Continuity and Differentiability

Since, lim (x→0) f (x)/x exist ⇒ f (0) = 0
Now, f' (x) = lim (h→0) (f (x+h)-f (x)/h = lim (h→0) (f (h)+xh²+x²h)/h (take y = h)
= lim (h→0) f (h)/h + lim (h→0) (xh) + x²
⇒ f' (x) = 1 + 0 + x²
⇒ f' (3) = 10

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)= (x- (1+x)ln (1+x)/ (x² (1+x). Let h (x)=x- (1+x)ln (1+x).
h' (x)=-ln (1+x). h' (x)>0 for x∈ (-1,0), <0 for x (0, ).
h (0)=0, so h (x)≤0. f' (x)≤0. f is decreasing.

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)

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.

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

Given

$\left(x\right)=\{\begin{array}{l}2sin\left(\frac{-\pi x}{2}\right),x<-1\\ \left|a{x}^2+x+b\right|,-1\le x\le 1\\ sin\pi x,x>1\end{array}$

If f (x) is continuous for all $x\in R$ then it should be continuous at x = 1 & x = -1

At x = -1, L.H.L = R.H.L. Þ 2 = |a + b - 1|

=>a + b – 3 = 0  OR  a + b + 1 = 0 . (i)

=>a + b + 1 = 0 . (ii)

           (i) & (ii), a + b =-1