If fx=ax2+b, b≠0, x≤1bx2+ax+c, x>1then fx is continuous and differentiable at x=1, if

LHD=Ltx→1f(x)−f(1)x−1=Ltx→1ax2+b−a−bx−1=Ltx→1ax2−1x−1=Ltx→1a(x+1)=2aRHD=Ltx→1f(x)−f(1)x−1=Ltx→1bx2+ax+c−a−bx−1==Ltx→1f(x)−f(1)x−1=Ltx→1bx2+ax+c−a−bx−1=Ltx→1[b(x+1)+a]+cx−1=2 b+a+cx−1=2 b+a if c=0 Since, f is diff. at x=1∴2a=2 b+a⇒a=2 b . Thus, result holds if a=2 b,c=0.

If fx=ax2+b, b≠0, x≤1bx2+ax+c, x>1then fx is continuous and differentiable at x=1, if

LHD=Ltx→1f(x)−f(1)x−1=Ltx→1ax2+b−a−bx−1=Ltx→1ax2−1x−1=Ltx→1a(x+1)=2aRHD=Ltx→1f(x)−f(1)x−1=Ltx→1bx2+ax+c−a−bx−1==Ltx→1f(x)−f(1)x−1=Ltx→1bx2+ax+c−a−bx−1=Ltx→1[b(x+1)+a]+cx−1=2 b+a+cx−1=2 b+a if c=0 Since, f is diff. at x=1∴2a=2 b+a⇒a=2 b . Thus, result holds if a=2 b,c=0.

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$\begin{array}{l} If f (x) = \{\begin{cases} a x^{2} + b, b \neq 0, x \leq 1 \\ b x^{2} + a x + c, x > 1 \end{cases} \\ t h e n f (x) i s c o n t i n u o u s a n d d i f f e r e n t i a b l e a t x = 1, i f \end{array}$

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c=0, a=2b

a=b and c is non zero constant

a=b, c any value

a=b,c=0

answer is A.

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Detailed Solution

$\begin{matrix} LHD = {Lt}_{x \to 1} \frac{f (x) - f (1)}{x - 1} \\ = {Lt}_{x \to 1} \frac{a x^{2} + b - a - b}{x - 1} \\ = {Lt}_{x \to 1} \frac{a (x^{2} - 1)}{x - 1} \\ = \underset{x \to 1}{Lt} a (x + 1) \\ = 2 a \\ R H D = {Lt}_{x \to 1} \frac{f (x) - f (1)}{x - 1} \\ = {Lt}_{x \to 1} \frac{b x^{2} + a x + c - a - b}{x - 1} \\ = = {Lt}_{x \to 1} \frac{f (x) - f (1)}{x - 1} \\ = {Lt}_{x \to 1} \frac{b x^{2} + a x + c - a - b}{x - 1} \\ = {Lt}_{x \to 1} [b (x + 1) + a] + \frac{c}{x - 1} \\ = 2 b + a + \frac{c}{x - 1} \\ = 2 b + a if c = 0 \\ Since, f is diff. at x = 1 ∴ 2 a = 2 b + a \\ \Rightarrow a = 2 b . Thus, result holds if a = 2 b, c = 0. \end{matrix}$