Explore topic-wise InterviewSolutions in Current Affairs.

1. Liability : Amount payable by the business for credit purchase of goods/asset or amount payable for borrowed money is known as liability. There are two types of liabilities in the business : (1) Internal liability and (2) External liability.

2. Internal liability : Amount payable by the business to the owner of the business is internal liability. Capital is an internal liability of business.

3. External liability : The amount payable by the business to the third party for credit purchase of goods / assets / services or payable amount for borrowed money is known as external liability. Bank loan, creditors, provident fund, etc. are external liabilities.

4. Current liability : A liability, which is to be paid within one year by the business, is known as current liability. Current liability is treated as short-term liability also. Creditor is treated as current (short¬term) liability.

Amount payable by the business to the owner is called internal liability of a business.

Correct option is (a) Less than 1 year

Discount received is an income.

Depreciation is non-cash expense of the business.

(b) Current asset

Correct option is (d) Current liability

Correct option is (d) capital reduction

Correct option is (c) To know non-financial transactions

Correct option is (d) Liability and Asset

Correct option is (d) Trade mark

Correct option is (b) Capital

Correct option is (d) Depreciation

Correct option is (d) Capital loss

Correct option is (c) Revenue profit

Correct option is (b) Expense

Correct option is (a) Voucher

Correct option is (c) Liquid assets

Correct option is (c) Six

Correct option is (a) Trade discount

Correct option is (d) Historical transactions

Correct option is (c) Fictitious asset

Correct option is (c) Accounting

Correct option is (b) Deshi Nama System

Correct option is (c) Trial balance

The error which is committed on one side of the ledger account compensates for an error committed on the other side of some other leader account is called compensating error.

Prepaid expenses refer to any expense or portion of expense paid in the current accounting year but the benefit or services of which will be received in the next accounting period. They are also called as unexpired expenses.

We have,

f(x) = 16x² – 16x + 28 on R 

= 16x² – 16x + 4 + 24 

= (4x – 2)² + 24 

Now, 

(4x – 2)² ≥ 0 for all x ∈ R 

= (4x – 2)² + 24 ≥  24 for all x ∈ R 

= f(x) ≥ f (\(\frac{1}{2}\))

Thus, the minimum value of f(x) is 24 at x = (\(\frac{1}{2}\)) 

Hence, 

f(x) can be made large as possibly by giving difference value to x. 

Thus, maximum values does not exist.

Given as f(x) = – (x – 1)³(x + 1)²

f’(x) = – 3(x – 1)²(x + 1)² – 2(x – 1)³(x + 1)

= – (x – 1)²(x + 1) (3x + 3 + 2x – 2)

= – (x – 1)²(x + 1) (5x + 1)

f’’(x) = – 2(x – 1)(x + 1)(5x + 1) – (x – 1)²(5x + 1) – 5(x – 1)²(x – 1)

For the maxima and minima, f'(x) = 0

– (x – 1)²(x + 1) (5x + 1) = 0

x = 1, – 1, – 1/5

f’’ (1) = 0

x = 1 is the inflection point

f’’(– 1) = – 4× – 4 = 16 > 0

x = – 1 is the point of minima

f’’ (– 1/5) = – 5(36/25) × 4/5 = – 144/25 < 0

x = – 1/5 is the point of maxima

Thus, local max value = f (– 1/5) = 3456/3125

The local min value = f (– 1) = 0

f(x) = – (x – 1)³(x + 1)² 

f’(x) = – 3(x – 1)²(x + 1)² – 2(x – 1)³(x + 1) 

= – (x – 1)²(x + 1)(3x + 3 + 2x – 2) 

= – (x – 1)²(x + 1)(5x + 1) 

f’’(x) = – 2(x – 1)(x + 1)(5x + 1) – (x – 1)²(5x + 1) – 5(x – 1)²(x – 1) 

For maxima and minima, 

f'(x) = 0 – (x – 1)²(x + 1)(5x + 1) = 0 

x = 1, – 1, – 1/5 

Now,

f’’(1) = 0 

x = 1 is inflection point 

f’’(– 1) = – 4× – 4 = 16 > 0 

x = – 1 is point of minima 

f’’(\(-\frac{1}{5}\)) = – 5(36/25) x 4/5 

= – 144/25 < 0 

x =\(-\frac{1}{5}\) is point of maxima 

Hence, 

local max value = f(\(-\frac{1}{5}\)) = \(\frac{3456}{3125}\)

local min value = f(–1) = 0

We know that, 

– |x + 1| ≤ 0 for every x ∈ R. 

Therefore, 

g(x) = – |x + 1| + 3 ≤ 3 for every x∈ R. 

The maximum value of g is attained when |x + 1| = 0 

|x + 1| = 0 

x = – 1 

Since, 

Maximum Value of g = g( – 1) 

= – | – 1 + 1| + 3 = 3 

Hence, function g does not have minimum value.

We have 

f(x) = 2x³ + 5 on R 

Here, 

We observe that the values of f(x) increase when the values of x are increased and f(x) can be made large, 

So, 

f(x) does not have the maximum value 

Similarly, 

f(x) can be made as small as we want by giving smaller values to x. 

So, 

f(x) does not have the minimum value.

f(x) = (x – 1)(x – 2)² 

f’(x) = (x – 2)² + 2(x – 1)(x – 2) 

= (x – 2)(x – 2 + 2x – 2) 

= (x – 2)(3x – 4) 

f’’(x) = (3x – 4) + 3(x – 2) 

For maxima and minima, 

f'(x) = 0 

(x – 2)(3x – 4) = 0 

x = 2, \(\frac{4}{3}\)

Now,

f’’(2) > 0 x = 2 is point of local minima 

f’’(\(\frac{4}{3}\)) = – 2 < 0 

x = \(\frac{4}{3}\)is point of local maxima 

Hence,

local max value = f(\(\frac{4}{3}\)) = \(\frac{4}{27}\)

local min value = f(2) = 0

Given as function is f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25 in [0, 3]

By differentiating we get

f’ (x) = 12x³ – 24x² + 24 x – 48

f’ (x) = 12 (x³ – 2x² + 2x – 4)

f’ (x) = 12 (x – 2) (x² + 2)

For the local minima and local maxima we have f'(x) = 0

x = 2 or x² + 2 = 0 for which there are no real roots.

So, we consider only x = 2 ∈ [0, 3].

Now, we evaluate of f at critical points x = 2 and at the interval [0, 3]

f (2) = 3 (2)⁴ – 8 (2)³ + 12 (2)² – 48 (2) + 25

f (2) = 48 – 64 + 48 – 96 + 25 = – 39

f (0) = 3 (0)⁴ – 8 (0)³ + 12 (0)² – 48 (0) + 25 = 25

f (3) = 3 (3)⁴ – 8 (3)³ + 12 (3)² – 48 (3) + 25 = 16

Thus, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the minimum value of f on [0, 3] is – 39 occurring at x = 2

Given function is :

f(x) = (x – 1)² + 3

∴f'(x) = 2(x – 1)

Now, 

f'(x) = 0 

2(x – 1) = 0 

x = 1 

Then, 

We evaluate of f at critical points x = 1 and at the interval [– 3, 1] 

f(1) = (1 – 1)² + 3 = 3 

f(– 3) = (– 3 – 1)² + 3 = 19 

Hence, 

We can conclude that the absolute maximum value of f on [ – 3, 1] is 19 occurring at x = – 3 and the minimum value of f on [ – 3, 1] is 3 occurring at x = 1

We have, 

f(x) = \(\frac{x}{2}\)+\(\frac{2}{x}\), x > 0

Differentiate w.r.t x, we get,

f'(x) = \(\frac{1}{2}\)+\(\frac{2}{x^2}\), x > 0

For the point of local maxima and minima,

f’(x) = 0

\(\frac{1}{2}\) - \(\frac{2}{x^2}\) = 0

= x² – 4 = 0

= x = \(\sqrt4\) , \(-\sqrt4\)

= x = 2, – 2 

At x = 2 f’(x) changes from –ve to + ve 

Since, x = 2 is a point of Minima 

Hence, local min value f (2) = 2

We have, 

f(x) = x³(2x – 1)³ 

Differentiate w.r.t x, we get, 

f‘(x) = 3x²(2x – 1)³ + 3x³(2x – 1)².2 

= 3x²(2x – 1)²(2x – 1 + 2x) 

= 3x²(4x – 1) 

For the point of local maxima and minima, 

f’(x) = 0 

= 3x²(4x – 1)= 0 

= x = 0, \(\frac{1}{4}\)

At x = \(\frac{1}{4}\) 

f’(x) changes from –ve to + ve 

Since, 

x = \(\frac{1}{4}\) is a point of Minima 

Hence, local min value f (\(\frac{1}{4}\)) = \(-\frac{1}{512}\)

We have,

f(x) = xe^x 

f'(x) = e^x + xe^x 

= e^x (x + 1) 

f''(x) = e^x(x + 1) + e^x 

= e^x (x + 2) 

For maxima and minima,

f'(x) = 0 e x (x + 1) = 0 

x = – 1 

Now, 

f''(– 1) = e – 1 = 1/e > 0 

x = – 1 is point of local minima 

Hence, local min = f(– 1) = – 1/e

local maximum value = f(4) 

= 4√(32 – 4²) 

= 4√(32 – 16) 

= 4√16 

= 16 

Local minimum at x = – 4 

Local minimum value = f(– 4) 

= – 4√(32 – (– 4)²) 

= – 4√(32 – 16) 

= – 4√16 

= – 16

We have, 

f(x) = x³ – 6x² + 9x + 15 

Differentiate w.r.t x, we get, 

f ‘(x) = 3x² – 12x + 9 

= 3(x² – 4x + 3) 

= 3(x – 3)(x – 1) 

For all maxima and minima, 

f ’(x) = 0 = 3(x – 3)(x – 1) = 0 

= x = 3, 1 

At x = 1 

f ’(x) changes from –ve to + ve 

Since, 

x = – 1 is a point of Maxima 

At x = 3 

f ‘(x) changes from –ve to + ve 

Since, 

x = 3 is point of Minima. 

Hence, local max value f(1) = (1)³ – 6(1)² + 9(1) + 15 = 19 

local min value f(3) = (3)³ – 6(3)² + 9(3) + 15 = 15

f(x) = sin x + √3 cos x 

f’(x) = cos x – √3 sin x 

Now, 

f’(x) = 0 

cos x – √3 sin x = 0 

cos x = √3 sin x 

cot x = √3 

x = \(\frac{\pi}{6}\) 

Differentiate f’’(x), we get 

f’’(x) = – sin x –√3 cos x

f’’(\(\frac{\pi}{6}\)) = - sin (\(\frac{\pi}{6}\)) - √3 cos(\(\frac{\pi}{6}\)) < 0

Hence, at x = \(\frac{\pi}{6}\) is the point of local maxima.

f(x) = x³ – 6x² + 9x + 15 

∴ f'(x) = 3x² – 12x + 9 

= 3(x² – 4x + 3) 

f''(x) = 6x – 12 

= 6(x – 2) 

For maxima and minima, 

f'(x) = 0 

3(x² – 4x + 3) = 0 

So roots will be x = 3, 1 

Now, 

f''(3) = 6 > 0 

x = 3 is point of local minima 

f''(1) = – 6 < 0 

x = 1 is point of local maxima 

local max value = f(1) = 19 

local min value = f(3) = 15

We have, 

f(x) = 2 sin x – x 

Differentiate w.r.t x, we get, 

f‘(x) = 2cos x – 1 = 0 

For, the point of local maxima and minima, 

f’(x) = 0 

cos x = \(\frac{1}{2}\) =  cos\(\frac{\pi}{3}\)

 = x = \(-\frac{\pi}{3}\),\(\frac{\pi}{3}\)

 At x = \(-\frac{\pi}{3}\) 

f’(x) changes from –ve to + ve

Since,

x = \(-\frac{\pi}{3}\) is a point of Minima with value = \(-\sqrt 3\) \(-\frac{\pi}{3}\)

At x = \(\frac{\pi}{3}\) 

f‘(x) changes from –ve to + ve

Since,

 x = \(\frac{\pi}{3}\) is point of local maxima with value = \(\sqrt 3\) \(-\frac{\pi}{3}\) 

Hence, local max value f(\(\frac{\pi}{3}\)) = \(\sqrt 3\) \(-\frac{\pi}{3}\) 

local min value  f(\(-\frac{\pi}{3}\)) = \(-\sqrt 3\) \(-\frac{\pi}{3}\) 

Option : (C)

In such type of questions 

find both the maximum and minimum value to compare the options.

So first,

f(x) = x + \(\frac{1}{x}\)

So,

f'(x) = 1 - \(\frac{1}{x^2}\) 

put f’(x) = 0,

1 - \(\frac{1}{x^2}\) = 0

1 = \(\frac{1}{x^2}\) 

⇒ x = ±1 

Hence by second derivative test 

f’’(x) > 0 or f”(x) < 0 

so it’s a point of minimum or maximum respectively.

f"(x) = \(\frac{2}{x^3}\),

f”(-1) = -20 

So x = 1 is a point of minimum and 

x = -1 is a point of maximum 

f(1) = 2 is minimum value. 

f(-1) = -2 is maximum value. 

Therefore, 

maximum value < minimum value.

Option : (C)

\(\displaystyle\sum_{r=1}^{5} (x-r)^2\) 

f(x) = (x - 1)² + (x - 2)² + (x - 3)² + (x - 4)² + (x - 5)²

f’(x) = 2 [5x - 15] 

f’(x) = 0 ; x = 3 

Hence by second derivative test 

f’’(x) > 0 so it’s a point of minimum. 

f”(x) = 1 > 0 so At x = 3 minimum value.

Option : (C)

f(x) = x² + x + 1

f’(x) = 2x + 1

f’(x) = 0 

⇒ 2x + 1 = 0 

⇒ x = -\(\frac{1}{2}\) 

At extreme points, 

f(0) = 0 

f(1) = 1 + 1 + 1 > 0  

So,

x = 1 is a least value.

Option : (C)

f(x) = sin x + \(\sqrt{3}\)cos x

Differentiating f(x) with respect to x, we get

f'(x) = cos x - \(\sqrt{3}\)sin x

Differentiating f’(x) with respect to x, we get

f''(x) = - sin x - \(\sqrt{3}\)cos x

For maxima at x = c, 

f’(c) = 0 and f’’(c) < 0

f’(x) = 0

⇒ tan x = \(\frac{1}{\sqrt 3}\) 

or, x = \(\frac{\pi}{6}\) or \(\frac{7\pi}{6}\)

f"(\(\frac{\pi}{6}\)) = - 2 < 0 and

f" (\(\frac{7\pi}{6}\)) = 2 > 0

Hence,  

x = \(\frac{\pi}{6}\) is a point of maxima for f(x).

Option : (B)

f(x) = x³ – 6x² + 9x, x ∈ [0,6] 

Differentiating f(x) with respect to x, we get 

f’(x)= 3x² - 12x + 9 = 3(x - 3)(x - 1) 

For extreme points, 

f’(x) = 0 

⇒ x = 1 or x = 3 

For least and greatest value of f(x) in [0,6], we will have to check at extreme points as well as interval extremes 

f(1) = 4 

f(3) = 0 

f(0) = 0 

f(6) = 54 

Hence the least value of f(x) in [0,6] is 0 and it’s greatest value is 54.