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Correct option is A) of

Drivers must not drive over the speed limt.

Correct option is A) on

(A) (2√2, -1)

In this case there will be always loss. The selling price is immaterial  

Hence, loss % = (common loss and gain%)² /10=(16/10)%=(64/25)%=2.56% 

Let the C.P of the whole be Rs x 

C.P of 3/4th = Rs 3x/4,C.P of 1/4th =Rs x/4 

total S.P=Rs [(120%of 3x/4)+x/4]=Rs(9x/10+x/4)=Rs 23x/20 

gain=Rs(23x/20-x)=Rs 3x/20 

gain%=3x/20*1/x*100)%=15%

(D) log(x¹⁰ + 10^x) + c

Option: (B) π.

Option: (A) tan^–1x 

(D) y = x logx + x

Option: (A) is correct

Given :

The average of 4-digit numbers 1x44, x345, 3356 and 41x3 is 2767

Formula used :

Average = Total sum of all observations/ Number of observations 

Calculations :

Average = Total value of all observations/4 

2767 = Total value of all observations/4 

Total value of all observations = 2767 × 4 = 11068 

1x44 + x345 + 3356 + 41x3 = 11068 

Now we should put value of x less than 3 as at x = 3 sum of these numbers will be more than 11068 

At x = 2, values get satisfied 

Now average of 6x, 5x + 3 and 7x + 3 

Average = (6x + 5x + 3 + 7x + 3)/3 

Average = (6 × 2 + 5 × 2 + 3 + 7 × 2 + 3)/3

⇒  42/3 = 14 

∴ The average will be 14

Given:

d + 2(2b + c) = 19 

Concept used :

Divisibility rule of 8 

A number is divisible by 8 if last 3 digits are divisible by 8 

Calculations :

abcd = 1000a + 100b + 10c + d

For abcd to be divisible by 8, 100b + 10c + d must be divisible by 8 

(100b + 10c + d)/8 = [100b + 10c + 19 - 2(2b + c)]/8            [d = 19 - 2(2b + c)]

⇒ (100b + 10c - 4b - 2c + 19)/8 

⇒ (96b + 8c + 19)/8 

⇒ (96b + 8c)/8)  + 19/8 

We can see that 96b + 8c is divisible by 8 and only 19 will give remainder 

Remainder = 3 

∴ Remainder will be 3 when abcd is divided by 8

(C) P(B/A) < P(B)

Given:

First Divisor = 52

First Remainder = 49

Second Divisor = 13

Second Remainder = x 

Concept Used:

Dividend = Divisor × Quotient + Remainder

Calculation:

let the dividend be D 

Quotient be Q₁ and Q₂

According to question,

D = Q₁ × 52 + 49     

D = Q₂ × 13 + x       

Since 13 is a multiple of 52, we can assume that D is divided by 13

it also leaves the remainder 49, but 49 can be further divided by 13

It leaves the remainder 10 

This remainder 10 will be equal to x

⇒ \(√{5x-1}\) = \(√{5 \times 10 - 1}\) 

⇒ √49 = 7

∴ The value \(√{5x-1}\) is 7

(B) symmetric

Given:

There is a digit at the 10th place of 6n.

Where n is a natural number that is greater than 1.

Calculation:

Let be n= 2, ,3, 4, 5,...

⇒ 6 × 2 = 12 = 1

⇒ 6 × 3 = 18 = 1

⇒ 6 × 4 = 24 = 2

⇒ 6 × 5 = 30 = 3

⇒ 6 × 6 = 36 = 3

⇒ 6 × 7 = 42 = 4

⇒ 6 × 8 = 48 = 4

⇒ 6 × 9 = 54 = 5

∴ It can be 1, 2, 3, 4, 5.

Calculation:

We know that number of terms in (a + b)ⁿ = n + 1

Then,

⇒ ((2x + y³)⁴)⁷ = (2x + y³)²⁸

Here, n = 28

∴ Number of terms = 28 + 1 = 29

Ans: x=3

Let f(x) = \(2x^3 -11x^2+12x+10\)

\(df/dx\)= \(6x^2-22x+12\)  And so roots of \(df/dx = 0\) are 3 and 2/3 by solving the quadratic eqn.

Check \(d^2f/dx^2 = 12x-22\) at x=3 and x=2/3. At x=3 \(d^2f/dx^2\) gives positive value and thus x=3 is a point of minima. 

(B) many-one into

(A) f(A) ⊂ B 

Option: (C) 16. 

Option: (C) 2 

Option: (B) abc

correct option:

(A) 1880 

correct option:

(B) |A| 

Option: (D) CB

correct option :

(C) 512 

Answer is (b) (x²/2) + tan^-1 (y/x) = k

Option: (C) π/3

Answer is (c) 1/x

Concept:

Logarithmic formula:

• \(\rm \log_{a}b = \frac{1}{\log_{b}a}\)
• log_a M + log_a N = log_a (MN)

Where a ≠ 1, a > 0 and b ≠ 1, b > 0 and M, N are arbitrary positive numbers and p is any real number.

Calculation:

Given: \(\log_{4}2 = a\)

To find: \(\log_{2}2 \)

Using property, \(\rm \log_{a}b = \frac{1}{\log_{b}a}\)

\(\Rightarrow \log_{4}2 = \frac {1}{\log_{2}4}= \frac{1}{\log_{2}(2\times2)}\)

Using property, log_a M + log_a N = log_a (MN)

\(\rm \Rightarrow a = \frac{1}{(\log_{2}2\ +\ log_{2}2)}\)

\(\rm \Rightarrow a = \frac{1}{2\log_{2}2}\)

\(\rm \Rightarrow 2\log_{2}2 = \frac{1}{a}\)

\(\rm \therefore \log_{2}2 = \frac{1}{2a}\)

Answer is (b) x = kye^x

Answer is (a) order = 2,degree = 3

Answer is (d) 1/2

Answer is (c) 1

Answer is (d) 0

Answer is (d) 0

SO₃ is non-polar molecule but contain polar bonds.
Explanation:
The reason is difference in electronegativity. If electronegativity difference lies between 0.4 and 2 then it contains polar bonds.
As we know that SO₃ is a trigonal planar structure and thus overall polarity is zero. But the electronegativity difference between S and O is more and thus exhibits polarity in bonds. O is more electronegative than S.

Also, there are no electrons left on S atom but O atom has two lone pairs. that describes the polarity of bonds with non-polar molecule.
H₂O is a polar molecule.

HCl is polar molecule.

BBr₃ is a non polar molecule  but electronegativity difference between its atom is less than o.4 so it contains non polar bonds.

We know that the co-ordinates of the mid-point of the line segment joining points (x₁, y₁) & (x₂, y₂) is given by \((\frac{x_1 + x_2}{2}, \frac{y_1+ y_2}{2})\)

Therefore, the co-ordinates of the mid-point of the line segment joining the points A(– 5, 4) and B(7,-8) is \(\big(\frac{-5+7}{2}, \frac{4+(-8)}{2}\big)\) = \(\big( \frac{2}{2}, \frac{-4}{2} \big)\) = (1,-2)

Hence, the mid-point of given line segment is (1,-2).

(a) \(\frac{x+2}{y+2} = \frac{1}{3}\)

⇒ 3(x +2) = y + 2 

⇒ 3x + 6 = y + 2 

⇒ 3x – y + 4 = 0 

⇒ 3x – y = – 4. ... (1) 

Hence, option (ii) is correct. 

(b) \(\frac{x+3}{y+3} = \frac{2}{5}\) 

⇒ 5(x+3) = 2(y+3) 

⇒ 5x + 15 = 2y + 6 

⇒ 5x – 2y + 9 = 0 

⇒ 5x – 2y = – 9. ... (2) 

Hence, option (ii) is correct. 

(c) Now, multiplying equation (1) by 2, we get 

6x – 2y = – 8.  ... (3)

Now, subtracting equation (2) from equation (3), we get 

(6x – 2y) – (5x – 2y) = – 8 – (– 9) 

⇒ 6x – 5x – 2y + 2y = – 8 + 9 

⇒ x = 1.

Hence, the value of x is 1. 

Hence, option (i) is correct. 

(d) By putting x = 1 in equation (1), we get 3 × 1 – y = – 4 

⇒ 3 – y = – 4 

⇒ y = 3 + 4 = 7. 

Hence, the value of y is 7. 

Hence, option (iii) is correct. 

(e) Required (original) fraction is \(\frac{x}{y} = \frac{1}{7}.\)

Hence, option (iv) is correct.

First of all y = cos²x = (cos x)²

Hence

y' = 2 cos x⋅(cos x)'

= 2 cos x⋅(−sin x)

= −2 cos x⋅ sin x

= −sin 2x

e^x = tan 2y----(1)

differentiate (1) w.r.t. x we get

e^x = 2 sec²2y.\(\frac{dy}{dx}\)----(2)

differentiate (2) w.r.t. x, we get

e^x = 2sec²2y \(\frac{d^2y}{dx^2}+\) 8sec 2y. sec 2y tan 2y \((\frac{dy}{dx})^2\) 

 = 2 sec²2y.\(\frac{d^2y}{dx^2}+\) 8 sec²2y tan 2y \((\frac{e^x}{2sec^22y})^2\) (From (2))

 = 2 sec²2y \(\frac{d^2y}{dx^2}+\) 2 \(\frac{tan2y}{sec^22y}e^{2x}\)

⇒ 2 sec²2y.\(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2tan2y}{sec^22y}e^{2x}\)

⇒ 2(1 + tan²2y)\(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2.e^x.e^{2x}}{1+tan^22y}\) (\(\because\) 1 + tan²\(\theta\) = sec²\(\theta\) and From (1))

⇒ 2(1 + e^2x) \(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2e^{3x}}{1+e^{2x}}\)

⇒ \(\frac{d^2y}{dx^2}\) = \(\frac{e^{3x}+e^x-2e^{3x}}{2(1+e^{2x})^2}\)

y = x^sinx + 2020

\(\therefore\) \(\frac{dy}{dx}=\frac{d}{dx}x^{sin x}\)---(1) \((\because\frac{d}{dx}constant=0)\)

Let x^sinx = z

Then sin log x = log z (by taking log on both sides)

⇒ \(\frac{sin x}x+log x cos x=\frac1z\frac{dz}{dx}\) (on differentiating both sides w.r.t. x)

\(\therefore\) \(\frac{dz}{dx}=z(\frac{sin x}x+log x\,cosx)\) 

⇒ \(\frac{d}{dx}x\,sin x\) = x^sinx(\(\frac{sinx}x\) + cos x log x) (From (1))

We have 2^x + 2^y = 2^{x + y}

By differentiating both sides w.r.to x, we get

2^x ln2 + 2^y ln2 \(\frac{dy}{dx}=\) 2^x+y (1 + \(\frac{dy}{dx}\)) ln2

(\(\because\) \(\frac{d}{dx}a^x=a^x \) ln a)

⇒ ln 2 (2^x + 2^y\(\frac{dy}{dx}\)) = 2^x+y + 2^x+y \(\frac{dy}{dx}\)

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

⇒ 2^y(2^x - 1) \(\frac{dy}{dx}\) = 2^x(1 - 2^y)

⇒ \(\frac{dy}{dx}\) = \(\frac{2^x(1-2^y)}{2^y(2^x-1)}\) = \(\frac{2^{x-y}(1-2^y)}{2^x-1}\)

 = \(\frac{2^{x-y}(2^y-1)}{1-2^x}\)

Rolle’s Theorem is a particular case of the mean value theorem which satisfies certain conditions. At the same time, Lagrange’s mean value theorem is the mean value theorem itself or the first mean value theorem.  In general, one can understand mean as the average of the given values. But in the case of integrals, the process of finding the mean value of two different functions is different. Let us learn Rolle’s theorem and the mean value of such functions and their geometrical interpretation.

Lagrange’s Mean Value Theorem

If a function f  is defined on the closed interval [a,b] satisfying the following conditions –

i) The function f is continuous on the closed interval [a, b]

ii)The function f  is differentiable on the open interval (a, b)

Then there exists a value  x =  c in such a way that

f'(c) = [f(b) – f(a)]/(b-a)

This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem.

f = \(\frac{x+9}{x+2}\) 

f¹ = \(\cfrac{(x+2)\frac{d}{dx}(x+9)-(x+9)\frac{d}{dx}(x+2)}{(x+2)^2}\)

 = \(\frac{(x+2)-(x+9)}{(x+2)^2}\) = \(\frac{2-9}{(x+2)^2}\) = \(\frac{-7}{(x+2)^2}\)

Correct option is C) of

Correct option is B) since