Q.Let $f(x) = \begin{cases} x^2 + ax + b, & x < 1 \\ x, & x \ge 1 \end{cases}$. If $f(x)$ is differentiable at $x = 1$, then $(a - b)$ is equal to

The expression is $\tfrac00$ at $x=2$; differentiate top and bottom (L'Hôpital) to get $f(2)-2f'(2)=-4$.

Evaluating a $\tfrac00$ limit that encodes a derivative is a CBSE/NCERT Class 12 limits and differentiability skill.

At $x=2$ the numerator is $2f(2)-2f(2)=0$ and the denominator is $0$, so it is $\tfrac00$. Differentiate numerator and denominator (with $f(2)=4$ a constant): …

Match value and slope of the two pieces at $x=1$: continuity gives $a+b=0$, differentiability gives $a=-1$; hence $a-b=-2$.

Continuity and differentiability of a piecewise function is a CBSE/NCERT Class 12 continuity and differentiability topic.

Continuity at $x=1$: the left piece value equals the right piece value:

$1^2 + a\cdot1 + b = 1 \Rightarrow 1 + a + b = 1 \Rightarrow a + b = 0.$

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This limit is exactly the definition of the derivative $G'(1)$.

By definition, $\displaystyle\lim_{x\to 1}\frac{G(x)-G(1)}{x-1} = G'(1)$.

Given $G(x) = -\sqrt{25-x^2} = -(25-x^2)^{1/2}$, differentiate using the chain rule:

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