Maths Continuity and Differentiability

(a + √2bcosx) (a - √2bcosy) = a² - b²
⇒ a² - √2abcosy + √2abcosx - 2b²cosxcosy = a² - b²
Differentiating both sides:
0 - √2ab (-siny dy/dx) + √2ab (-sinx) - 2b² [cosx (-siny dy/dx) + cosy (-sinx)] = 0
At (π/4, π/4):
ab dy/dx - ab - 2b² (-1/2 dy/dx + 1/2) = 0
⇒ dy/dx = (ab+b²)/ (ab-b²) = (a+b)/ (a-b); a, b > 0

y² + ln (cos² x) = y x ∈ (-π/2, π/2)
for x = 0 y = 0 or 1
Differentiating wrt x
⇒ 2y' - 2tan x = y'
At (0,0)y' = 0
At (0,1)y' = 0
Differentiating wrt x
2yy' + 2 (y')² - 2sec² x = y'
At (0,0)y' = -2
At (0,1)y' = 2
∴ |y' (0)| = 2

f (x) = {ae? +be? , -1For continuity at x=1
Lim (x→1? )f (x) = Lim (x→1? )f (x)
⇒ ae+be? ¹=c ⇒ b=ce-ae²
For continuity at x=3
Lim (x→3? )f (x) = Lim (x→3? )f (x)
⇒ 9c=9a+6c ⇒ c=3a
f' (0)+f' (2)=e
(ae? -be? ) at x=0 + (2cx) at x=2 = e
⇒ a-b+4c=e
From (1), (2) and (3)
a-3ae+ae²+12a=e
⇒ a (e²+13-3e)=e
⇒ a=e/ (e²-3e+13)

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

Using LMVT

$f^{\mathrm{\text{'}}}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

$3c^2-8c+8=\frac{16-11}{1-0}$

$3c^2-8\mathrm{c}+3=0$

$c=\frac{4-\sqrt[]{7}}{3}\in \left(\mathrm{0,1}\right)$

f (x) is discontinuous at integers x=1,2,3. P= {1,2,3}.
f (x) is not differentiable at integers and where x- [x]=1+ [x]-x ⇒ 2 (x- [x])=1 ⇒ {x}=1/2.
So at x=0.5, 1, 1.5, 2, 2.5.
Q= {0.5, 1, 1.5, 2, 2.5}. Sum of elements is not asked.
Number of elements in P=3, in Q=5. Sum = 8.
Let's check the solution. Q= {1/2, 1, 3/2, 5/2}.
The sum of number of elements: 3+5=8.

P (x) = a (x-2)² + b (x-2) + c.
lim (x→2) P (x)/sin (x-2) = lim (x→2) P (x)/ (x-2) = P' (2) = 7.
P' (x) = 2a (x-2) + b. P' (2) = b = 7.
P' (x) = 2a.
P (3) = a (1)² + b (1) + c = a+b+c = 9.
Continuity at x=2 means lim f (x) = f (2).
lim (x→2) (a (x-2)²+b (x-2)+c)/ (x-2) = P' (2) = b=7. This is given.
The problem states f (2)=7.
P (x) = (x-2) (ax+b) form used in solution. Let's use this.
lim (x→2) (x-2) (ax+b)/sin (x-2) = lim (x→2) ax+b = 2a+b = 7.
P (3) = (3-2) (3a+b) = 3a+b=9.
Solving: a=2, b=3.
P (x) = (x-2) (2x+3).
P (5) = (5-2) (2*5+3) = 3 * 13 = 39.

For x>2, f (x) = ∫? ¹ (5+1-t)dt + ∫? ² (5+t-1)dt + ∫? (5+t-1)dt
= ∫? ¹ (6-t)dt + ∫? ² (4+t)dt + ∫? (4+t)dt
= [6t-t²/2]? ¹ + [4t+t²/2]? ² + [4t+t²/2]?
= (6-1/2) + (8+2 - (4+1/2) + (4x+x²/2 - (8+2)
= 5.5 + 5.5 + 4x+x²/2 - 10 = 4x+x²/2 + 1.
f (2? ) = 8+2+1 = 11. f (2? ) = 5 (2)+1 = 11. Continuous.
f' (x) = 4+x for x>2. f' (2? ) = 6.
For x<2, f' (x)=5. f' (2? )=5.
Not differentiable at x=2.

L.H.L = lim (x→0? ) (1 + |sin x|)³? /|sin x| = lim (h→0) (1 + sinh)³? /sinh = e³?
R.H.L = lim (x→0? ) e^ (cot 4x / cot 2x) = lim (x→0? ) e^ (tan 2x / tan 4x) = e¹/².
f (0) = b.
For continuity, e³? = e¹/² = b.
3a = 1/2 ⇒ a = 1/6. b = e¹/².
6a + b² = 6 (1/6) + (e¹/²)² = 1 + e