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u = log (tan x + tan y + tan z)

\(\cfrac{du}{dy}\) = \(\cfrac{1}{tan x + tan y + tan z}\). \(\cfrac{d}{dy}\) tan y

\(\cfrac{du}{dy}\) = \(\cfrac{sec^2y}{tanx+tany+tanz}\)

Electrons/protons

Correct Answer is: (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

Correct answer is (b) -6.8 eV

Correct answer is (b) Half life

Correct Answer is: (b) Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

(d) In magnetic and electric field both

Correct Answer is: (b) Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

Let the numbers be x and y, such that x > y 

Then, 3x - 4y = 5 ...(i) and (x + y) - 6 (x - y) = 6 

=> -5x + 7y = 6 …(ii) Solving (i) and (ii), we get: x = 59 and y = 43.

Hence, the required numbers are 59 and 43. 

Let the ten's digit be x. 

Then, unit's digit = (x + 3). 

Sum of the digits = x + (x + 3) = 2x + 3. 

Number = 10x + (x + 3) = 11x + 3. 

11x +3 / 2x + 3 = 4 / 1 

=> 11x + 3 = 4 (2x + 3) 

=> 3x = 9 

=> x = 3. 

 Hence, required number = 11x + 3 = 36.

Let the two parts be x and (50 - x). 

Then, 1 / x + 1 / (50 – x) = 1 / 12 

=> (50 – x + x) / x ( 50 – x) = 1 / 12 

 => x² – 50x + 600 = 0 

=> (x – 30) ( x – 20) = 0 

=> x = 30 or x = 20. 

So, the parts are 30 and 20.

Let the required fraction be x/y. Then, 

x+1 / y+1 = 2 / 3 

=> 3x – 2y = - 1 …(i) and x – 1 / y – 1 = 1 / 2 

2x – y = 1 …(ii) 

Solving (i) and (ii), we get : 

x = 3 , y = 5 therefore, 

Required fraction= 3 / 5.

Let the numbers be x, y and z. Then, 

x + y = 10 ...(i) y + z = 19 ...(ii) x + z = 21 …(iii) 

Adding (i) ,(ii) and (iii), we get: 

2 (x + y + z ) = 50 or (x + y + z) = 25. 

Thus, x= (25 - 19) = 6; y = (25 - 21) = 4; z = (25 - 10) = 15.

Hence, the required numbers are 6, 4 and 15.

Let the ten's digit be x. Then, unit's digit = (9 - x). 

Number = 10x + (9 - x) = 9x + 9. 

Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x. therefore, (9x + 9) - 63 = 90 - 9x 

=> 18x = 144 

=> x = 8. 

 So, ten's digit = 8 and unit's digit = 1. 

 Hence, the required number is 81. 

Let the two digit number is xy = 10x + y. 

Given that product of required number is 35. 

i.e.,

xy = 35 … (1) 

(∵Unit digit of number is y and second digit of number is x.) 

Also,

Given that if 18 added to number (10x + y), 

The digits interchange their places i.e., 

It becomes yx = 10y + x. 

i.e., 10x + y + 18 = 10y + x 

⇒ 9x − 9y + 18 = 0 

⇒ 9(x − y) = −18 

⇒ y − x = 2 … (2) 

Now,

(y + x)² = (y − x)² + 4xy = 4 + 4×35 

(using equation (1) & (2)) 

= 4 +140 = 144 

⇒ y + x = 12 … (3) 

Now,

Adding equation (2) and (3), we get 

(y – x) + (y + x) = 2 + 12 

⇒ 2y = 14 

⇒ y = 7. 

Now, 

Putting y = 7 in equation (3), we get 

7 + x = 12 

⇒ x = 12 – 7 = 5. 

∴ The required number is xy = 10x + y 

= 10 × 5 + 7 

= 57.

Calculations:

My present age = 6

Present age of my brother = 6/2 = 3

The age gap between me and my brother = 6 – 3 = 3

Then my age = 60

My brother’s age = 60 – 3 = 57 years.

∴ My brother’s age is 57 years 

Given:

Efficiency ratio of A and B = 2 ∶ 1

Condition = A takes 60 days less than B

Formula:

W = E × T

Here,

W = work

E = Efficiency

T = Time

Calculation:

We know that –

E ∝ 1/T       (w = constant)

Efficiency Ratio of A & B = 2 : 1

Days Ratio of A & B = 1 : 2

Now,

2 – 1 = 60

1 = 60

Now,

A = 1 ×60

A = 60 days

B = 2 ×60

B = 120

Now,

LCM (60, 120) = 120 unit

Efficiency of A and B together = 2 + 1 = 3

Required time = 120/3

Required time = 40 days

∴ They completed the work together in 40 days.

Given:

3 years ago age ratio of son and father = 2 ∶ 3

3 years after age ratio of son and father = 5 ∶ 7

Calculation:

Let be 3 years ago the age of son and father be 2x and 3x respectively

Now,

Total the difference of age between son and father = 3 – (– 3)

⇒ 6 years

Now,

(2x + 6)/(3x + 6) = 5/7

⇒ 14x + 42 = 15x + 30

⇒ 42 – 30 = 15x – 14x

⇒ 12 = x

Now,

The age of son 3 years ago = 2x

⇒ 2 × 12

⇒ 24 years

Now,

The present age of son = The Age of son 3years ago + 3

⇒ 24 + 3

⇒ 27

∴ The Present age of son will be 27 years

Given,

A alone can complete a work in 30 days.

B alone can complete the work in 50 days.

Formula:

Total work = Efficiency × Time

Calculation:

Total work = 150 units     [LCM of 30 and 50]

Efficiency of A = 5 units/day

Efficiency of B = 3 units/day

Total efficiency of A and B = 5 + 3 = 8 units/day

100% of the total work = 150 × (100/100) = 150 units

∴ Time taken by A and B to complete 100% of the total work = 150/8 = 18.75 days

Calculation:

Let present age of father, mother and daughter be m years, n years and p years respectively.

⇒ (p + 10) = (n + 10) - 35

⇒ p = n - 35

Then,

⇒ m + n = 96

⇒ (m - 5) = 10(p - 5) - 4

⇒ m - 5 = 10p - 54

⇒ m = 10p - 49

⇒ 96 - n = 10p - 49

⇒ n = 145 - 10p

Solving,

m = 51 years, n = 45 years and p = 10 years

∴ Present age of mother is 45 years.

Given:

Age ratio of Rina and Babita (5 years before) = 1 : 2

Ratio of their present age = 7 : 9

Calculations:

Let the present ages of Rina and Babita be 7x and 9x

So, the ratio of their ages 5 years ago:

(7x – 5) ∶ (9x – 5) = 1 ∶ 2

⇒  14x  – 10 = 9x – 5

⇒  x = 1

∴ The present age of Babita is 9 × 1 = 9 years

Correct option is

In bacterial chromosome  There is one origin of replication. 

Root formation in callus culture is induced by Auxin.

It is now possible to breed plants and animals with desired characters through  Genetic engineering.

‘itai-itai’ disease is caused due to Cadmium.

Integrated pest management (IPM) combines the use of biological, cultural and chemical practices to control insect pests in agricultural production.It seeks to use natural predators or parasites to control pests, using selective pesticides for backup only when pests are unable to be controlled by natural means. IPM should not be confused with organic practices.It does not discourage spraying chemicals; it promotes spraying with selective pesticides only when the crop needs it, which generally means that less pesticide is used.

Correct answer is (i) Rain water harvesting 

Correct answer is (i) Family

Correct answer is (i) Educational institution 

Correct answer is (ii) Family 

Answer: The substrate concentration that gives one half V_max