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Question
Find the values of p and q so that f(x) = `{{:(x^2 + 3x + "p"",", "if" x ≤ 1),("q"x + 2",", "if" x > 1):}` is differentiable at x = 1
Solution
Given that: f(x) = `{{:(x^2 + 3x + "p"",", "if" x ≤ 1),("q"x + 2",", "if" x > 1):}` at x = 1.
L.H.L. f'(c) = `lim_(x -> 1^-) ("f"(x) - "f"("c"))/(x - "c")`
⇒ f'(1) = `lim_(x -> 1^-) ("f"(x) - "f"(1))/(x - 1)`
= `lim_(x -> 1^-) ((x^2 + 3x + "p") - (1 + 3 + "p"))/(x - 1)`
= `lim_("h" -> 0) ([(1 - "h")^2 + 3(1 - "h") + "p"] - [4 + "p"])/(1 - "h" - 1)`
= `lim_("h" -> 0) ([1 + "h"^2 - 2"h" + 3 - 3"h" + "p"] - [4 + "p"])/(-"h")`
= `lim_("h" -> 0) (["h"^2 - 5"h" + 4 + "p"] - [4 + "p"])/(-"h")`
= `lim_("h" -> 0) ("h"^2 - 5"h" + 4 + "p" - 4 - "p")/(-"h")`
= `lim_("h" -> 0) ("h"^2 - 5"h")/(-"h")`
= `lim_("h" -> 0) ("h"["h" - 5])/(-"h")`
= 5
R.H.L. f'(1) = `lim_(x -> 1^+) ("f"(x) - "f"(1))/(x - 1)`
= `lim_(x -> 1^+) (("q"x + 2) - (1 + 3 + "p"))/(x - 1)`
= `lim_("h" -> 0) (["q"(1 + "h") + 2] - [4 + "p"])/(1 + "h" - 1)`
= `lim_("h" -> 0) ("q" + "qh" + 2 - 4 - "p")/"h"`
= `lim_("h" -> 0) ("qh" + "q" - 2 - "p")/"h"`
For existing the limit
q – 2 – p = 0
⇒ q – p = 2
⇒ `lim_("h" -> 0) ("qh" - 0)/"h"` = q
If L.H.L. f'(1) = R.H.L. f'(1) then q = 5.
Now putting the value of q in equation (i)
5 – p = 2
⇒ p = 3.
Hence, value of p is 3 and that of q is 5.
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