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JEE MainMathematicsModerateContinuity and derivability

Let f: R→ R be a function defined as
$f(x) = \left\{ \begin{matrix}
5, & \ if\ & x \leq 1 \\
a + bx, & \ if\ & 1 < x < 3 \\
b + 5x, & \ if\ & 3 \leq x < 5 \\
30, & \ if\ & x \geq 5
\end{matrix} \right.\ $
Then, f is

Show answer and solution
$f(x) = \left\{ \begin{matrix}
5 &; & x \leq 1 \\
a + bx &; & 1 < x < 3 \\
b + 5x &; & 3 \leq x < 5 \\
30 &; & x \geq 5
\end{matrix} \right.\ $
for function to be continuous at x=1
f(1)=lim⁡h→0\mspace2muf(1+h)=lim⁡h→0\mspace2muf(1−h)f(1) = \lim_{h \rightarrow 0}\mspace{2mu} f(1 + h) = \lim_{h \rightarrow 0}\mspace{2mu} f(1 - h)
f(1)=5f(1) = 5
lim⁡h→0\mspace2muf(1+h)=lim⁡h→0\mspace2mua+b(1+h)=a+b\lim_{h \rightarrow 0}\mspace{2mu} f(1 + h) = \lim_{h \rightarrow 0}\mspace{2mu} a + b(1 + h) = a + b
lim⁡h→0\mspace2muf(1−h)=lim⁡h→0\mspace2mu5=5\lim_{h \rightarrow 0}\mspace{2mu} f(1 - h) = \lim_{h \rightarrow 0}\mspace{2mu} 5 = 5
hence a+b=5.........(i)
for function to be continuous at x=3
f(3)=lim⁡h→0\mspace2muf(3+h)=lim⁡h→0\mspace2muf(3−h)f(3) = \lim_{h \rightarrow 0}\mspace{2mu} f(3 + h) = \lim_{h \rightarrow 0}\mspace{2mu} f(3 - h)
f(3)=b+15f(3) = b + 15
lim⁡h→0\mspace2muf(3+h)=lim⁡h→0\mspace2mub+5(3+h)=b+15\lim_{h \rightarrow 0}\mspace{2mu} f(3 + h) = \lim_{h \rightarrow 0}\mspace{2mu} b + 5(3 + h) = b + 15
lim⁡h→0\mspace2muf(3−h)=lim⁡h→0\mspace2mua+b(3−h)=a+3b\lim_{h \rightarrow 0}\mspace{2mu} f(3 - h) = \lim_{h \rightarrow 0}\mspace{2mu} a + b(3 - h) = a + 3b
 hence b+15=a+3b\ hence\ b + 15 = a + 3b ⇒ a+2b=15a + 2b = 15.... (2)
for function to be continuous at x=5
f(5)=lim⁡h→0\mspace2muf(5+h)=lim⁡h→0\mspace2muf(5−h)f(5)=30\begin{matrix} f(5) = \lim_{h \rightarrow 0}\mspace{2mu} f(5 + h) = \lim_{h \rightarrow 0}\mspace{2mu} f(5 - h) \\ f(5) = 30 \end{matrix}

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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