Explore topic-wise InterviewSolutions in Current Affairs.

c) ensures gender equity

Correct option is A) got into

Correct option is A) Get with it

a. The given statement is defended; as it is difficult to measure the monetary value of the services performed by a woman (homemaker). Therefore, these activities may not be considered as an economic activity. 

b. Compensation given to the victims of a cyclone is an example of a social welfare measure taken by the government. However, it is not included in estimation of national income as it is a transfer payment which does not lead to corresponding flow of goods and services.

Correct option is C) get it

c. The strands can be separated and recombined in any number to achieve differing thickness and effects. 

a. Allocation of resources: Government seeks to allocate resources with a view to balance the goals of all sections of the society. Production of goods and services which are injurious to health may be discouraged through taxation policy. Similarly, production of goods of socially beneficial nature may be encouraged through subsidies. If the private sector does not take initiative in certain activities (Public Goods), the government may directly control them like water supply and sanitation etc. 

b. Government budget can be used as an effective tool in the process of employment generation in various ways. Investment in infrastructural projects like construction of flyovers, bridges, expansion of roads etc. creates jobs for different sections of the workforce. In rural/urban areas government can provide jobs through various employment generation schemes like MGNREGA, SJSRY, PMRY etc.

Correct option is C) don’t / don’t

Correct answer is a. Aari 

Complete the following statement. The credit provisions of SHGs are generally referred to as Microcredit programmes

Co-operative marketing has failed to gain adequate success in India because of inadequate coverage of farmer members and lack of appropriate link between marketing and processing management.

(i) The visitor said that he wanted to speak to me.

(ii) The student told the teacher that they were sorry as they were very late.

(iii) My wife said that she would answer the phone for me.

(iv) His neighbour said that I had an excellent cook.

(v) The captain said that he would have to cancel the match if it rained.

(i)unless;

(ii) otherwise;

(iii) even if;

(iv)even if

(v) if;

(vi) unless;

(vii) Unless

Correct answer is (iv) All the above

Correct answer is a. 3-5 inches 

Correct answer is (iii) 14

After the initiation of reforms, the growth rate of agriculture sector has declined to 2.3% during the 1990s which was lower than earlier years. The reasons for this decline are inadequate infrastructure, alternate employment opportunities in the industry or service sector, increase in casualisation of employment in agriculture, etc.

Correct answer is (ii) Fish

Correct answer is (iii) Milk

The growth of horticulture sector at a spectacular speed is called ‘Gloden Revolution’. The period between 1991 – 2003 is known as the period of golden revolution. During this period, the investment in horticulture became highly productive and the sector emerged as a sustainable livelihood option. Thus India could become leading producer of mangoes, bananas, coconuts etc.

Correct answer is (i) Horticulture

(i) should;

(ii) may;

(iii) shall;

(iv) may;

(v) dare not;

(vi) must;

(vii) could;

(viii) might;

(ix) could;

(x) shall; 

(xi) May;

(xii) must, ought;

(xiii) could;

(xiv) must not;

(xv) should, must.

Correct answer is (ii) horticulture

The correct option: (b) I, II and III

Nation X has registered a GDP growth rate of 100% and has performed better on the front of GDP rise as compared to Nation Y that has registered a GDP growth rate of 66.67%.

The correct option: (b) I and II

Answer is (b) 40,000

d) Assertion (A) is false but Reason (R) is true.

a) Both the statements are true

(c) holds the rank of a cabinet minister 

The correct option is: (c) 1 and 3

(d) None of the above

Muslin is a handy fabric for clothing designers and seamstresses since it can be used for drafting designs. But there’s more to muslin than just that. Designers can utilize the fabric for many other pieces other than rough designs. So if you’re wondering what is muslin, then keep reading.

The correct option is: (a) 1 and 4

∵ 1 + \(\frac x{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!}+...\) = e^x

^{∴ \(\int\frac{x^n}{e^x}dx=\int x^ne^{-x}dx\)}

 = xⁿ ∫e^-xdx - ∫(nx^n-1(-e^-x))dx (∵∫e^-x  = -e^-x)

= -xⁿe^-x + ∫nx^n-1e^-xdx

= -xⁿ e^-x - nx^n-1e^-x + ∫n(n - 1) x^n-2 e^-x dx

= -xⁿ e^-x - nx^n-1e^-x - n(n-1)x^n-2e^-x-....-n! e^-x

 = -(xⁿ + nx^n-1 + n(n-1)x^n-2+.....+n!)e^-x

Answer : 5 

3 - π  electron

1 Ring

1 C=O  Bond

Let I = \(\int\frac{x^7dx}{(1+x^2)^5}\)

Put 1 + x² = t

⇒ 2xdx = dt

\(\therefore\) I = \(\frac12\int\frac{(t-1)^3dt}{t^5}\) 

= \(\frac12\int\frac{t^3-3t^2+3t-1}{t^5}dt\) 

= \(\frac12\int[t^{-2}-3t^{-3} +3t^{-4}-t^{-5}]dt\)

= \(\frac12\left[\frac{-1}{t}+\frac32\frac1{t^2}-\frac33\frac1{t^3}+\frac14\frac1{t^4}\right]+C\)

= \(\frac12\left[\frac{-1}{1+x^2}+\frac3{2(1+x^2)^2}-\frac{1}{(1+x^2)^3}+\frac14\frac{1}{(1+x^2)^4}\right]+C\)

(i) SMTP : Simple Mail Transfer Protocol 

XML: Extensible Mark Up Language

(ii) Wired- optical fibre 

Wireless – microwave

(a) One record 

(b) tuple

Given that function f(x) is derivable at x = c

i.e. f'(x) \(=\underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}\) exists ........(1)

Now, we have to prove that f(x) is continuous function at x = c

i.e. we have to prove that \(\underset{x\rightarrow c}{lim}\) f(x) = f(c)

Now,  \(\underset{x\rightarrow c}{lim}\) f(x) - f(c) =  \(\underset{x\rightarrow c}{lim}\) (f(x) - f(c)) \((\frac{x-c}{x-c})\)

(By multiply numerator and denominator by x - c)

\(=(\underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c})\) \(\underset{x\rightarrow c}{lim}\) (x - c)  [∵ \(\underset{x\rightarrow c}{lim}\) f₁(x) f₂(x) = {\(\underset{x\rightarrow c}{lim}\) f₁(x)} {\(\underset{x\rightarrow c}{lim}\) f₂(x)}]

= f'(c) \(\underset{x\rightarrow c}{lim}\) (x - c)  (From equation (1))

= f'(c) x (c - c) = 0  (By taking limit)

Hence, (\(\underset{x\rightarrow c}{lim}\) f(x) - f(c)) = 0

⇒ \(\underset{x\rightarrow c}{lim}\) f(x) - \(\underset{x\rightarrow c}{lim}\) f(c) = 0

⇒  \(\underset{x\rightarrow c}{lim}\) f(x) - f(c) = 0

⇒  \(\underset{x\rightarrow c}{lim}\) f(x) = f(c)

Which implies that function f(x) is continuous at x = c.

Thus, we can say that if a function is derivable at a point then it is continuous at that point.

(∵ c is an orbitrary point)

the answer of the above qiestion is in the picture

Let f(x)  = \(\int\limits_{a-1}^{a+1}e^{-(x-1)^2}dx\)

∴ f'(x) = 0 gives

-2(x - 1) \(e^{-(x-1)^2}\) = 0

⇒ x - 1 = 0 (∵ e^{-(x - 1)2} \(\neq0\))

⇒ x = 1

Now, f"(x) = \(e^{-(x-1)^2}\)(-2 + 4(x - 1)²)

f"(1) = e⁰(-2 + 0) = -2 < 0

∴ x = 1 is point of maxima

Hence, maximum value of \(\int\limits_{a-1}^{a+1}e^{-(x-1)^2}dx \) is

attained at x = 1

Let I = \(\int\frac{dx}{\sqrt{x\sqrt x-x^2}}\) = \(\int\frac{dx}{x\sqrt{\frac1x-1}}\) 

Let \(\frac1{\sqrt x}-1=t^2\) ⇒ \(\frac1{\sqrt x}=1+t^2\) ⇒ \(\sqrt x = \frac{1}{1+t^2}\) 

\(\therefore\) \(-\frac12x^{-3/2}dx\) = 2tdt

⇒ \(\frac1{x\sqrt x}dx=-4t dt\)

⇒ \(\frac1xdx\) = -ut\(\sqrt x\)dt

 = \(\frac{-4t}{1+t^2}dt\)

\(\therefore\) I = \(\int\frac{-4tdt}{(1+t^2).t}\)

= -4\(\int\frac{dt}{1+t^2}\) 

= -4 tan^-1 t + c

= -4 tan^-1(\(\sqrt{\frac1{\sqrt x}-1}+c\)) (\(\therefore t = \sqrt{\frac1{\sqrt x}-1}\))

\(\therefore\) \(\int\frac{dx}{\sqrt{x\sqrt x-x^2}}\) = -4 tan^-1\(\sqrt{\frac1{\sqrt x}-1}+c\)

\(\int\frac{1+x^2+ln\,x}{x+x^2ln\,x}dx\)

Let x+ x²ln x = t

⇒ (1 + \(\frac{x^2}x\) + 2x ln x) dx = dt

⇒ (1 + x (1 + 2 ln x))dx = dt

\(\therefore\) \(\int\frac{[1+x(1+2ln\,x)]dx}{x+x^2ln\, x}\) = \(\int\frac{dt}t=ln\,t + c\) 

= ln|x + x²ln x| + c (\(\because\) t = x + x² ln x)

x - \(\frac{x^3}2+\frac{x^5}{2.4}-\frac{x^7}{2.4.6}+....\)

 = x (1 - \(\frac{x^2}2 + \frac{x^4}{2^2(1.2)}-\frac{x^6}{2^3(1.2.3)}+....\) )

 = x(1 - \(\frac{x^2}2+\cfrac{(\frac{x^2}2)^2}{2!}-\cfrac{(\frac{x^2}2)^3}{3!}+....\))

 = x e^{\(-\frac{x^2}2\)} (\(\because\) e^-x = 1 - x + \(\frac{x^2}{2!} - \frac{x^3}{3!}+...\))

\(\therefore\) \(\int\limits_0^{\infty}(x - \frac{x^3}2+\frac{x^5}{2.4}-\frac{x^7}{2.4.7}+...)\) e^{\(\frac{x^2}4\)} dx

 = \(\int\limits_0^{\infty}xe^{-\frac{x^2}2}.e^{\frac{x^2}4}dx\) 

 = \(\int\limits_0^{\infty}xe^{\frac{x^2}4-\frac{x^2}2}\)dx

 = \(\int\limits_0^{\infty}xe^{-\frac{x^2}4}\)dx

Let \(\frac{x^2}4=t\)

Then \(\frac{2xdx}4dt\)

⇒ xdx = 2dt

limit converts into from t = \(\frac{0^2}4 = 0\) t = \(\frac{\infty^2}4=\infty\)

 = \(2\int\limits_0^{\infty}e^{-t}dt\) 

 = \(-2[e^{-t}]_0^{\infty}\)

 = -2(e^-\(\infty\) - e⁰)

 = -2 (\(\frac{1}{e^{\infty}}-1\))

 = -2 (\(\frac1{\infty}\) - 1)

 = -2(0 - 1)

 = -2(0 - 1)

 = -2 (-1)

 = 2

Hence, \(\int\limits_0^{\infty}(x - \frac{x^3}2+\frac{x^5}{2.4}-\frac{x^7}{2.4.7}+...)\)e^{\(\frac{x^2}4\)} dx = 2

\(\int\limits_{-3}^0\frac{dx}{(x+7)\sqrt{x+5}}\) 

Let x + 5 = t²

∴ dx = 2tdt

x + 7 =  t² + 2

Let I = \(\int\frac{dx}{(x+7)\sqrt{x+5}}\)

Then I = \(\int\frac{2tdt}{t(t^2+2)}\) = 2\(\frac{dt}{t^2+2}\) 

 = \(\sqrt2 tan^{-1}(\sqrt{\frac{x+5}{2}})\) 

∴ \(\int\limits_{-3}^0\frac{dx}{(x+7)\sqrt{x+5}}\) = \(\left[\sqrt2tan^{-1}\sqrt{\frac{x+5}2}\right]_{-3}^0\) 

= √2 tan^-1\(\sqrt{\frac52}-\sqrt2 tan^{-1}1\) 

= √2 tan^-1\(\left(\cfrac{\sqrt{\frac52}-1}{1+\sqrt{\frac52}}\right)\)

= √2 tan^-1 \((\frac{\sqrt5-\sqrt2}{\sqrt5+\sqrt2})\)

1. Simple random sampling. 

2. Systematic sampling 

3. Stratified sampling.

A random variable which assumes all the possible values in its range is a continuous random variable. 

Ex: Height or weight of children.

Random variable is a function which assigns a real number-to every sample point in the sample space.