Recall that surface area of a sphere with radius r is given by S(r) = 4\(\pi \)r³. Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) - S(5) ≈ S'(5)(0.2)
= 8\(\pi \)(5)(0.2)
= 8\(\pi \) cm²
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16\(\pi \)-100\(\pi \) = cm².
Percentage error = relative error \(\times\)100 = \(\frac { 8.16\pi -8\pi }{ 8.16\pi } \)\(\times\)100 = 1.9607%

\(\underset { x\rightarrow 0 }{ lim } \frac { 1-cosx }{ { x }^{ 2 } } =\frac { 1-cos0 }{ 0 } =\frac { 1-1 }{ 0 } =\frac { 0 }{ 0 } \)
Indeterminate form, Applying L' Hopital rule we get,
\(\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ 2x } =\frac { 1 }{ 2 } \underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } \)\(\left[ \because ​​​​​​\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } =1 \right] \)
\(=\frac { 1 }{ 2 } \times 1\)
= \(\frac12\)

Given f(x) = x² − x, x ∈ [0, 1] 
(i) f(x) is continuous in [0, 1]
(ii) f(x) is differentiable in (0, 1)
(iii) f(0) = 0² - 0 = 0
f(1) = 12-1 = 1-1 = 0
∴ f(0) = f(1)
By Rolle's theorem, there exists C ∈ [0, 1] such that
f'(c) = 0
⇒ 2c - 1 = 0
⇒ 2c = 1
⇒ c = \(\frac12\) ∈ [0, 1]

Here ∝ = 1, β = 1 and ૪ = -2
∴ The required cubic equation is
x³-(1+1-2)x²(1-2-2)x-(1)(1)(-2) = 0
x³- 0x²-3x+2 = 0
x³-3x-2 = 0

tan^-1\((tan(\frac{3\pi}{4})\)
Observe that \(\frac{3\pi}{4}\) is not in the interval \(\left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \), the principal range of tan^-1 x.
So, we write \(\frac{3\pi}{4}=\pi-\frac{\pi}{4}\)
Now, \(tan\left( \frac { 3\pi }{ 4 } \right) =tan\left( \pi -\frac { \pi }{ 4 } \right) =-tan\frac { \pi }{ 4 } =tan\left( -\frac { \pi }{ 4 } \right) and-\frac { \pi }{ 4 } \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
Hence, \({ tan }^{ -1 }\left( tan\left( \frac { 3\pi }{ 4 } \right) \right) ={ tan }^{ -1 }\left( tan\left( -\frac { \pi }{ 4 } \right) \right) =-\frac { \pi }{ 4 } ,since-\frac { \pi }{ 4 } \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

\({ cos }^{ -1 }\left( cos\frac { \pi }{ 7 } cos\frac { \pi }{ 17 } -sin\frac { \pi }{ 17 } sin\frac { \pi }{ 17 } \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 7 } +\frac { \pi }{ 17 } \right) \right) \)
\(\left[ \therefore cosA\ cosB-sinA\ sinB=cos(A+B) \right] \)
= \({ cos }^{ -1 }\left( cos\left( \frac { 24\pi }{ 119 } \right) \right) \)    \(\left[ \therefore \frac { 24\pi }{ 119 } \varepsilon \left[ 0,\pi \right] \right] \)
= \(\frac { 24\pi }{ 119 } \) 

Let A = \(\left[ \begin{matrix} 4 & 3 \\ -3 & -1 \\ 6 & 7 \end{matrix}\begin{matrix} 1 & -2 \\ -2 & 4 \\ -1 & 2 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 4. So ρ(A)  ≤ min {3, 4} = 3.
The highest order of minors of A is 3. We search for a non-zero third-order minor of A. But we find that all of them vanish. In fact, we have
\(\left| \begin{matrix} 4 & 3 & 1 \\ -3 & -1 & -2 \\ 6 & 7 & -1 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 3 & -2 \\ -3 & -1 & 4 \\ 6 & 7 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 1 & -2 \\ -3 & -2 & 4 \\ 6 & -1 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 3 & 1 & -2 \\ -1 & -2 & 4 \\ 7 & -1 & 2 \end{matrix} \right| \) = 0.
So, ρ(A) < 3. Next, we search for a non-zero second-order minor of A.
We find that \(\left| \begin{matrix} 4 & 3 \\ -3 & -1 \end{matrix} \right| \) = -4 + 9 = 5 ≠ 0. So, ρ(A) = 2.

If the position vector of the given point is \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \), then the equation of the plane passing through a point and normal to a vector is given by \((\vec { r } -\vec { a } ).\vec { n } =0\) or \(\vec { r } .\vec { n } =\vec { a } .\vec { n } \)
Substituting \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \) in the above equation, we get
\(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )=(4\hat { i } +2\hat { j } -3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
Thus, the required vector equation of the plane is \(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )\)= 3. If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) then
we get the Cartesian equation of the plane 2x − y + z = 3.

Let sin⁻¹= \(\theta\). Then, x = sin\(\theta\) and x \(\neq\) 0, we get \(\theta \in\left[\frac{-\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]\)
Hence, \(\cos \theta \geq 0 \text { and } \)\(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\)\( \sqrt { 1-x^{ 2 } } \)
Thus, \(cot(sin^{ -1 }x)=cot\theta =\frac { \sqrt { 1-x^{ 2 } } }{ x } ,|x|\le1\ and\ x\neq0\)

|z-4| = 16
Given z = x + iy
|z - 4| = 16
⇒ |x + iy - 4| = 16
⇒ |(x- 4) + iy| = 16
⇒ \(\\ \sqrt { (x-4)^{ 2 }+{ y }^{ 2 } } \) = 16
⇒ (x - 4)² + y² = 16²
[Squaring both sides]
⇒ x²-8x + 16 + y² = 256
⇒ x²-8x + y²+ 16-256 = 0
⇒ x²-8x + y²-240 = 0 Which is the required Cartesian equation.
The locus of the point is a circle.

By definition, cot^-1x\(\in(0,\pi)\)
Therefore, cot^-1\((\frac{1}{7})\) = \(\theta\) implies \(\theta \in(0,\pi)\)
But cot^-1\((\frac{1}{7})\) = \(\theta\) implies cot \(\theta\) = \(\frac{1}{7}\) and hence tan \(\theta\) = 7 and \(\theta\) is acute.
Using tan \(\theta\) = \(\frac{7}{1}\), We construct a right triangle as shown .
Then, we have, cos \(\theta\) = \(\frac{1}{5\sqrt2}\).

Since \(2+i\sqrt { 3 } \) is a root of the polynomial equation, its conjugate 2-i\(\sqrt3\) is also a root of the equation:
∴ Sum of the roots \(=2+i\sqrt { 3 } +2-i\sqrt { 3 } =4\)
Product of the roots \(=(2+i\sqrt { 3 } )(2-i\sqrt { 3 } )\)
\(={ 2 }^{ 2 }+{ (\sqrt { 3 } })^{ 2 }\)
\([\because (a+ib)(a-ib)={ a }^{ 2 }+{ b }^{ 2 }]\)
= 4 + 3 = 7
Hence, the polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x(4)+7=0\)
\(\Rightarrow { x }^{ 2 }-4x+7=0\)

Given r = 5 cm

Since the circle touches the x axis, its centre is (0, ±5)
Equation of the circle is (x - h)² + (y - k)² = r²
⇒ (x - 0)² + (y ± 5)² = 25
\(\Rightarrow x^{2}+y^{2}+\not 25 \pm 10 y=\not 25\)
⇒ x²+y²+10y = 0

First, we find |adj (A)| = \(\left| \begin{matrix} 7 & 7 & -7 \\ -1 & 11 & 7 \\ 11 & 5 & 7 \end{matrix} \right| \) =  7(77 - 35) - 7(-7 - 77) - 7(-5 - 121) = 1764 > 0
So, we get
A = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(adj A) = \(\pm \frac { 1 }{ \sqrt { 1764 } } { \left[ \begin{matrix} +\left( 77-35 \right) & -\left( -7-77 \right) & +\left( -5-121 \right) \\ -\left( 49+35 \right) & +\left( 49+77 \right) & -\left( 35-77 \right) \\ +\left( 49+77 \right) & -\left( 49-7 \right) & +\left( 77+7 \right) \end{matrix} \right] }^{ T }\)
= \(\pm \frac { 1 }{ 42 } { \left[ \begin{matrix} 42 & 84 & -126 \\ -84 & 126 & 42 \\ 126 & -42 & 84 \end{matrix} \right] }^{ T }=\pm \left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2 \end{matrix} \right] \).

\(\frac { 1 }{ 2 } \)

If x and y are numbers such that x = y, then x² = y²
Converse statement :
If x and y are numbers such that x² = y² then x = y
Inverse statement :
If x and y are numbers such that x ≠ y then x² ≠ y²
Contrapositive statement :
If x and y are numbers such that x²≠ y² then x ≠ y

Let p: 19 is a prime number.
q: All the angles of a triangle are equal be two simple statements.
(i) 19 is not a prime number and all the angles of a triangle are equal.
~p ∧ q
(ii) 19 is a prime number or all the angles of a triangle are not equal.
p ∧ ~q
(iii) 19 is a prime number and all the angles of a triangle are equal.
p ∧ q
(iv) 19 is not a prime number.
~p.

Since × is binary operation on Z, a,b ∈ Z⇒ a × b = ab∈Z and b × b = b²∈Z ...(1)
The fact that + is binary operation on Z and (1) ⇒ 3ab = (ab + ab + ab) ∈Z and 5b²= (b²+b²+b²+b²+b²)∈Z     ...(2)
Also a∈Z and 3ab ∈Z implies a+3ab∈Z   ...(3)
(2),(3), the closure property of -on Z yield a * b = (a+3ab-5b²)∈Z. Since a * b belongs to Z, * is a binary operation on Z.

Given f(x, y) = x²y + 6x³ + 7
f(λx, λy) = λ²x²λy + 6λ³x³ + 7
= λ³ x² y + 6λ³ x³ + 7
≠ λf(x, y)
There is no common λ in this equation.
It is not homogeneous

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } \)
\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } =\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| +c\right] \)
\(=\frac { 1 }{ 4 } \left[ log\left( \frac { 4-2 }{ 4+2 } \right) -log\left( \frac { 3-2 }{ 3+2 } \right) \right] \)
\(=\frac { 1 }{ 4 } log\left[ \left( \frac { 2 }{ 6 } \right) - log \ \frac { 1 }{ 5 } \right] \\ =\frac { 1 }{ 4 } log\left( \frac { 1 }{ 3 } \times 5 \right) \)
\(=\frac { 1 }{ 4 } log\left( \frac { 5 }{ 3 } \right) \)

Given y = _ex^2_-5x+7 cos (x² - 1)
Taking differentials,
dy = (_ex^2_-5x+7 (-sin (x² - 1)(2x)) +cos (x² - 1) _ex^2_-5x+7 (2x - 5) dx
= _ex^2_-5x+7 [(2x - 5)cos (x² - 1) - 2x sin (x² - 1)]dx

The equation of the family of parabolas is given by y² ax = 4, a is an arbitrary constant. ... (1)
Differentiating both sides of (1) with respect to x , we get 2y\(\frac{dy}{dx}=4a\Rightarrow a=\frac{y}{2}\frac{dy}{dx}\) 
Substituting the value of a in (1) and simplifying, we get \(\frac{dy}{dx}=\frac{y}{2x}\) as the required differential equation.

Given v = x³
Differentiating with respect to x we get,
\(\frac { dv }{ dt } \) = 3x²
When x = 5, \(\frac { dv }{ dt } \) = 3(52) = 75
∴ \(\frac { dv }{ dt } \) when x = 5 is 75 units.

y''− 5y' + 6y = 0 ......(1)
Given y = e^mx  .....(2)
Differentiating cquation (2) w.r.t 'x', we get
\(\frac{dy}{dx} = em^x . m\)
To find the value of m:
Given y" - 5y' + 6y = 0 
e^mx . m² -5e^mx+ 6e^mx = 0
e^mx [m- 5m +6] = 0
m - 5m + 6 = 0
(m - 3) (m - 2) = 0
m = 3, 2

(i) If at any time t, The amount of Radium present is Q. The rate at which Q is decreasing \(\frac { dQ }{ dt } \).
This rate of decrease or decay is found to be proportional to Q itself. Hence we have the law, \(\frac { dQ }{ dt } = kQ\). where k is the dt constant of proportionality. Which is a required differential equation.
(ii) The rate of change of population Solution increases with respect to time t, is \(\frac { dp }{ dt } \) & the rate of population is proportional| the product of population is \(\frac { dp }{ dt } \) = kP & the also the difference between 5,00,000 & the population is \(\frac { dp }{ dt } \) = kP (5,00,000 - P) is a  required differential equation.
(iii) The rate of change of vapor pressure P with respect to time t is \(\frac { dp }{ dt } \)& the rate of dt increase vapor pressure is P at time T is proportional to the vapor pressure and also is inversely proportional to the square of the temperature is \(\frac { dp }{ dt } \)\(\infty\) P and \(\frac { dp }{ dt } ​​\infty\frac{1}{T^2}\)
Combining the two, we get
\(\frac { dp }{ dt } ​​\infty\frac{p}{T^2} \Rightarrow \frac { dp }{ dt }= k(\frac{p}{T^2})\), where 'k' is constantof proportionality
(iv) Let x be the amount. Amount varies from every year. (ie) Amount varies with respect to time t is \(\frac { dp }{ dt } \) & in addition the income from other source credited Rs. 400 continuously for every year.
\(\frac { dx }{ dt } = \frac{8}{100}\times x + 400\)
\(\Rightarrow\frac{dx}{dt} = \frac{2x}{25}+400\) is a required differential equation.

\(\sqrt { \frac { dy }{ dx } } -4\frac { dy }{ dx } -7x=0\)
The given differential equation is 
\(\sqrt { \frac { dy }{ dx } } =4\frac { dy }{ dx } +7x\)
Squaring both sides,
\(\frac { dy }{ dx } =\quad { \left( 4\frac { dy }{ dx } +7x \right) }^{ 2 }\)
\(16{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ 49 }x^{ 2 }+56x{ \left( \frac { dy }{ dx } \right) }\)
The highest derivative is 1 and its maximum power is 2.
∴ Order 1, degree 2.

\(\left| \left( \overline { 1+i } \right) \left( 2+3i \right) \left( 4i-3 \right) \right| =\left| \left( \overline { 1+i } \right) \right| \left| 2+3i \right| \left| 4i-3 \right| \) (\(\because \) |z₁z₂z₃|=|z₁|z₂||z₃|)
= |1+i| |2+3i| |-3+4i|  \(\left( \because |z|=\left| \overline { z } \right| \right) \)
= \(\left( \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } \right) \left( \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 } } \right) \left( \sqrt { \left( 3 \right) ^{ 2 }+{ 4 }^{ 2 } } \right) \).
\(=(\sqrt{2})(\sqrt{13})(\sqrt{25})=5 \sqrt{26}\)

\(\frac { 2-i }{ 1+i } +\frac { 1-2i }{ 1-i } \)
Let z = \(\frac { 2-i }{ 1+i } +\frac { 1-2i }{ 1-i } \)
= \(\frac { (2-i)(1-i)+(1-2i)(1+i) }{ (1+i)(1-i) } \)
= \(\frac { 2-2i-i+{ i }^{ 2 }+1+i-2i-2i^{ 2 } }{ { 1 }^{ 2 }-{ i }^{ 2 } } \)
= \(\\ \frac { 2-i-1+1-i+2 }{ 2 } =\frac { 4-4i }{ 2 } \)
= \(\frac { 2(2-2i) }{ 2 } \) = 2 - 2i
∴ |z| = \(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)

\({ i }^{ 59 }+\frac { 1 }{ { i }^{ 59 } } \)
i^{4 \(\times\) 14 + 3} + i^{-(4 \(\times\) 14 + 3)}
= (i⁴)¹⁴.i³ + (i⁴)^-14.i^-3
= 1.i³+1.i^-3  [∵ i⁴ = 1]
= -i + i  [∴ i³ = -i and i^-3= i]
= 0

Length of perpendicular from \(\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right) \) to the plane.
\(ax+by+cz-p=0\left| \frac { { ax }_{ 1 }+{ by }_{ 1 }+{ cz }_{ 1 }-p }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \right| \)
\(\therefore \) Length of perpendicular from (1, -2, 3) to the plane
\(x-y-z-5=0\ is\ \delta =\left| \frac { 1-(-2)+3-5 }{ \sqrt { { 1 }^{ 2 }+\left( -1 \right) ^{ 2 }+{ 1 }^{ 2 } } } \right| \)
= \(\left| \frac { 1+2+3-5 }{ \sqrt { 1+1+1 } } \right| =\left| \frac { 1 }{ \sqrt { 3 } } \right| \)
= \(\frac { 1 }{ \sqrt { 3 } } \)

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Here A = 1, B = 0, C = 1, D = 1, E = -1
Here A = C and B = 0 there i no xy term.
Hence, the given equation represent a circle.

Let A = \(\left[ \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3
The third order minor |A| = \(\left| \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right| \) = (2)(3)(1) = 6 ≠ 0. So, ρ(A) = 3.
Note that there are three non-zero rows.

Using the given value for z₁ and z₂ the value of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { 3-2i }{ 6+4 } =\frac { 3-2i }{ 6+4i } \times \frac { 6-4i }{ 6-4i } \)
= \(\frac { \left( 18-8 \right) +i\left( 12-12 \right) }{ { 6 }^{ 2 }+{ 4 }^{ 2 } } =\frac { 10-24i }{ 52 } =\frac { 10 }{ 52 } =\frac { 24i }{ 52 } \)
= \(\frac { 5 }{ 26 } -\frac { 6 }{ 13 } i\)

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

Given n = 12
P = 0.9
(i) P(X = 10) = 12C₁₀(0.9)¹⁰(1 - 0.9)²
= 12C₁₀(0.9)¹⁰ (0.1)²
(ii) P(X 2: 11) = R(X = 11) + P(X = 12)
= 12C₁₁(0.9)¹¹ (0.1) + 12C₁₂(0.9)¹²(0.1)⁶
= 12C (0.9)¹¹ (0.1) + (0.9)¹²
= 12(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)11 (1.2 + 0.9) = (0.9)11 (2.1)
(iii) P( at least 2 will not have a useful life of atleast 600 hours) = 1 - P( atleast 11 will have a useful life of atleast 600 hours)
= 1 - P(X ≥11)
= 1-2.1 (0.9)¹¹