$Let f : R \to R satisfies | f (x) | \leq x^{2}, \forall x \in R and g : R \to R satisfies$
$g (x + y) = g (x) - g (y) + 2 x y - 1 and g^{'} (0) = \sqrt{3 + a + a^{2}} . Now match the entries from the following two columns$
Column I Column II
P. $At x = 0, f (x) is necessarily$ 1. continuous
Q. $At x = 0, g (x) is necessarily$ 2. differentiable
R. $The number of roots of the equation g (x) = f^{'} (0) is$ 3. 2
S. $If f (t) can be a non-zero root of the equation$ $g (x) = 0 then the least integral value of t can be$ 4. 1

Moderate

Solution

$(P) | f (0) | \leq 0 \Rightarrow f (0) = 0$
$Also, f^{'} (0) = lim_{x \to 0} \frac{f (x) - f (0)}{x} = lim_{x \to 0} \frac{f (x)}{x}$
$and |\frac{f (x)}{x}| \leq | x | \Rightarrow lim_{x \to 0} |\frac{f (x)}{x}| \leq 0 \Rightarrow f^{'} (0) = 0$
$(Q) Put x = y = 0, we get g (0) = - 1$
$\begin{array}{l} g^{'} (x) = lim_{h \to 0} \frac{g (x + h) - g (x)}{h} \\ = lim_{h \to 0} \frac{g (x) - g (h) + 2 x h - 1 - g (x)}{h} \\ = 2 x - lim_{x \to 0} \frac{g (h) - g (0)}{h} = 2 x - g^{'} (0) \\ ∴ g (x) = x^{2} - \sqrt{3 + a + a^{2}} x + C and g (0) = - 1 \Rightarrow C = - 1 \\ ∴ g (x) = x^{2} - \sqrt{3 + a + a^{2}} x - 1 \end{array}$
So, g(x) is continuous and differentiable at x = 0
$(R) g (x) = f^{'} (0) \Rightarrow x^{2} - \sqrt{3 + a + a^{2}} x - 1 = 0 which clearly has two distinct roots.$
$(S) g (- 1) g (1) = - (3 + a + a^{2}) \leq 0 (∵ 3 + a + a^{2} \geq 0)$
$Thus g (x) = 0 has at least one root in [- 1, 1] .$
$Also - 1 \leq f (t) \leq 1 . So, f (1) can be a root of g (x) = 0 [But not necessarily]$

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Column I		Column II
P.	$At x = 0, f (x) is necessarily$	1.	continuous
Q.	$At x = 0, g (x) is necessarily$	2.	differentiable
R.	$The number of roots of the equation g (x) = f^{'} (0) is$	3.	2
S.	$If f (t) can be a non-zero root of the equation$ $g (x) = 0 then the least integral value of t can be$	4.	1