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DSSSB TGT Maths Female Subject Concerned - 18 Nov 2018 Shift 3

Option 2 : 0.5, 1.5

**Concept:**

f(x) is said to be differentiable at a point a if f(x) is continuous at that point which means left hand limit and right hand limit at point a of function is equal.

And also if f(x) is differentiable at point a then left hand derivative at point a is equal to right hand derivative at that point.

__Calculation:__

As function f(x) is differentiable at x = 1 so f(x) is also continuous at x = 1.

So, (L.H.L at x = 1)=\(\lim\limits_{x \to 1^- }f(x)\)

\(=\lim\limits_{h \to 0 }f(1-h)\)

\(=\lim\limits_{h \to 0 }[a(1-h)^2-b]\)

L.H.L = a - b

So, (R.H.L at x = 1)=\(\lim\limits_{x \to 1^+ }f(x)\)

\(=\lim\limits_{h \to 0 }f(1+h)\)

\(=\lim\limits_{h \to 0 }-\frac{1}{{\left|1-h \right|}}\)

R.H.L = -1

As function is continuous. So, L.H.L = R.H.L

We have, a - b = -1 ....... (i)

Now, derivatives

(L.H.D at x = 1)

\(=\displaystyle\frac{d}{dx}(ax^2-b)\)

\(=(2ax)_{x=1}\)

= 2a

(R.H.D at x = 1)

\(=\displaystyle\frac{d}{dx}(-\frac{1}{x})\)

\(=(\frac{1}{x^2})_{x=1}\)

= 1

Now, f(x) is differentiable at x = 1. So, L.H.D = R.H.D

2a = 1

a = 0.5

Putting value of a in equation (i) we get,

b = 1.5

**Hence, the value of a and b for which the function is differentiable is 0.5 and 1.5 respectively.**