Given: f(x) = [x]

\(\operatorname{Rf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\)

\(=\lim _{h \rightarrow 0} \frac{[1+h]-[1]}{h}=\lim _{h \rightarrow 0} \frac{(1-1)}{h}=0\)

\(L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{[1-h]-[1]}{-h}\)

\(=\lim _{h \rightarrow 0} \frac{0-1}{-h}=\infty\)

\(R f^{\prime}(1)^{1} L f^{\prime}(1) .\)

So, \(f^{\prime}(1)\) does not exist.

Hence, the correct option is (D).

We have \(f(x+y)=f(x)\) f \((y)\), for all \(x, y\) Î \(R\). 

Putting \(x=y=0\), we get

f(0)=f(0)f(0)

Now f'(0)=2 (Given)

We know,

\(\lim _{x \rightarrow 0} \frac{f(0+h)-f(0)}{h}=2\)

\(\Rightarrow \lim _{h \rightarrow 0} \frac{f(0) f(h)-f(0)}{h}=2\)

\(\Rightarrow \lim _{h\rightarrow 0} \frac{f(h)-1^{n}}{h}=2\) [using \(\left.f(0)=1\right]\)

Again, \(f(4)=\frac{f(4+h)-f(4)}{h}\)

\(f'(x)=\lim _{1 \rightarrow 0} \frac{((4)+(h)-f(4)}{h}\)

\(=\left(\lim _{n} \frac{f(h)-1}{h}\right) f(4)=2 f(4)\) 

\(\therefore\ f'(x)= 2×4=8\) (since f(4)=4)

Hence , options (B ) is the correct answer.

Here the correct option is (C). 

We know that, If A, B and C are subsets of a set X. Then,

I. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

II. A ∪ A = A, A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A and A ∩ A = A

III. (A ∩ B) ∪ C = (A ∩ C) ∪ (B ∩ C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Now, checking the options,

(A): A ∪ (A ∩ B)

= (A ∪ A) ∩ (A ∪ B)    ---- (Using property I)

= A ∩ (A ∪ B)

= A    ---- (Using property II)

So, option (A) is not correct.

(B): A ∩ (A ∪ B)

= (A ∩ A) ∪ (A ∩ B)    ---- (Using property I)

= A ∪ (A ∩ B)

= A    ---- (Using property II)

So, option (B) is correct.

(C): (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)       ---- (Using property III)

So, option (C) is correct.

(D): (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)       ---- (Using property IV)

So, option (D) is correct.

\(\mathrm{P}=(6,8,10)\)

\(\mathrm{Q}=(3,4,8)\)

\(\mathrm{P}^{1}=\mathrm{x}=6 \quad \mathrm{P}_{1}^{1} \mathrm{x}=3\)

\(\mathrm{P}^{2}=\mathrm{y}=8 \quad \mathrm{P}_{2}^{1} \mathrm{y}=4\)

\(\mathrm{P}^{3}=\mathrm{z}=10 \quad \mathrm{P}_{3}^{1}=\mathrm{z}=8\)

\(\mathrm{S} 0\)

\(\mathrm{I}=6-3=3\)

\(\mathrm{b}=8-4=4\)

\(\mathrm{h}=10-8=2\)

diagonal \(=\sqrt{1^{2}+\mathrm{b}^{2}+\mathrm{b}^{2}}\)

\(=\sqrt{3^{2}+4^{2}+2^{2}}\)

\(=\sqrt{9+16+4}\)

\(=\sqrt{29}\)

We know that,

\(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)    ---- (1)

and \(\lim _{x \rightarrow 0} \frac{\tan x}{x}=1\)    ---- (2)

We have to find the value of \(\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{\tan 3 x^{\circ}}\).

We also know that \(180^{\circ}=\pi\) radian

So, if \( 1^{\circ}=\frac{\pi}{360}\)

\( \therefore x^{\circ}=\frac{\pi x}{180}\)

Therefore, the given expression will become,

\(\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{\tan 3 x^{\circ}}=\lim _{x \rightarrow 0} \frac{\sin \frac{\pi x}{180}}{\tan \frac{3 \pi x}{180}}\)

\(=\lim _{x \rightarrow 0} \frac{\frac{\sin \frac{\pi x}{180}}{\frac{\pi x}{180}} \times \frac{\pi x}{180}}{\frac{\tan \frac{3 \pi x}{180}}{\frac{3 \pi x}{18}} \times \frac{3 \pi x}{180}}\)

\(=\lim _{x \rightarrow 0} \frac{\frac{\pi x}{180}}{\frac{3 \pi x}{180}}\)    [From equation (1) and (2)]

\(=\frac{1}{3}\)