-x+(-x+1)=-2x+1 \quad \text{for } x\in(-1,0)

\displaystyle \text{or}

\displaystyle f(x)=x+(-x+1)=1 \quad \text{for } x\in(0,1)

\displaystyle \text{or}

\displaystyle f(x)=x+(x-1)=2x-1 \quad \text{for } x\in(1,2)

\displaystyle \text{Continuity:}

\displaystyle \text{LHL at } x=0=\lim_{x\to 0^-} f(x) =\lim_{x\to 0^-} (-2x+1) =1

\displaystyle \text{RHL at } x=0=\lim_{x\to 0^+} f(x) =\lim_{x\to 0^+} 1 =1

\displaystyle \text{Hence, at } x=0,\ \text{LHL}=\text{RHL}

\displaystyle \text{Again,}

\displaystyle \text{LHL at } x=1=\lim_{x\to 1^-} f(x) =\lim_{x\to 1^-} 1 =1

\displaystyle \text{RHL at } x=1=\lim_{x\to 1^+} f(x) =\lim_{x\to 1^+} (2x-1) =1

\displaystyle \text{Hence, at } x=1,\ \text{LHL}=\text{RHL}

\displaystyle \text{Differentiability:}

\displaystyle f(x)=-2x+1 \Rightarrow f'(x)=-2 \quad \text{for } x\in(-1,0)

\displaystyle \text{or}

\displaystyle f(x)=1 \Rightarrow f'(x)=0 \quad \text{for } x\in(0,1)

\displaystyle \text{or}

\displaystyle f(x)=2x-1 \Rightarrow f'(x)=2 \quad \text{for } x\in(1,2)

\displaystyle \text{Now,}

\displaystyle \text{LHD at } x=0=\lim_{x\to 0^-} f'(x) =\lim_{x\to 0^-} (-2) =-2

\displaystyle \text{RHD at } x=0=\lim_{x\to 0^+} f'(x) =\lim_{x\to 0^+} 0 =0

\displaystyle \text{Since, at } x=0,\ \text{LHD} \ne \text{RHD}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=0.

\displaystyle \text{Again,}

\displaystyle \text{LHD at } x=1=\lim_{x\to 1^-} f'(x) =\lim_{x\to 1^-} 0 =0

\displaystyle \text{RHD at } x=1=\lim_{x\to 1^+} f'(x) =\lim_{x\to 1^+} 2 =2

\displaystyle \text{Since, at } x=1,\ \text{LHD} \ne \text{RHD}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=1.

\displaystyle \text{Question 6: } \text{Find whether the following function is differentiable at } x=1 \text{ and } x=2 \text{ or not:} 

\displaystyle f(x)= \begin{cases} x, & x\le 1,\\ 2-x, & 1\le x\le 2,\\ -2+3x-x^2, & x>2 \end{cases} \hspace{7cm}\text{[CBSE 2015]}

\displaystyle \text{Answer:}

\displaystyle f(x)= \begin{cases} x, & x\le 1 \\ 2x-x^2, & 1<x\le 2 \\ 1, & x>2 \end{cases}

\displaystyle \Rightarrow f'(x)= \begin{cases} 1, & x<1 \\ 3-2x, & 1<x<2 \\ -1, & x>2 \end{cases}

\displaystyle \text{Now,}

\displaystyle \text{LHL at } x=1 =\lim_{x\to 1^-} f'(x) =\lim_{x\to 1^-} 1 =1

\displaystyle \text{RHL at } x=1 =\lim_{x\to 1^+} f'(x) =\lim_{x\to 1^+} (3-2x) =1

\displaystyle \text{Since, at } x=1,\ \text{LHL}=\text{RHL}

\displaystyle \text{Hence, } f(x) \text{ is differentiable at } x=1.

\displaystyle \text{Again,}

\displaystyle \text{LHL at } x=2 =\lim_{x\to 2^-} f'(x) =\lim_{x\to 2^-} (3-2x) = -1

\displaystyle \text{RHL at } x=2 =\lim_{x\to 2^+} f'(x) =\lim_{x\to 2^+} (-1) =-1

\displaystyle \text{Since, at } x=2,\ \text{LHL}=\text{RHL}

\displaystyle \text{Hence, } f(x) \text{ is differentiable at } x=2.

\displaystyle \text{Question 7: } \text{Show that the function } f(x)= \begin{cases} x^{m}\sin\left(\dfrac{1}{x}\right), & x\ne 0,\\ 0, & x=0 \end{cases} \text{ is}

\displaystyle (i)\ \text{differentiable at } x=0,\ \text{if } m>1. \\ (ii)\ \text{continuous but not differentiable at } x=0,\ \text{if } 0<m<1. \\ (iii)\ \text{neither continuous nor differentiable, if } m\le 0.

\displaystyle \text{Answer:}

\displaystyle \text{Given:}

\displaystyle f(x)= \begin{cases} x^{m}\sin\left(\dfrac{1}{x}\right), & x\ne 0 \\ 0, & x=0 \end{cases}

\displaystyle \text{(i) Let } m=2. \text{ Then the function becomes}

\displaystyle f(x)= \begin{cases} x^{2}\sin\left(\dfrac{1}{x}\right), & x\ne 0 \\ 0, & x=0 \end{cases}

\displaystyle \text{Differentiability at } x=0:

\displaystyle \lim_{x\to 0} \frac{f(x)-f(0)}{x-0} =\lim_{x\to 0} \frac{x^{2}\sin\left(\frac{1}{x}\right)}{x} =\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)

\displaystyle \text{Now, } \lim_{x\to 0} x\sin\left(\frac{1}{x}\right)=0

\displaystyle \left|x\sin\left(\frac{1}{x}\right)-0\right| =\left|x\sin\left(\frac{1}{x}\right)\right| =|x|\left|\sin\left(\frac{1}{x}\right)\right| \le |x|

\displaystyle \text{Given } \epsilon>0,\ \text{choose } \delta=\epsilon.

\displaystyle \text{Then } |x-0|<\delta \Rightarrow \left|x\sin\left(\frac{1}{x}\right)-0\right|<\epsilon

\displaystyle \therefore \lim_{x\to 0} x\sin\left(\frac{1}{x}\right)=0

\displaystyle \Rightarrow f'(0)=0

\displaystyle \text{Hence, the given function is differentiable at } x=0.

\displaystyle \text{(ii) Let } m=\dfrac{1}{2},\ 0<m<1.

\displaystyle \text{Then the function becomes}

\displaystyle f(x)= \begin{cases} x^{\frac{1}{2}}\sin\left(\dfrac{1}{x}\right), & x\ne 0 \\ 0, & x=0 \end{cases}

\displaystyle \text{Continuity at } x=0:

\displaystyle \text{LHL at } x=0 =\lim_{x\to 0^-} f(x) =\lim_{h\to 0} (-h)^{\frac{1}{2}}\sin\left(\frac{1}{-h}\right)

\displaystyle \text{(Not defined for real values)}

\displaystyle \text{RHL at } x=0 =\lim_{x\to 0^+} f(x) =\lim_{h\to 0} h^{\frac{1}{2}}\sin\left(\frac{1}{h}\right) =0

\displaystyle f(0)=0

\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=0.

\displaystyle \text{Differentiability at } x=0 \text{ when } 0<m<1:

\displaystyle \text{LHD at } x=0 =\lim_{x\to 0^-} \frac{f(x)-f(0)}{x-0} \text{ does not exist}

\displaystyle \text{Hence, } f(x) \text{ is not differentiable at } x=0 \text{ when } 0<m<1.

\displaystyle \text{Question 8: } \text{Find the values of } a \text{ and } b \text{ so that the function } f(x)= \begin{cases} x^{2}+3x+a, & \text{if } x\le 1,\\ bx+2, & \text{if } x>1 \end{cases} \\ \text{ is differentiable at each } x \in R.

\displaystyle \text{Answer:}

\displaystyle \text{Given:}

\displaystyle f(x)= \begin{cases} x^2+3x+a, & x\le 1 \\ bx+2, & x>1 \end{cases}

\displaystyle \text{It is given that } f(x) \text{ is differentiable for all } x\in\mathbb{R}.

\displaystyle \text{Since every differentiable function is continuous, } f(x) \text{ is continuous at } x=1.

\displaystyle \text{Continuity at } x=1:

\displaystyle \lim_{x\to 1^-} f(x)=\lim_{x\to 1^+} f(x)=f(1)

\displaystyle \Rightarrow \lim_{x\to 1^-} (x^2+3x+a)=\lim_{x\to 1^+} (bx+2)

\displaystyle \Rightarrow (1)^2+3(1)+a=b(1)+2

\displaystyle \Rightarrow a+4=b+2 \qquad \ldots (i)

\displaystyle \text{Differentiability at } x=1:

\displaystyle \text{Since } f(x) \text{ is differentiable at } x=1,

\displaystyle \text{LHD at } x=1=\text{RHD at } x=1

\displaystyle \lim_{x\to 1^-} \frac{f(x)-f(1)}{x-1} =\lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1^-} \frac{x^2+3x+a-(a+4)}{x-1} =\lim_{x\to 1^+} \frac{bx+2-(a+4)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1^-} \frac{x^2+3x-4}{x-1} =\lim_{x\to 1^+} \frac{bx-b}{x-1} \quad [\text{Using (i)}]

\displaystyle \Rightarrow \lim_{x\to 1} \frac{(x+4)(x-1)}{x-1} =\lim_{x\to 1} \frac{b(x-1)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1} (x+4)=\lim_{x\to 1} b

\displaystyle \Rightarrow 5=b

\displaystyle \text{From (i):}

\displaystyle a+4=b+2

\displaystyle \Rightarrow a+4=5+2

\displaystyle \Rightarrow a=3

\displaystyle \text{Hence, } a=3 \text{ and } b=5.

\displaystyle \text{Question 9: } \text{Show that the function } f(x)= \begin{cases} |2x-3|\,[x], & x\ge 1,\\ \sin\!\left(\dfrac{\pi x}{2}\right), & x<1 \end{cases} \\ \text{is continuous but not differentiable at } x=1.

\displaystyle \text{Answer:}

\displaystyle \text{Given:}

\displaystyle f(x)= \begin{cases} |2x-3|\,[x], & x\ge 1 \\ \sin\left(\dfrac{\pi x}{2}\right), & x<1 \end{cases}

\displaystyle \text{Continuity at } x=1:

\displaystyle \text{(LHL at } x=1\text{)}

\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0} f(1-h) =\lim_{h\to 0} \sin\left(\dfrac{\pi(1-h)}{2}\right) =\sin\left(\dfrac{\pi}{2}\right) =1

\displaystyle \text{(RHL at } x=1\text{)}

\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0} |2(1+h)-3|\,[1+h] =\lim_{h\to 0} |2h-1|\cdot 1 =1

\displaystyle \therefore \lim_{x\to 1^-} f(x)=\lim_{x\to 1^+} f(x)

\displaystyle \text{Also, } f(1)=|2(1)-3|\,[1]=1

\displaystyle \text{Hence, } f(x) \text{ is continuous at } x=1.

\displaystyle \text{Differentiability at } x=1:

\displaystyle \text{(LHD at } x=1\text{)}

\displaystyle \lim_{x\to 1^-} \frac{f(x)-f(1)}{x-1} =\lim_{h\to 0} \frac{f(1-h)-1}{-h} =\lim_{h\to 0} \frac{\sin\left(\frac{\pi(1-h)}{2}\right)-1}{-h}

\displaystyle =\lim_{h\to 0} \frac{\cos\left(\frac{\pi h}{2}\right)-1}{-h} =-\frac{\pi}{2} \lim_{h\to 0} \frac{\cos\left(\frac{\pi h}{2}\right)-1}{\frac{\pi h}{2}} =0

\displaystyle \text{(RHD at } x=1\text{)}

\displaystyle \lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1} =\lim_{h\to 0} \frac{|2(1+h)-3|-1}{h} =\lim_{h\to 0} \frac{(1-2h)-1}{h} =\lim_{h\to 0} \frac{-2h}{h} =-2

\displaystyle \text{Since LHD} \ne \text{RHD}

\displaystyle \text{Hence, } f(x) \text{ is continuous but not differentiable at } x=1.

\displaystyle \text{Question 10: } \text{If } f(x)= \begin{cases} ax^{2}-b, & |x|<1,\\ \dfrac{1}{|x|}, & |x|\ge 1 \end{cases} \text{ is differentiable at } x=1,\ \text{find } a,\ b.

\displaystyle \text{Answer:}

\displaystyle \text{Given: }f(x)= \begin{cases} ax^2+b, & |x|<1 \\ \dfrac{1}{x}, & |x|\ge 1 \end{cases}

\displaystyle \text{Equivalently,}

\displaystyle f(x)= \begin{cases} -\dfrac{1}{x}, & x<-1 \\ ax^2+b, & -1<x<1 \\ \dfrac{1}{x}, & x\ge 1 \end{cases}

\displaystyle \text{It is given that } f(x) \text{ is differentiable at } x=1.

\displaystyle \text{Since every differentiable function is continuous, } f(x) \text{ is continuous at } x=1.

\displaystyle \text{Continuity at } x=1:

\displaystyle \lim_{x\to 1^-} f(x)=\lim_{x\to 1^+} f(x)

\displaystyle \Rightarrow \lim_{x\to 1^-} (ax^2+b)=\lim_{x\to 1^+} \frac{1}{x}

\displaystyle \Rightarrow a(1)^2+b=\frac{1}{1}

\displaystyle \Rightarrow a+b=1 \qquad \ldots (i)

\displaystyle \text{Differentiability at } x=1:

\displaystyle \text{Since } f(x) \text{ is differentiable at } x=1,

\displaystyle \text{LHD at } x=1=\text{RHD at } x=1

\displaystyle \lim_{x\to 1^-} \frac{f(x)-f(1)}{x-1} =\lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1^-} \frac{ax^2+b-1}{x-1} =\lim_{x\to 1^+} \frac{\frac{1}{x}-1}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1} \frac{a(x^2-1)}{x-1} =\lim_{x\to 1} \frac{1-x}{x(x-1)} \quad [\text{Using (i)}]

\displaystyle \Rightarrow \lim_{x\to 1} a(x+1) =\lim_{x\to 1} \left(-\frac{1}{x}\right)

\displaystyle \Rightarrow 2a=-1

\displaystyle \Rightarrow a=-\frac{1}{2}

\displaystyle \text{From (i):}

\displaystyle a+b=1

\displaystyle \Rightarrow -\frac{1}{2}+b=1

\displaystyle \Rightarrow b=\frac{3}{2}

\displaystyle \text{Hence, } a=-\frac{1}{2} \text{ and } b=\frac{3}{2}.

\displaystyle \text{Question 11: } f(x)= \begin{cases} x^{2}+3x+a, & x\le 1,\\ bx+2, & x>1 \end{cases} \text{ is differentiable at } x=1.

\displaystyle \text{Answer:}

\displaystyle \text{Given that } f(x) \text{ is differentiable at } x=1. \text{ Therefore, } f(x) \text{ is continuous at } x=1.

\displaystyle \lim_{x\to 1^-} f(x)=\lim_{x\to 1^+} f(x)=f(1)

\displaystyle \Rightarrow \lim_{x\to 1^-} (x^2+3x+a)=\lim_{x\to 1^+} (bx+2)

\displaystyle \Rightarrow 1^2+3(1)+a=b(1)+2

\displaystyle \Rightarrow 4+a=b+2

\displaystyle \Rightarrow a-b+2=0 \qquad \ldots (1)

\displaystyle \text{Again, } f(x) \text{ is differentiable at } x=1.

\displaystyle \text{Hence, LHD at } x=1=\text{RHD at } x=1

\displaystyle \lim_{x\to 1^-} \frac{f(x)-f(1)}{x-1} =\lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1^-} \frac{(x^2+3x+a)-(4+a)}{x-1} =\lim_{x\to 1^+} \frac{(bx+2)-(4+a)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1^-} \frac{x^2+3x-4}{x-1} =\lim_{x\to 1^+} \frac{bx-b}{x-1} \quad [\text{Using (1)}]

\displaystyle \Rightarrow \lim_{x\to 1} \frac{(x+4)(x-1)}{x-1} =\lim_{x\to 1} \frac{b(x-1)}{x-1}

\displaystyle \Rightarrow \lim_{x\to 1} (x+4)=\lim_{x\to 1} b

\displaystyle \Rightarrow 5=b

\displaystyle \text{Putting } b=5 \text{ in (1), we get}

\displaystyle a-5+2=0

\displaystyle \Rightarrow a=3

\displaystyle \text{Hence, } a=3 \text{ and } b=5.

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