GATE CSE 1996 | Question: 3

Question

Let $f$ be a function defined by

$$f(x)=\begin{cases}
x^2 & \text{for } x \le 1 \\
ax^2 + bx + c & \text{for } 1 &lt x \le 2\\

x + d & \text{for } x > 2
\end{cases}$$
Find the values for the constants $a, b, c$ and $d$ so that $f$ is continuous and differentiable everywhere on the real line.

Comments

For continuity

Critical points are x=1 and x=2 , so we will check there.

1)   $f(1^{+})$=$f(1)$=$f(1^{-})$

 $1=a+b+c$  ----------$equation 1$

2)   $f(2^{+})$=$f(2)$=$f(2^{-})$

$4a+2b+c=2+d$ -------$equation 2$

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For derivability

1)  $f'(1^{+})$=$f'(1^{-})$

$\frac{\mathrm{d} (x^{2})}{\mathrm{d} x}$=$\frac{\mathrm{d} (ax^{2}+bx+c)}{\mathrm{d} x}$    

so $2x=2ax+b$   // put $x=1$

$2=2a+b$ ------$equation 3$

2)   $f'(2^{+})$=$f'(2^{-})$

$\frac{\mathrm{d} (ax^{2}+bx+c)}{\mathrm{d} x}$=$\frac{\mathrm{d} (x+d)}{\mathrm{d} x}$

so $2ax+b=1$    at $x=2$

 $4a+b=1$   --------$equation 4$

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This is a very clear explanation. Thanx  a lot !

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@Arjun  sir  whenever I enable dark mode,

the color of question in pip mode became invisible type.

Please do some modification in the color of pip mode when dark mode is active.

Answer 1

$f$ is differentiable at $1$ if

$\lim\limits_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim\limits_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$

$\Rightarrow2 = 2a+b - (1)$

$f$ is differentiable at $2$ if

$\lim\limits_{h \to 0^{-}} \frac{f(2+h)-f(2)}{h}=\lim\limits_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}$

$\Rightarrow 4a+b = 1 - (2)$

Solving $(1)$ and $(2),$ we get

$a = -0.5, b = 3$

Now, $f$ has to be continous on $1$ also, so

$\lim\limits_{x \to 1^{-}} f(x)=\lim\limits_{x \to 1^{+}}(x)=f(1)$

$\Rightarrow 1 = a + b + c$

$\Rightarrow c = -1.5$

Similarly, $f$ has to be continous on $2$ also, so

$\lim\limits_{x \to 2^{-}} f(x)=\lim\limits_{x \to 2^{+}}(x)=f(2)$

$\Rightarrow 4a+2b+c = 2+d$

$\Rightarrow d = 0.5$

So, $a = -0.5, b = 3, c = -1.5, d = 0.5$

Comments

@Sachin Mittal 1 sir I have the same doubt as @ankitgupta.1729

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bohot accha hai -->questoion

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@ankitgupta.1729 @chokostar both are same.

one subtracts a small positive number $h\rightarrow0^{+}$ from $x_{0}$,

while another adds a small negative number $h\rightarrow0^{-}$ to $x_{0}$.

Answer 2

Sorry we can also differentiate all function one at a time 

Comments

Here we correct my answer

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By solving (1) and (2)

a=0.5and b=0

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nope a = -½  , b= 3, c  = -3/2 , d= ½

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