Given, circumference of a circle \(S=2 \pi r=56\) 
\(\Rightarrow r=\frac{28}{\pi}\) 
\(\therefore \text { Error } \delta S=2 \pi \delta r=0.02\) 
\(\Rightarrow \delta r=\frac{0.02}{2 \pi}\) 
Let area of circle, \(A=\pi r^{2}\) 
\( \therefore \text { Percentage error in } A=\frac{\delta A}{A} \times 100 \)
\(=2 \times \frac{\delta r}{r} \times 100 \)
\(=2 \times \frac{0.02 \times \pi}{2 \pi \times 28} \times 100=\frac{1}{14} \)

Let r be the radius of the sphere and \(\Delta r\) be the error in measuring radius 
Then, \(r=7 \mathrm{~m} \text { and } \Delta r=0.02 \mathrm{~m}\) 
Now, volume of a sphere is given by \(V=\frac{4}{3} \pi r^{3}\) 
On differentiate W.r.t. r, we get \(\frac{d V}{d r}=\left(\frac{4}{3} \pi\right)\left(3 r^{2}\right)=4 \pi r^{2}\) 
\( \therefore \Delta V =\left(\frac{d V}{d r}\right) \Delta r=\left(4 \pi r^{2}\right) \Delta r \)
\(=4 \pi \times 7^{2} \times 0.02=3.92 \pi \mathrm{m}^{3} \) 
Hence, the approximate error in calculating the volume is \(3.92 \pi \mathrm{m}^{3}\).

We have, \(y=x^{4}-10 \Rightarrow d y / d x=4 x^{3}\) 
and \(\Delta x=2.00-1.99=0.01\) 
\( \therefore \quad \Delta y =\frac{d y}{d x} \times \Delta x=4 x^{3} \times \Delta x \\ =4 \times 2^{3} \times 0.01=32 \times 0.01=0.32 \) 
So, the approximate change in y is 0.32.

We have, \(a y+x^{2}=7 \text { and } x^{3}=y\) 
On differentiating w.r.t. X in both equations, we get
\( \quad a \cdot \frac{d y}{d x}+2 x=0 \quad \text { and } 3 x^{2}=\frac{d y}{d x} \\ \Rightarrow \quad \frac{d y}{d x}=-\frac{2 x}{a} \text { and } \frac{d y}{d x}=3 x^{2} \\ \Rightarrow \quad\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{-2}{a}=m_{1} \\ \text { and }\left(\frac{d y}{d x}\right)_{(1,1)}=3 \cdot 1=3=m_{2} \)
Since, the curves cut orthogonally at (1, 1).
 \(\begin{array}{ll} \therefore & m_{1} \cdot m_{2}=-1 \\ \Rightarrow & \left(\frac{-2}{a}\right)+3=-1 \\ \therefore & a=6 \end{array}\)

We have,\(y=x(x-3)^{2}\) 
\( \therefore \quad \frac{d y}{d x} =x \cdot 2(x-3) \cdot 1+(x-3)^{2} \cdot 1 \\ =2 x^{2}-6 x+x^{2}+9-6 x=3 x^{2}-12 x+9 \\ =3\left(x^{2}-3 x-x+3\right)=3(x-3)(x-1) \) 
So, y = x(x - 3)² decreases for (1, 3).
[since, y' < 0 for all x E (1,3), hence y is decreasing on (1, 3)]

As  \(\int _{ 0 }^{ a }{ x \ dx} \) ≤ a + 4
⇒ \(\frac{d^2}{2}\)≤ a + 4
⇒ a² – 2a — 8 ≤ 0
⇒ (a – 1)² ≤ (3)²
⇒ -3 ≤ a – 1 ≤ 3
⇒ -2 ≤ a ≤ 4

As \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+cos\left( \frac { \pi }{ 2 } -x \right) } } \)
= \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } { sec }^{ 2 }\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } .{ \left[ \frac { tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) }{ -\frac { 1 }{ 2 } } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=-tan\left( \frac { \pi }{ 4 } -\frac { \pi }{ 4 } \right) +tan\left( \frac { \pi }{ 4 } -0 \right) =1\)

∫sec²(7 – 4x)dx =\(\frac { tan(7-4x) }{ -4 } +C=-\frac { 1 }{ 4 } \)tan(7 - 4x) + C

As y’ = 3x² – 3, for a point of absolute maximum or minimum y’=0 ⇒ x = ± 1.
y]_x=0 = 2,
y]_x=1 = 1 – 3 + 2 = 0,
y]_x=-1 = -1 +3+ 2 = 4,
y]_x=2 = 8 – 6 + 2 = 4

As for y² = x, 2yy^' = 1⇒ y']_(1,1) = \(\frac12\)
for x² = y, y' = 2x ⇒ y']_(1,1)= 2
\(\therefore tan\theta =\left| \frac { \frac { 1 }{ 2 } -2 }{ 1+\frac { 1 }{ 2 } .2 } \right| \) \(\Rightarrow tan\theta =\left| \frac { -\frac { 3 }{ 2 } }{ 2 } \right| \)
\(\Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \) or \({ cot }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)

As 2y \(\frac{dy}{dx}\) = 1
⇒\(\frac{dy}{dx}=\frac{1}{2y}\) (slope of tangent)
⇒ \(\frac{1}{2y}\) = tan(-45^o)
⇒ y = \(-\frac{1}{2}\) ⇒ x = \(\frac14\)
∴ Point as \(\left( \frac { 1 }{ 4 } ,-\frac { 1 }{ 2 } \right) \)