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JEE MainMathematicsToughContinuity and derivability

For what triplets of real numbers (a, b, c) with a ≠\neq 0 the
function
$f(x) = \left\{ \begin{matrix}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ x \leq 1 \\
ax^{2} + bx + c:\ \ \ \ \ otherwise\
\end{matrix} \right.\ $

Show answer and solution
$f(x) = \left\{ \begin{matrix}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x;\ \ \ \ x \leq 1 \\
ax^{2} + bx + c:\ \ \ \ \ \ \ \ \ \ \ \ \ otherwise\
\end{matrix} \right.\ $
At x =1
LHL=lim⁡x→1−\mspace2muf(x)=1LHL = \lim_{x \rightarrow 1^{-}}\mspace{2mu} f(x) = 1
RHL=lim⁡x→1+\mspace2muf(x)=a+b+cRHL = \lim_{x \rightarrow 1^{+}}\mspace{2mu} f(x) = a + b + c
∴\therefore f (x) is different a bleat x = 1
⇒f(x) \Rightarrow f(x)\ must be continuous at x = 1
Now f (1) = 1;
∴f(1−)=f(1+)=f(1)\therefore f\left( 1^{-} \right) = f\left( 1^{+} \right) = f(1)
⇒1=a+b+c\Rightarrow 1 = a + b + c
 Also f' (x) = $\left\{ \begin{matrix}
1;\ \ \ \ \ \ \ \ \ \ \ x \leq 1 \\
2ax + b:\ \ \ \ \ \ otherwise\
\end{matrix} \right.\ $
At x =1 LHD=lim⁡x→1−\mspace2muf′(x)=1LHD = \lim_{x \rightarrow 1^{-}}\mspace{2mu} f^{'}(x) = 1
RHD=lim⁡x→1+\mspace2muf′(x)=2a+b,RHD = \lim_{x \rightarrow 1^{+}}\mspace{2mu} f^{'}(x) = 2a + b, for
derivability at x = 1
LHD = RHD
⇒(2a+b=1)\Rightarrow (2a + b = 1)
∴a+b+c−1⇒a+(1−2a)+c=1⇒c=a\therefore a + b + c - 1 \Rightarrow a + (1 - 2a) + c = 1 \Rightarrow c = a
∴a=a,b=1−2a,c=a\therefore a = a,b = 1 - 2a,c = a
∴{(a,1−2a,a);a∈R}\therefore\{(a,1 - 2a,a);a \in R\} are the tripletspossible

Source: Grass-G / M-09 Mathematics Paper 26 October.docx

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