Question

Determine all real values of p and q that ensure the function

f(x) `{:( = "p"x + "q"",", "for"  x ≤ 1),(= tan ((pix)/4)",", "for"  1 < x < 2):}` is differentiable at x = 1

Answer 1

f is differentiable at x = 1.

∴ Lf'(1) = Rf'(1)    ...(1)

f(x) = px + q, for x ≤ 1

∴  f(1) = p(1) + q = p + q

Now, Lf'(1) = `lim_("h" -> 0^-) ("f"(1 + "h") - "f"(1))/"h"`

= `lim_("h" -> 0) (["p"(1 + "h") + "q"] - ["p" + "q"])/"h"`  ...[∵ f(x) = px + q, for x ≤ 1]

= `lim_("h" -> 0) ("p" + "ph" + "q" - "p" - "q")/"h"`

= `lim_("h" -> 0) "ph"/"p"`

= `lim_("h" -> 0) "p"`    ...[∵ h → 0, ∴ h ≠ 0]

= p

Rf'(1) = `lim_("h" -> 0^+) ("f"(1 + "h") - "f"(1))/"h"`

= `lim_("h" -> 0) (tan[(pi(1 + "h"))/4] - ["p" + "q"])/"h"   ...[because "f"(x) = tan((pix)/4),  "for"  1 < x < 2]`

∵ Rf'(1) exists, we must have

p + q = 1   ...(2)

∴ Rf'(1) = `lim_("h" -> 0) (tan[pi/4 + (pi"h")/4] - 1)/"h"`

= `lim_("h" -> 0) (tan[pi/4 + (pi"h")/4] - tan  pi/4)/"h"`

= `lim_("h" -> 0) (tan[pi/4 + (pi"h")/4 - pi/4][1 + tan (pi/4 + (pi"h")/4) tan  pi/4])/"h"`   ...[∵ tan A – tan B = tan(A – B) (1 + tan A tan B)]

= `lim_("h" -> 0) {[(tan  (pi"h")/4)/(((pi"h")/4))][1 + tan (pi/4 + (pi"h")/4) tan  pi/4] xx pi/4}`

= `pi/4[lim_("h" -> 0) tan((pi"h")/4)/((pi"h")/4)] xx  lim_("h" -> 0) [ 1 + tan(pi/4 xx (pi"h")/4) tan  pi/4]`

= `pi/4 xx 1 xx [1 + tan(pi/4 + 0) tan  pi/4]   ...[because "h" -> 0, (pi"h")/4 -> 0  "and" lim_(theta -> 0) tan theta/theta = 1]`

= `pi/4 xx [1 + 1 xx 1]`

= `pi/2`

∴ p = `pi/2`   ...[By (1)]

Substituting the value of p in (2), we get

∴ `pi/2 + "q"` = 1

∴ q = `1 - pi/2 = (2 - pi)/2`

∴ p = `pi/2`, q = `(2 - pi)/2`