Let OABCDEFG be a unit cube.
Keeping O as origin.
Let \(\overrightarrow{OA}=\hat{i},\overrightarrow{OC}=\hat{j}\) and \(\overrightarrow{OG}=\hat{k}\)
Consider the diagonals OE and BG.

\(\overrightarrow{OE}=\overrightarrow{OB}+\overrightarrow{BE}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BE}\)
\(=\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OG}=\hat{i}+\hat{j}+\hat{k}\)       \([\because \overrightarrow{AB}= \overrightarrow{OC}, \overrightarrow{BE}= \overrightarrow{OG}]\)
and  \(\overrightarrow{GB}=\overrightarrow{GO}+\overrightarrow{OB}=-\hat{k}+\overrightarrow{OA}+\overrightarrow{AB}=\)\(\hat{i}+\hat{j}-\hat{k}\)
Let \(\theta\) be the smaller angle between the diagonals OE and GB, then
cos \(\theta\)=\({\overrightarrow{OE}.\overrightarrow{GB}\over|\overrightarrow{OE}||\overrightarrow{GB}|}=\)\({1(1)+1(1)+1(-1)\over \sqrt{1^2+1^2+1^2}.\sqrt{1^2+1^2+(-1)^2}}={2-1\over \sqrt{3}\sqrt{3}}={1\over3}\)
Thus \(\theta=\)cos^-1 \(({1\over3})\)

Let A, B, C be the event of winning FIDE cup respectively by X,Y, and Z this year.
Given that X is thrice as likely to win as Y.
A : B :: 3 : 1        (1)
Y is twice as likely as to win Z
B : C :: 2 : 1     (2)
From (1) and (2)
A : B : C :: 6 : 2 : 1
A = 6k, B = 2k, C = k, where k is proportional constant.
Probability to win the cup by X is \(P(A)=\frac { 6k }{ 9k } =\frac { 2 }{ 3 } \)
Probability to win the cup by Y is \(P(B)=\frac { 2k }{ 9k } =\frac { 2 }{ 9 } \) and
Probability to win the cup by Z is \(P(C)=\frac { k }{ 9k } =\frac { 1 }{ 9 } \)

Let be the event of speaks the truth, be the event of speaks the truth
∴ \(\bar { A } \) is the event of X not speaking the truth and \(\bar { B } \) is the event of Y not speaking the truth.
Let C be the event that they will contradict each other.
Given that
P(A) = 0.70 ⇒ P(\(\bar { A } \)) = 1 - P(A) = 0.30
P(B) = 0.90 ⇒ P(\(\bar { B } \)) = 1 - P(B) = 0.10
C = (A speaks truth and B does not speak truth or B speaks truth and A does not speak truth)
C =\(\left[ (A\cap \bar { B } )\cup (\bar { A } \cap B) \right] \) (see figure)
since \((A\cap \bar { B } )\) and \((\bar { A } \cap B)\) are mutually exclusively,
P(C) = \((A\cap \bar { B } )+(\bar { A } \cap B)\)
= P(A)P(\(\bar { B } \)) + P(\(\bar { A } \))P(B)
( Since A, B are independent event A, \(\bar { B } \) are also independent events
= (0.70) (0.10) + (0.30) (0.90)
= 0.070 + 0.270 = 0.34
P(C) = 0.34
 

Let A₁, A₂ and A₃ be the event of A, B, C becoming managers of the company respectively. Let X be the event that the office canteen will be improved.
Then, \(P\left(A_1\right)=\frac{5}{10}=0.5 \)
\(P\left(A_2\right)=\frac{3}{10}=0.3 \)
\(P\left(A_3\right)=\frac{2}{10}=0.2 \)
\(P\left(X / A_1\right)=0.4 \)
\(P\left(X / A_2\right)=0.5\)
\(P\left(X / A_3\right)=0.3\)
\(P\left(A_2 / X\right)=\frac{P\left(A_2\right) P\left(X / A_2\right)}{P\left(A_1\right) P\left(X / A_1\right)+P\left(A_2\right) P\left(X / A_2\right)}+P\left(A_3\right) P\left(X / A_3\right)\)
\(=\frac{0.3(0.5)}{0.5(0.4)+0.3(0.5)+0.2(0.3)} \)
\(=\frac{0.15}{0.2+0.15+0.06} \)
\(=\frac{0.15}{0.41}=\frac{15}{41}\)

Let A₁ be the event of selecting balls from urn I and A₂ be the event of selecting urn II.
Let B be the event of a black ball is drawn from it.
Then we have \(P\left(A_1\right)=P\left(A_2\right)=\frac{1}{2}\)
\(P\left(B / A_1\right)=\frac{6 C_1}{10 C_1} ; P\left(B / A_2\right)=\frac{2 C_1}{1 C_1}\)
\(\text {(i) } P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right): P\left(B / A_2\right)\)
\(=\frac{1}{2} \cdot \frac{6 C_1}{10 C_1}+\frac{1}{2} \cdot \frac{2 C_1}{4 C_1} \)
\(=\frac{1}{2}\left[\frac{6}{10}+\frac{2}{4}\right]=\frac{1}{2}\left[\frac{3}{5}+\frac{1}{2}\right]=\frac{1}{2}\left[\frac{6+5}{10}\right]\)
\(=\frac{11}{20} \)
(ii) P(I/B)
\(=\frac{P(B/I)P(I)}{P(B)}\)
\(=\frac{\frac{6}{10}\times \frac{1}{2}}{\frac{11}{20}}=\frac{6}{20}/\frac{11}{20}=\frac{6}{11}\)

Let I = \(\int{2x+3\over \sqrt{x^2+x+1}}dx\)
\(2x+3=A{d\over dx}(x^2+x+1)+B\)
\(2x+3=A(2x+1)+B\)
Comparing the coefficients of like terms, we get
2A = 2 \(\Rightarrow \ A=1; A+B=3 \Rightarrow B=2\)
I =\(\int{(2x+1)+2\over \sqrt{x^2+x+1}}dx\)
I = \(\int{2x+1\over \sqrt{x^2+x+1}}dx+2\int{1\over \sqrt{x^2+x+1}}dx\)
\(=2\sqrt{x^2+x+1}+2\int{1\over \sqrt{(x+{1\over 2})^2}+({\sqrt{3}\over2})^2}dx\)
\(=2\sqrt{x^2+x+1}+2log|x+{1\over2}+\sqrt{(x+{1\over2})^2+({\sqrt{3}\over2})^2}|+c\)
Therefore, I = \(2\sqrt{x^2+x+1}+2log|x+{1\over2}+\sqrt{x^2+x+1}|+c\)

Let I = \(\int{3x+5\over x^2+4x+7}dx\)
3x + 5 = A\({d\over dx}(x^2+4x+7)+B\)
3x + 5 = A(2x+4)+ B
Comparing the coefficients of like terms, we get
2A = 3\(\Rightarrow\)A = \({3\over2};4A+B=5\Rightarrow B=-1\)
\(I=\int{{3\over2}(2x+4)-1\over x^2+4x+7}dx\)
\(I={3\over2}\int{2x+4\over x^2+4x+7}dx-\int {1\over x^2+4x+7}dx\)
\(={3\over2}log|x^2+4x+7|-\int {1\over (x+2)^2+(\sqrt{3})^2}dx\)
\(={3\over2}log|x^2+4x+7|-{1\over \sqrt{3}tan^{-1}}({x+2\over \sqrt{3}})+c\)

Let I =\(\int tan^{-1}({2x\over 1-x^2})dx\)
Putting x = \(tan \theta \Rightarrow dx=sec^2 \theta d \theta\)
Therefore, \(I=\int tan^{-1}({{2tan \theta}\over 1-tan^2 \theta})sec^2 \theta d\theta\)
\(=\int tan^{-1}(tan 2\theta)sec^2\theta d \theta\)
\(=\int 2\theta sec^2\theta d \theta\)
\(=2\int (\theta )(sec^2\theta d \theta)\)

Applying integration by parts
I = 2[\(\theta tan \theta- \int tan \theta d \theta\)]   \(tan \theta =x\)
= 2\((\theta tan \theta- log|sec \theta | )+c\)    \(sec \theta =\sqrt{1+x^2}\)
\(\int tan^{-1}({2x\over 1-x^2})dx=2x tan ^{-1}x-2log|\sqrt{1+x^2}|+c\)

The rate of change of height h with respect to time t is the derivative of h with respect to t.
Therefore,\({dh\over dt}={18\over \sqrt{t}}=18t^{-{1\over2}}\)
So, to get a general expression for the height, integrating the above equation with respect to t.
\(h=\int {18t^{-{1\over 2}}}dt=18(2t^{1\over2})+c=36\sqrt{t}+c\)
Given that when t = 0, the height h = 5 cm.
\(5=0+c \Rightarrow c=5\)
\(h=36\sqrt{t}+5\).
(i) To find the height of the tree after 4 years.
When t = 4 years,
\(h=36\sqrt{t}+5\Rightarrow h=36\sqrt{4}+5=77\)
The height of the tree after 4 years is 77 cm .
(ii) When h = 149 cm
\(h=36 \sqrt{t}+5 \Rightarrow 149=36 \sqrt{t}+5\)
\(\sqrt{t}=\frac{149-5}{36}=4 \Rightarrow t=16\)
Thus after 16 years the height of the tree will be 149 cm.

Given \(y_1=\frac{\sin ^{-1} x}{\sqrt{1-x^2}} .....(1)\)
\(\sqrt{1-x^2} y=\left(\sin ^{-1} x\right)\)
Squaring on both sides, (1 - x²) y² = (sin-1 x)²
Differentiate W.R.T x 
\(\left(1-x^2\right)\left(2 y y_1\right)+y^2(-2 x)=2 \sin ^{-1} x \frac{1}{\sqrt{1-x^2}}\) (using (1))
\(\left(1-x^2\right)\left(2 y y_1\right)-2 x y^2=2 y\)
The above equation divided by 2y.
(1 - x²) y₁ - ay = 1.
Diffrentiate W. R. To x
\( \left(1-x^2\right) y_2+y_1(-2 x)-x y_1-y(1)=0 \)
\(\left(1-x^2\right) y_2-2 x y_1-x y_1-y=0 \)
\(\left(1-x^2\right) y_2-3 x y_1-y=0\)
Hence proved.

We have x⁴ + y⁴ = 16.
Differentiating implicitly, 4x³ +4y³y' = 0
Solving for y' gives
\(y'=-{x^3\over y^3}\)
To find y'' we differentiate this expression for y' using the quotient rule and remembering that y is a function of x.
\( y^{\prime \prime}=\frac{d}{d x}\left(\frac{-x^3}{y^3}\right) =\frac{-\left[y^3 \frac{d}{d x}\left(x^3\right)-x^3 \frac{d}{d x}\left(y^3\right)\right]}{\left.\left(y^3\right)^2\right]} \\ \)
\(=-\frac{\left[y^3 \cdot 3 x^2-x^3\left(3 y^2 y^{\prime}\right)\right]}{y^6} \)
\( =-\frac{3 x^2 y^3-3 x^3 y^2\left(-\frac{x^3}{y^3}\right)}{y^6} \)
\( =-\frac{3\left(x^2 y^4+x^6\right)}{y^7}=\frac{-3 x^2\left[x^4+y^4\right]}{y^7} \)
\( = \frac{-3 x^2(16)}{y^7}=\frac{-48 x^2}{y^7} . \)

Given \(f(x)={x^3+64\over x+4}\)
The given function does not exist at x = - 4.
Hence, f(x) has a removable discontinuity at x = -4
\(lim_{x\rightarrow -4}f(x)=lim_{x\rightarrow -4}{x^3+64\over x+4}\)

\(=lim_{x\rightarrow -4}\) x²-4x+16 = (-4)²-4(-4)+ 16 = 16+ 16+ 16
=48
\(\therefore\) The continuous function g(x) can be written as
g(x) = {\(\begin{matrix} \frac { { x }^{ 3 }+64 }{ x+4 } & if\ x\ \neq -4 \\ 48 & if\ x=-4 \end{matrix}\)

Given \(f(x)= \begin{cases}\frac{x^{4}-1}{x-1}, & \text { if } x \neq 1 \\ \alpha, & \text { if } x=1\end{cases}\)
\(lim_{x\rightarrow 1^-}f(x)=lim_{x\rightarrow 1^-}{x^4-1\over x-1}=lim_{x\rightarrow1^-}{(x^2+1)(x^2-1)\over (x-1)}\)

\(=lim_{x\rightarrow1^-}(x^2+1)(x+1)=(2)(2)=4\)
\(lim_{x\rightarrow 1^+}f(x)=lim_{x\rightarrow 1^+}{x^4-1\over x-1}=lim_{x\rightarrow1^+}{(x^2+1)(x+1)}=4\)
\(\therefore lim_{x\rightarrow 1^-}f(x)=lim_{x\rightarrow 1^+}f(x)=4\)
Since f(x) is continuous at x = 1, one sided limits = value of the function.
\(\Rightarrow \therefore f(1)=\alpha=4\)
\(\alpha =4\)

Let f(x) = \(\begin{cases}\frac{x^{3}-1}{x-1}, & \text { if } x \neq 1 \\ 3, & \text { if } x=1\end{cases}\)
\(lim_{x\rightarrow 1^-}f(x)=lim_{x-1^-}{(x-1)(x^2+x+1)\over x-1}=lim_{x\rightarrow 1^-}x^2+x+1=3\)
\(lim_{x\rightarrow 1^+}f(x)=lim_{x\rightarrow 1^+}x^2+x+1=3\)
and f(1) = 3
\(lim_{x\rightarrow 1^-}f(x)=lim_{x\rightarrow 1^+}f(x)=f(1)=3\)
\(\therefore f(x)\) is continuous on \((-\infty,\infty)\)

\(f(x)=\frac{x\lfloor x\rfloor}{\sin |x|} = \begin{cases}\frac{-x}{\sin (-x)} & \text { if }-1<x<0 \\ \frac{x \cdot 0}{\sin x} & \text { if } 0<x<1\end{cases} \)
\(= \begin{cases}\frac{x}{\sin x} & \text { if }-1<x<0 \\ 0 & \text { if } 0<x<1\end{cases} \)
Therefore, \(lim_{x\rightarrow 0^-}f(x)=+1\)
\(lim_{x\rightarrow 0^+}f(x)=0\).
Hence the limit does not exist.

\(\lim _{v \rightarrow c^{-}}(m)=\lim _{v \rightarrow c^{-}} \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{m_0}{\sqrt{\lim _{v \rightarrow c^{-}}\left(1-\frac{v^2}{c^2}\right)}}\)
For h > 0, c- h < v < c . This implies,(c-h)² \(<v^2<c^2\)
That is, \(\frac{(c-h)^2}{c^2}<\frac{v^2}{c^2}<1\) 
That is, \(\lim _{h \rightarrow 0} \frac{(c-h)^2}{c^2}<\lim _{h \rightarrow 0} \frac{v^2}{c^2}<\lim _{h \rightarrow 0} 1\)
That is, \(1<\lim _{h \rightarrow 0} \frac{v^2}{c^2}<1\)
That is  \(1<\lim _{v \rightarrow c^{-}} \frac{v^2}{c^2}<1\). By Sandwich theorem, \(lim_{v\rightarrow c^-}=1.\)
Therefore, \(lim_{v-c^-}(m)\rightarrow \infty.\)
That is, the mass becomes very very large (infinite) as \(v\rightarrow c^-.\)
The left hand limit is necessary.
Otherwise as\(v \rightarrow c^+\) makes \(1-{v^2\over c^2}<0\) and consequently we cannot find the mass.

\(lim_{x\rightarrow2}{2-\sqrt{x+2}\over 3\sqrt{2}-3\sqrt{4-x}}\)\(=lim_{x\rightarrow2}{\sqrt{x+2}-2\over 3\sqrt{4-x}-3\sqrt{2}}=lim_{x\rightarrow2} {(x+2)^{1/2}-(2^2)^{1/2}\over (4-x)^{1/3}-(2)^{1/3}}\)
Multiplying and dividing by (x - 2) we get,
\(=lim_{x\rightarrow2} {(x+2)^{1/2}-(4)^{1/2}\over x-2}\times {x-2\over (4-x)^{1/3}-2^{1/3}}\)
\(=lim_{x\rightarrow2} {(x+2)^{1/2}-(4)^{1/2}\over x+2-4}\times {-(2-x)\over (4-x)^{1/3}-2^{1/3}}\)
\(=[lim_{x\rightarrow2+4} {(x+2)^{1/2}-(4)^{1/2}\over x+2-4}]\times [{-1\over lim_{x\rightarrow2} {(4-x)^{1/3}-2^{1/3}\over 4-x-2}}]\)
\(={1\over2}\times 4^{{1\over2}-1}\times [{-1\over {1\over3}\times 2^{{1\over3}-1}}]\)\(=-{1\over2}(4^{-1\over2})\times {1\over {1\over3}2^{-2\over3}}\)
\(=-{1\over2}\times 4^{{1\over2}-1}\times [{-1\over {1\over3}\times 2^{{1\over3}-1}}]=-{1\over2}(4^{-1/2})\times {1\over {1\over3}2^{-2\over3}}\)
\(=-{1\over2}\times {3\over1}\times {1\over \sqrt{4}}\times 2^{2/3}={-3\over4}\times (2^2)^{1/3}=-{3\over4}{\sqrt[3]{4}}\)

\(|A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{array}\right|\)
Put x = Y
\(|A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y & y & z \\ y^2 & y^2 & z^2 \end{array}\right|=0 \quad\left(C_1 \cong C_2\right)\)
\(\therefore(x-y) \text { is a factor of }|A|\)
The given determinant is in cyclic symmetric form in x, y and z.
\(\therefore\) (y - 2) and (z - x) are also factors.
The degree of the product of the factors (x - y)y - z)(z - x) is 3 and the degree of the product of the leading diagonal elements 1 - y , z² = 3
The other factor is k
\(\therefore\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{array}\right|=k(x-y)(y-z)(z-x)\)
Put x = 1, y = -1, z = 0
\(\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{array}\right|=k(2)(-1)(-1)\)
1(1 + 1) = 2k
2 = 2k
k = 1
Put in (1)
\(|A|=(x-y)(y-z)(z-x)\)
Hence proved.

Taking p = 0, we get | A | = \(\begin{vmatrix} (q+r)^2& 0&0 \\ q^2 & r^2 & q^2 \\ r^2 &r^2 & q^2 \end{vmatrix}=0\)
Therefore, (p - 0) is a factor. That is, p is a factor.
Since | A | is in cyclic symmetric form in p, q, r and hence q and r also factors.
Putting p + q + r = 0 \(\Rightarrow\) q + r = - p ; r + p = - q ; and p + q = - r.
| A | = \(\begin{vmatrix} p^2& p^2&p^2 \\ q^2 & q^2 & q^2 \\ r^2 &r^2 & r^2 \end{vmatrix}=0\) since 3 columns are identical.
Therefore, (p + q + r)² is a factor of | A | .
The degree of the obtained factor pqr (p + q + r)² is 5. The degree of | A | is 6.
Therefore, required factor is k (p + q + r).
\(\begin{vmatrix} (q+r)^2& p^2 &p^2 \\ q^2 & (r+p)^2 & q^2 \\ r^2 &r^2 & (p+q)^2 \end{vmatrix}\) = k(p + q + r) (p + q + r)² x pqr
Taking p = 1, q = 1, c = 1, we get
\(\begin{vmatrix} 4 &1 &1 \\ 1 & 4 &1 \\ 1 & 1 & 4 \end{vmatrix}\) = k(1 +1 + 1)³ (1) (1) (1).
4(16 - 1) - 1(4 - 1) + 1(1 - 4) = 27k
60 - 3 - 3 = 27 k \(\Rightarrow\) k = 2.
| A | = 2pqr (p + q + r)³.

Given f(x) = \(\begin{bmatrix} cos \ x & -sin \ x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
f(x) \(\times\) f(y) = \(\begin{bmatrix} cos \ x & -sin \ x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} cosxcosy-sinxcosy & -cosxsiny-sinxcosy & 0 \\ sin \ x cos \ y+cos \ xsin \ y & -sin \ x \ sin \ y+cos \ x \ cos \ y & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
[\(\therefore\)cos(x + y) = cos x cos y - sin x sin y]
sin(x + y) = sin x cos y + cos x sin y]
= f(x + y)

Let I be the event of getting State Government service and C be the event of getting Central Government job.
Given that P(I) = 0.12, P(C) = 0.25, and \(P(I \cap C)\) = 0.07
(i) P( at least one of the two jobs) \(=P(I \text { or } C)=P(I \cup C)\)
\(=P(I)+P(C)-P(I \cap C) \)
\(=0.12+0.25-0.07=0.30\)
(ii) P(only one of the two jobs) = P [only I or only C]
=\(P\left( I\cap \overline { C } \right) +P\left( \overline { I } \cup C \right) \)
\(=\{0.12-0.07\}+\{0.25-0.07\}\)
\(=0.23 \)

Let u = log x
Then e^u = x and
\(\frac{d u}{d x}=\frac{1}{x}\)
\(\therefore \frac{d u}{d x}=\frac{1}{e^u}\)
\(\therefore e^u d u=d x\)
\( \int \frac{\log x}{(1+\log x)^2} d x =\int \frac{u}{(1+u)^2} \times e^u d u\)
\(=\int e^u\left(\frac{(1+u)-1}{(1+u)^2}\right) d u \)
\(=\int e^u\left(\frac{1}{1+u}-\frac{1}{(1+u)^2}\right) d u \)
\(=e^u\left(\frac{1}{1+u}\right)+c \)
Putting u = log x and e^u = u
\(=x\left(\frac{1}{1+\log x}\right)+c \)
\(=\frac{x}{1+\log x}+c\)

\(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=\int \sin ^{-1} x \times \frac{x}{\sqrt{1-x^2}} d x \)
\(Let\ u=\sin ^{-1} x\)
Differentiating on both sides
\(\frac{d u}{d x}=\frac{1}{\sqrt{1-x^2}}\)
\(d u=\frac{1}{\sqrt{1-x^2}} d x \)
\(d v=\frac{x}{\sqrt{1-x^2}} \)
\(v=\int \frac{x}{\sqrt{1-x^2}} d x\)
Take \(t=\sqrt{1-x^2}\)
\( \mathrm{t}^2 =1-x^2\)
\(2 \mathrm{td} t =-2 x \mathrm{~d} x \)
\(\mathrm{tdt} =-x \mathrm{~d} x \)

\(x \mathrm{~d} x =-\mathrm{t} \mathrm{dt}\)
\( v =\int \frac{1-1 d }{1} \)
\(=-\int d=-1 \)
\(v =-\sqrt{1-x^2} \)
\(\therefore \int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x =\sin ^{-1} x\left(-\sqrt{1-x^2}\right)- \int\left(-\sqrt{1-x^2}\right) \times \frac{1}{\sqrt{1-x^2}} d x \ {\left[\because \int u d v=u v-\int v d u\right] } \)
\(=-\sqrt{1-x^2} \sin ^{-1} x+\int d x\)
\( =-\sqrt{1-x^2} \sin ^{-1} x+x+c\)

Let x = tan \(\theta\)
Then \({1+x\over 1-x}={1+tan \theta \over 1- tan \theta}=tan({\pi\over 4}+\theta)\).
\(tan^{-1}({1+x\over 1-x})=tan^{-1}[tan({\pi\over 4}+\theta)]={\pi\over 4}+\theta={\pi\over 4}+tan^{-1}x\)
\(y={\pi\over 4}+tan^{-1}x\)
\(y'={1\over 1+x^2}\).

\(\therefore\) f'(x) = \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { f(x)-f(0) }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { 3x-0 }{ x } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { 3x }{ x } = 3\) [f(x) = 3x]  .....(1)
\(\therefore\) f'(0⁺) = \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { f(x)-f(0) }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { -4x-0 }{ x } \) 
= \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { -4x }{ x } \) = -4 .........(2)
by (1) & (2), f'(0^-) ≠ f'(0⁺)
\(\therefore\) It is not differentiable.