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Calculation:

For Statement I:

Let the CP of each of two cheapest articles = x

CP of the costliest article = x + 2

∴ x + x + x + 2 = 50

⇒ 3x = 48

⇒ x = 16

For statement II and III:

We can say that the CP of 2 articles is the same and the cost price of the costliest is 12.5% more than the CP of the cheapest.

So, combining both we can find the answer.

Statement I alone or Statement II and III together are sufficient.           

Given∶

Ratio of ages of Agam, Bony and Chetna = 1 ∶ 2 ∶ 4.

Difference in ages of Agam and Bony = 17 years.

Calculation∶

let the ages of Agam, Bony and Chetana be x, 2x and 4x respectively

According to the question,

⇒ 2x – 1x = 17

⇒ x = 17 yr

Also, the age difference between Bony and Chetna,

⇒ 4x – 2x = 2x         ---(1)

Putting the value of x = 17 in equation (1)

⇒ 2 × 17 = 34 years

\(tan \,x = \frac{sin\alpha - cos\alpha}{sin\alpha + cos\alpha}\)

\(= - \left(\frac{1- tan\,\alpha}{1+ tan\alpha}\right)\)

\(= - tan(\frac\pi4 - \alpha)\)

\(= tan(\alpha - \frac\pi4)\)

\(\therefore x = \alpha - \frac\pi4\)

\(\therefore cos \,x= cos\left(\alpha - \frac\pi4\right) = cos\left(\frac\pi4- \alpha\right)\)

\(= cos\frac\pi4 cos\,\alpha + sin \frac\pi4 sin \,\alpha\)

\(= \frac1{\sqrt2} (cos\, \alpha + sin\,\alpha)\)

⇒ \(\sqrt2 \, cos\,x = cos\,\alpha + sin\,\alpha\)

Hence Proved.

heyy! hope it helps

\(f(x) = sin x + cosx - 1\; \text{in}\; [0, \frac\pi2]\)

\(f'(x) = cos x - sin x\)

\(\therefore f'(x) = 0\) given 

\(cos x - sin x = 0\)

⇒ \(tan x = 1\)

⇒ \(x = \frac{\pi}4\)

\(f''(x) = - sin x - cos x \)

\(f''(\frac\pi4) = -\frac{2}{\sqrt2} = -\sqrt 2 < 0\)

∴ \(x = \frac\pi4\) is point of maxima of f(x).

∴ Maximum value = \(f(\frac\pi4) = sin\frac\pi4 + cos\frac\pi 4 - 1\)

\(= \frac1{\sqrt2} + \frac1{\sqrt2} - 1 \)

\(=\sqrt2 -1\)

Calculation:

30 = 2 × 3 × 5

25 = 5 × 5

The common factor is 5

∴ The HCF is 5

\(y = \frac{(x - a) (x - b)}{(x-c)(x - d)}\)

\(log \,y = log(x -a) + log(x - b) - log(x - c) - log(x - d)\)

\(\frac1y \frac{dy}{dx} = \frac1{x - a} + \frac1 {x - b} -\frac1{x - c} - \frac1{x - d}\)

\(\therefore \frac{dy}{dx} = \frac{(x - a) (x - b)}{(x-c)(x - d)}\left( \frac1{x - a} + \frac1 {x - b} -\frac1{x - c} - \frac1{x - d}\right)\)

For a given number x^y, divide y by 5 to find the remainder

Divide the power of 2 by 4 i.e 31/4 

⇒ 4 × 7 = 28

∴ Remainder 31 – 28 = 3

∴ A cycle of 3 can be used

⇒ 2³/5 = 8/5 or remainder 3      (5 × 1 = 5, 8 – 5 = 3)

∴ The remainder when 231 is divided by 5 is 3

\(log(x^2 + y^2) = 2tan^{-1} \left(\frac yx\right)\)

Differentiate both sides w.r.t. x, we get

\(\therefore \frac1{x^2 + y^2} \left(2x + 2y \frac{dy}{dx}\right) = \frac2{1 + \left(\frac yx\right)^2} \left(\frac{x \frac{dy}{dx} - y}{x^2}\right)\)

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

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

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

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

\((x + y)^2 = 2axy\)

Differentiate both sides w.r.t. x, then

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

⇒ \((2x + 2y - 2ax) \frac{dy}{dx} = 2ay - 2x - 2y\)

⇒ \(\frac{dy}{dx} = \frac{2(ay - y - x)}{2(x + y - ax)} = \frac{y(a -1)-x}{x(1 -a)}\)

\(log_2(0.0625) = x\)

⇒ \(x = log_2\left(\frac{625}{10000}\right) = log_2\left(\frac{25\times25}{100\times 100}\right)\)

\(= log_2\left(\frac1{16}\right) = - log_2 16\)

\(= - log_22^4\)

\(= -4log_22\)

\(= -4\)

The boy loves nature. He enjoys the skylark's company and likes to hear birds sing. The boy gets entertained by the company of the hunter who blows his horn from a distant field and the sweet lullabies of skylarks.

Correct option is B) check

i. (d)

ii. (c) 

iii. (a)

 iv. (b)

i.(c) 

ii. (b) 

iii. (c) 

iv. (b)

i.(a) 

ii. (b) 

iii. (b) 

iv. (d)

i – a ; 

ii – a ; 

iii – c ; 

iv – c

Given l + m + n = 0 ——— (1)

mn – 2nl – 2lm = 0 ——— (2)

From (1), l = –(m + n)

Substituting in (2),

mn + 2n(m + n) + 2m(m + n) = 0

mn + 2mn + 2n² +2m² + 2mn = 0

2m² + 5mn + 2n² = 0

(2m + n) (m + 2n) = 0

2m = –n or m = –2n

Case (i) : 2m₁ = –n₁

From l₁ = –m₁ – n₁

= –m₁ + 2m₁ = m₁

l₁/1 = m₁/1 = x₁/-2

D.Rs of the first line are 1, 1, –2

D.Cs of this line are 1/√6,1/√6, - 2/√6

Case (ii) : m₂ = –2n₂

From (1) l₂ = –m₂ – n₂ = +2n₂ – n₂ = n₂ 

l₂/1 = m₂/-2 = n₂/1

d.cs of the second line are 1, –2, 1

d.cs of this line are 1/√6, -2/√6, 1/√6

Correct option is B) would help

Both A and B are correct :

(a) Focusing human efforts for maintaining a healthy body and mind. 

(b) Capable of better withstanding stressful situations.

Correct option is B) half gallon

Correct option is B) chicken