Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is  C) 20

Correct option is  B) 3.92

Correct option is  A) 2√3

Correct option is  D) 5

Correct option is  C) 36

Correct option is   D) πr

Correct option is  D) 100

Correct option is  B) 16

Correct option is C) 25

Correct option is  D) 21

Correct option is  C) 75.46

Correct option is  A) 10,000

Correct option is  A) 17

Correct option is  C) 80

Correct option is  C) 1/2 bh

A) \(\frac{xº}{360º}\)× 2 π r 

Correct option is  A) 56

Correct option is  B) 154

Correct option is  A) 49

Correct option is D) 180°

Correct option is  B) 10:1

Correct option is  C) 4.1

Correct option is  B) 5784

Correct option is  A) 45

Correct option is  A) 64

Correct option is  C) 360°

Correct option is A) πr² 

Correct option is  D) 35

Correct option is    C) \(\frac{l+r}{2}\)

Base of the triangle shaped lawn = 12 m. 

Height = 7 m. 

Area of triangle shaped lawn 1

= \(\frac {1}2\) × b × h

= \(\frac {1}2\) × 12 × 7

= 6 × 7 = 42 Sq.m

Cost of grass for laying in lawn per 1 

Sq.m = ₹ 300 

Cost of grass for laying in lawn for 42 

Sq.m = ₹ 300 × 42 = ₹ 12,600

Given points are A (3, 4, 5) and B (-1, 3, -7). 

Let P = (x, y, z) 

Given: PA²  + PB²  = k² 

⇒ (x - 3)² + (y - 4)²  + (z - 5)²  +(x + 1)²  + (y – 3)²  + (z + 7)²  =k² 

⇒ 2x¹  – 6x + 9 + y²  -8y + 16 + z²  -10z + 25 + x²  + 2x + 1 + y²  – 6y + 9 + z²  +14z + 49 = k²

⇒ 2x²  + 2y²  + 2z²  – 4x – 14y + 4z +109 = k² 

∴ Required equations of the set of points P is, 

2x²  + 2y²  +2z² – 4x – 14y + 4z + 109 – k²  =0

Answer is (c) sec^-1 x 

(dy/dx) - y.tan x = - y sec² x

(dy/dx) + y sec²x - y tan x = 0

(dy/dx) + (sec²x - tan x)y = 0

(dy/dx) = -(sec² x - tan x)y

(dy/y) = (tan x - sec² x)dx

Integrating both sides,

∫dy/dx = ∫(tan x - sec² x) dx

log y = log|sec x| - tan x + c

Here, y = tan^-1(2x/(1 - x²)) = 2tan^-1 x

[∵ tan^-1(2x/(1 - x²)) = 2tan^-1 x]

Differentiating w.r.t. x, we have

dy/dx = 2(d(tan^-1 x)/dx) = 2 x (1/(1 + x²))

local max. value is 19 at x = 1 and local min. value is 15 at x = 3

F’(x) = 3x² - 12x+9 = 0

⇒ 3(x -3)(x -1) = 0

⇒ x = 3,1

F’’(x) = 6x -12

F’’(3) = 18 -12 = 6>0, 3 is the of local min.

F’’(1)<0, 1 is the point of local max.

F(3) = 15

F(1) = 19

local max. value is 0 at x = −2 and local min. value is −4 at x = 0

f’(x) = (x -1)2(x+2)+(x+2)² = 0

x = 0,-2

f’’(0)>0, 0 is the point of local min.

f’’(-2)<0, -2 is the point of local max.

f(0) = -4

f(-2) = 0

local max. value is 0 at each of the points x = 1 and x = −1 and local min. value is -3456/3125 at x = -1/5

F’(x) = -(x -1)³2(x+1) - 3(x -1)²(x+1)² = 0

⇒ x = 1, -1, -1/5

Since, f || (1) and f || (-1)<0, 1 and -1 are the point of local max.

F || (-1/5)>0, -1/5 is the point of local min.

F(1) = f(-1) = 0

Also, f(-1/5) = -3456/3125

local max. value is 68 at x = 1 and local min. values are−1647 at x = −6 and −316 at x = 5

F’(x) = 4x³ - 124x+120 = 0

⇒ 4(x³ - 31x+30) = 0

For x = 1, the given eq is 0

x -1 is a factor,

4(x -1)(x+6)(x -5) = 0

⇒ X = 1, -6,5

F’’(1))<0, 1 is the point of max.

F’’(-6) and f ’’(5) 0, -6 and 5 are point of min.

F(1) = 68

F(-6) = -1647

F(5) = -316

Answer is : C. (e, ∞)

⇒ f(x) = \(\frac{2x}{log\,x}\)

⇒ f'(x) = \(\frac{2.log\,x-2}{log^2x}\)

Put f’(x) = 0

We get

⇒ \(\frac{2.log\,x-2}{log^2x}\) = 0

⇒ 2.log x = 2

log x = 1

⇒ x = e

We only have one critical point

So, we can directly say x > e f(x) would be increasing

∴ f(x) will be increasing in (e, ∞)

f'(x) = 2x - 1

f'(x) > 0, ∀ x ∈ (1/2 ,1)

f'(x) < 0 , ∀ x ∈ (-1, 1/2)

. ^.. f(x) is neither increasing nor decreasing in (-1, 1)

f(x) is strictly increasing on (1/2 , 1)

and f(x) is strictly decreasing on (-1, 1/2).

It is given that

f(x) = log_a x

(a) By differentiating w.r.t x

f’(x) = 1/x log a > 0 for x ∈ (0, ∞) and a > 1

Here f’(x) > 0 for x ∈ (0, ∞) and a > 1

Therefore, f’(x) is strictly increasing function.

(b) We know that

1/x log a < 0 for x ∈ (0, ∞) where 0 < a < 1

So f’(x) < 0

Therefore, f’(x) is strictly decreasing on (0, ∞) and 0 < a < 1.

It is given that

f(x) = log_e x

By differentiating w.r.t x

f’(x) = 1/x

Here 1/x > 0 for x ∈ (0, ∞)

Similarly f’(x) > 0 for x ∈ (0, ∞) where f’(x) is strictly increasing on] 0, ∞ [.

Domain of the function is R

finding derivative i.e f’(x) = 2e^x

As we know e x is strictly increasing its domain

f’(x) > 0

hence f(x) is strictly increasing in its domain

(a) Narmada Bachao Andolan 

(b) Bhartiya Kisan Union 

(c) Mazdur Kisan Sakti Sanghatan

Given x is π/2

Value of π is 22/7

22/14 is π/2

Hence there will be no change.

Social Movement

Domain of sin^-1 x is [-1, 1].

As, tan^-1(tan x) = x

Provided x ∈ \((\frac{-\pi}2,\frac{\pi}2)\)

⇒ tan^-1(tan 1) = 1

As, tan^-1(tan x) = x

Provided x ∈ \((\frac{-\pi}2,\frac{\pi}2)\)

Here our x is 2 which does not belong to our range

We know tan (π - θ) = -tan(θ)

\(\therefore\) tan (θ - π) = tan(θ)

\(\therefore\) tan (2 - π) = tan(2)

Now 2 - π is in the given range

\(\therefore\) tan^-1(tan 2) = 2 - π

Given \(sin^{-1}\{cos(sin^{-1}\frac{1}{2})\}\) 

=\(sin^{-1}\{cos(sin^{-1}(sin\frac{\pi}{3}))\}\) 

= \(sin^{-1}\{cos(\frac{\pi}{3})\}\)

= sin^-1 (1/2)

\(=sin^{-1}(sin\frac{\pi}{3})\)

= \(\frac{\pi}{3}\)