Explore topic-wise InterviewSolutions in Current Affairs.

Correct option  (d) 25 kHz

Correct option (b) 70.7

Correct option (b) 1/√2

(b) – 6 dB/octave

a. True. Below 100 keV, scatter and attenuation within the patient become problematic, e.g. Tl-201. 

b. False. The pulse height analyser filters out scatter by accepting gamma rays that fall within a small energy window; therefore mono energetic gamma rays are preferable. 

c. False. High-energy gamma rays are difficult to collimate and a proportion will also pass straight through the crystal of the gamma camera without being detected. 

d. False. Low-energy gamma rays are absorbed by the patient and do not reach the gamma camera. 

e. True. Gamma rays result in the formation of the image, whereas the beta particles deposit an unnecessary dose in the patient. 

a. False. It should only accumulate in the target tissue. 

b. False. The biological half-life should be suited to the duration of the test but should not be unduly long in order to reduce the dose to the patient. 

c. True. 

d. False. The function of the pharmaceutical agent is to concentrate in the target tissue, whereas the radio nuclide will emit gamma rays. 

e. True.

(d) only (a) and (b).

(d) all of the above

(c) odd cosine terms

(b) odd cosine terms

(d) both (b) and (c).

(d) any of the above.

Concept:

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

A differential equation is said to be linear when

• Dependent variable and its derivative should have power ‘1’.
• Dependent variable and its derivatives can have the product with the independent variable.
• Dependent variable and its derivatives can’t have the product.

Calculation:

Given the differential equation is

\(\rm x{dy\over dx} + {d^2y\over dx^2} = y\)

Here x is the independent variable

y is the dependent variable

Highest derivate is \(\rm {d^2y\over dx^2}\) dydxdydx

So, the order of the given differential equation = 2

The power of the highest derivate = 1

So, the degree of the given differential equation = 1

As the multiplication of dependent variable derivative with the independent variable, so non-linear

Concept:

The solution of the differential equation can be found out by separating the variables and integrating them individually.

Calculation:

The given differential equation is 

\(\rm {dy\over dx} = e^{x-y}\)

⇒ \(\rm {dy\over dx} = {e^x\over e^y}\)

⇒ e^y dy = e^x dx

Integrating both sides 

⇒ ∫ ey dy = ∫ ex dx

⇒ ey = ex + c

⇒ ey - ex = c

Concept:

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

A differential equation is said to be linear when

• Dependent variable and its derivative should have power ‘1’.
• Dependent variable and its derivatives can have the product with the independent variable.
• Dependent variable and its derivatives can’t have the product.

Calculation:

Given differential equation is y''' - sin(y') + y = 0

∵ In the given equation the polynomial equation cannot be formed in y', so order and degree cannot be defined

Concept:

The equation of parabola having their vertices at the origin and foci on the x-axis is \(\rm y^2 = 4ax \)

Calculations:

The equation of parabola having their vertices at the origin and foci on the x-axis is

 \(\rm y^2 = 4ax \)              ....(1)

Differentiating on both side, we get

\(\rm 2y \dfrac {dy}{dx} = 4a\)

⇒ \(\rm y \dfrac {dy}{dx} = 2a\)

⇒ \(\rm a = \dfrac y2 \dfrac {dy}{dx} \)

Put this value in equation (1), we get

\(\rm y^2 = 4\dfrac y2 \dfrac {dy}{dx} x \)

⇒ \(\rm y^2 - 2xy\dfrac {dy}{dx} = 0\)

⇒ y(y - 2xy') = 0, Here y' = \(\rm \frac{dy}{dx}\)

∴ y-2xy' = 0

Hence,  the differential equation to the family of parabolas having their vertices at the origin and foci on the x-axis is y - 2xy' = 0

Concept:

Homogenous equations are of the form \(\frac{{di}}{{dt}} = \frac{{f\left( {x,y} \right)}}{{\emptyset \left( {x,y} \right)}}\)

Where \(f\left( {x,y} \right)and\;\emptyset \left( {x,y} \right)\) Homogenous functions of the same degree in x and y.

To solve a homogenous equation

1. Put i = vt, then \(\frac{{di}}{{dt}} = v + t\frac{{dv}}{{dt}}\)
2.  Separate the variable v and x and integrate.
3.  Put \(v = \frac{i}{t}\)

Calculation:

\(\frac{{di}}{{dt}} = 1 + \frac{i}{t} + {\left( {\frac{i}{t}} \right)^2}\)

Put i = vt, then \(\frac{{di}}{{dt}} = v + t\frac{{di}}{{dt}}\)

\(v + t\frac{{dv}}{{dt}} = 1 + v + {v^2}\)

\(\frac{{dv}}{{1 + {v^2}}} = \frac{{dt}}{t}\)

By integrating both sides we get

\(\smallint \frac{{dv}}{{1 + {v^2}}} = \smallint \frac{{dt}}{t}\)

\({\tan ^{ - 1}}v = \ln ct\)

Put v = 1/t in the above equation we get,

The differential equation is:

\(\left( {y - x} \right)\frac{{dy}}{{dx}} = 1\)

The above equation can be written as:

\(\frac{{dx}}{{dy}} + x = y\)

\(I.F. = {e^{\smallint dy}} = {e^y}\)

Then solution is

x.e^y = ∫ ye^y dy + c

x.e^y = (y - 1) e^y + c

so x = (y - 1) + ce^-y

Curve passing through (0, 0) so,

0 = 0 – 1 + c ⇒ c = 1

Now, x = (y - 1) + e^-y

If curve passes through (α, 1), then

\(a = \left( {1 - 1} \right) + {e^{ - 1}} = \frac{1}{e}\)

Concept:

To form the differential equation of the given equation

• Differentiate the equation, the number of times as many as the constants are there.
• Find out the constants in terms of the variables.
• Substitute the variables in the original equation.

Calculation:

Given equation is y = e4x(a + bx)

There are 2 constants a and b so differentiate 2 times

⇒ y' = 4ae^4x + be^4x + 4bxe^4x

⇒ y' = 4e^4x(a + bx) + be^4x

⇒ y' = 4y + be4x 

⇒ be^4x = y' - 4y

Differentiating one more time 

⇒ 4be4x = y'' - 4y'

⇒ 4(y' - 4y) = y'' - 4y'

⇒ y'' - 8y' + 16y = 0

Concept:

The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x.

The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(x\frac{{dy}}{{dx}} + y = {x^3}\)

\(\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}\)

It is form of \(\frac{{dy}}{{dx}} + Py = Q\)

Here, P = 1/x and Q = x²

\(I.F. = {e^{\smallint pdx}}\)

\(I.F. = {e^{lnx}} = x\)

The solution of the linear equation is given by

\(y\left( x \right) = \smallint {x^2}\left( x \right)dx + c\)

\(xy = {\frac{x}{4}^4} + c\)

Put initial condition x = 1 and y = 1 / 4 in the above equation we get ⇒ c = 0

\(xy = {\frac{x}{4}^4}\)

Concept:

If in an equation it is possible to collect all function of x and dx on one side and all function of y and dy on the other side, then the variables are said to be separable. Thus the general form of such an equation is \(f\left( y \right)dy = \emptyset \left( x \right)dx\)

Integrating both sides, we get \(\smallint f\left( y \right)dy = \smallint \emptyset \left( x \right)dx + c\) as its solution.

Calculation:

\(dy = \left( {{e^y} + {x^2}{e^y}} \right)dx\)

\(dy = {e^y}\left( {1 + {x^2}} \right)dx\)

\({e^{ - y}}dy = \left( {1 + {x^2}} \right)dx\)

Integrate both side

\(\smallint {e^{ - y}}dy = \smallint \left( {1 + {x^2}} \right)dx\)

\( - {e^{ - y}} = x + {\frac{x}{3}^3} + c_1\)

Concept:

If in an equation it is possible to collect all function of x and dx on one side and all function of y and dy on the other side, then the variables are said to be separable. Thus the general form of such an equation is \(f\left( y \right)dy = \emptyset \left( x \right)dx\)

Integrating both sides, we get \(\smallint f\left( y \right)dy = \smallint \emptyset \left( x \right)dx + c\) as its solution.

Calculation:

\(\left( {y + x\frac{{dy}}{{dx}}} \right) = a\left( {{y^2} - 2\frac{{dy}}{{dx}}} \right)\)

\(ydx + xdy = a{y^2}dx - 2ady\)

\(\left( {x + 2a} \right)dy = \left( {a{y^2} - y} \right)dx\)

\(\frac{{dy}}{{a{y^2} - y}} = \frac{{dx}}{{x + 2a}}\)

\(\left( {\frac{a}{{y - 1}} - \frac{1}{y}} \right)dy = \frac{{dx}}{{x + 2a}}\)

Integrate both side

\(\smallint \left( {\frac{a}{{y - 1}} - \frac{1}{y}} \right)dy = \smallint \frac{{dx}}{{x + 2a}}\)

\(\ln \left( {ay - 1} \right) - \ln y = \ln \left( {x + 2a} \right) + \ln c\)

\(\ln \frac{{ay - 1}}{y} = \ln c\left[ {x + 2a} \right]\)

Concept:

If in an equation it is possible to collect all function of x and dx on one side and all function of y and dy on the other side, then the variables are said to be separable. Thus the general form of such an equation is \(f\left( y \right)dy = \emptyset \left( x \right)dx\)

Integrating both sides, we get \(\smallint f\left( y \right)dy = \smallint \emptyset \left( x \right)dx + c\) as its solution.

Calculation:

\(\frac{{dx}}{{dy}} = \frac{{1 + {x^2}}}{y}\)

\(\frac{{dx}}{{1 + {x^2}}} = \frac{{dy}}{y}\)

Integrate both side

\(\int \frac{{dx}}{{1 + {x^2}}} = \int \frac{{dy}}{y}\)

\({\tan ^{ - 1}}x = \ln y + \ln c\)

\({\tan ^{ - 1}}x = \ln yc\)

\(yc = {e^{{{\tan }^{ - 1}}x}}\)

Put x = 0, y = 1

\(c = {e^{{{\tan }^{ - 1}}0}} = 1\)

Concept:

The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x.

The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(\sin x\frac{{dy}}{{dx}} + \frac{y}{{\sin x}} = x\sin x{e^{\cot x}}\)

\(\frac{{dy}}{{dx}} + \frac{y}{{{{\sin }^2}x}} = x.{e^{\cot x}}\)

It is form of \(\frac{{dy}}{{dx}} + Py = Q\)

\(I.F. = {e^{\smallint pdx}}\)

\(I.F. = {e^{\smallint cose{c^2}xdx}} = {e^{ - \cot x}}\)

The solution of the linear equation is given by

\(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c\)

\(y{e^{ - \cot x}} = \smallint x{e^{\cot x}}.{e^{ - \cot x}}dx + c\)

\(y{e^{ - \cot x}} = \smallint xdx + c\)

Concept:

For first-order differential equation, separate the variable and integrate accordingly.

Put the given condition to find out the integration constant

Calculation:

Given differential equation 

\(\rm \frac{\mathrm{d} y}{\mathrm{d x}} -4y = 0\)

⇒ \(\rm \frac{\mathrm{d} y}{y} =4dx\)

Integrating both sides

⇒ \(\rm \int\frac{\mathrm{d} y}{y} =\int4dx\)

⇒ ln y = 4x + c

⇒ y = e^{4x + c}

Now y(0) = 1

⇒ 1 = e^{0 + c}

⇒ c = 0

∴ y = e^4x

Concept:

In first order linear differential equation;

\(\rm {dy\over dx}+Py=Q\), where P and Q are function of x

Integrating factor (IF) = e^{∫ P dx}

y × (IF) = ∫ Q(IF) dx

Calculation:

Linear differential equation is of first order

\(\rm x^2 {dy\over dx}+4xy=4{\ln⁡x\over x^3}\)

⇒ \(\rm {dy\over dx}+4{y\over x}=4{\ln⁡x\over x^5}\)

IF = e^{∫ \(\rm 4\over x\) dx}

⇒ IF = e^{4 ln x}

⇒ IF = x⁴

Now, y × (IF) = ∫ Q (IF) dx

⇒ y × x⁴ = ∫ 4 \(\rm \ln x\over x^5\) × x⁴ dx

⇒ yx⁴ = ∫ 4 \(\rm \ln x\over x\) dx

Integrating,

⇒ yx⁴ = 2 (ln x)² + c (where c is integration constant)

Given y(1) = 1

⇒ (1)(1)⁴ = 2 (ln 1)² + c

⇒ c = 1

∴ yx⁴ = 2 (ln x)2 + 1

For y(e)

y(e)⁴ = 2 (ln e)² +1

⇒ y(e⁴) = 3

⇒ y = \(\boldsymbol{\rm 3\over e^4}\)

Concept:

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

A differential equation is said to be linear when

• Dependent variable and its derivative should have power ‘1’.
• Dependent variable and its derivatives can have product with independent variable.
• Dependent variable and its derivatives can’t have product.

Calculation:

Given differential equation is

\(\rm {\left(dy\over dx\right)}^2 - xy = x^3y^4\)

Here x is the independent variable

y is the dependent variable

Highest derivate is \(\rm dy\over dx\)

So, the order of the given differential equation = 1

The power of the highest derivate = 2

So, the degree of the given differential equation = 2

Median of 2, 3, 4, 5, 6 is 4.

(a) (2 units)

(d) non real

let slope of lines l₁ and l₂ are respectively m₁ and m₂.

we know that the the angle between two lines whose slopes are m₁ & m₂ is given by

tanθ = \(\frac{m_1\,-\,m_2}{1\,+\,m_1\,m_2},\).....(1) 

where θ is angle between lines l₁ and l₂.

for lines l₁ & l₂ to be perpendicular,

we have θ = 90°

\(\Rightarrow\) tanθ = tan90° = ∞ = \(\frac{1}{0}\)

therefore, equation (i), we have 

\(\frac{m_1\,-\,m_2}{1\,+\,m_1\,m_2}=\frac{1}{0}\)

\(\Rightarrow\) 1 + m₁ m₂ = 0

\(\Rightarrow\) m₁ m₂ = -1.

Hence, two straight line are perpendiculer is the product of there slopes is equal to -1.

Answer is (c) π²/192

Answer is (a) (1/3)(1 + log x)³ + c

Answer is (b) 1

Answer is (d) None of these

(A) x – α  = y – β = z – γ

Answer is (a) x + y

Answer is (a) 1

Answer is (b) √420