Answer :

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) = x^{2}+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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