# Find the values of p and q so thatIs differentiable at x = 1.

Given that, is differentiable at x = 1.

We know that,f(x) is differentiable at x =1 Lf’(1) = Rf’(1).

Lf’(1) =

=

= ( f(x) = x2+3x+p, if x≤ 1)

=

=

=

=

= 5

Rf’(1) =

=

= ( f(x) = qx+2, if x > 1)

=

=

=

=q

Since, Lf’(1) = Rf’(1)

5 = q (i)

Now, we know that if a function is differentiable at a point,it is necessarily continuous at that point.

f(x) is continuous at x = 1.

f(1-) = f(1+) = f(1)

1+3+p = q+2 = 1+3+p

p-q = 2-4 = -2

q-p = 2

Now substituting the value of ‘q’ from (i), we get

5-p = 2

p = 3

p = 3 and q = 5

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