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Concept:

Number of ways to choose r things out of n things is given by ⁿC_r

Number of ways to arrange r things out of n things is given by ⁿP_r

Calculation:

Number of large cages = m - n

Number of big animals = p

Here, ways to arrange big animals(p) in m, n cages = ^m-nP_p 

Number of remaining cages = n - p

Now, remaining animals can be arranged as, ^n-pP_m-p

So, total number of ways = \(\{^{m - n}P_p\}\{^{m - n}P_{m - p}\}\)

Let the rational number be \(\frac ab\)

According to given conditions, we obtain 

a =  b - 3      ......(i)

And

\(\frac{a - 2}{b + 2} = \frac13\)

⇒ 3a - 3 = b + 2

⇒ 3(b - 3) - 3 = b + 2     (From(i))

⇒ 3b - 12 = b + 2

⇒ 2b = 14

⇒ b = 7

∴ a = 7 - 3 = 4

Hence, required rational number is \(\frac47\).

\(\frac{dy}{dx}-\frac{3y}{2x}=\frac{2x}{y}\)

⇒ \(\frac{dy}{dx}=\frac{2x}{y}+\frac{3}{2}\frac{y}{x}\)

⇒ \(\cfrac{dy}{dx}=\cfrac{2}{\frac yx}+\frac{3}{2}\frac{y}{x}\)   ----(i)

let y = vx

then \(\frac{dy}{dx}=V+x\frac{dv}{dx}\)

∴ v + \(x\frac{dv}{dx}=\frac{2}{v}+\frac{3}{2}v\)  ----(from i)

⇒ \(x\frac{dv}{dx}=\frac{2}{v}+\frac{3}{2}v-v=\frac{2}{v}+\frac{1}{2}v=\frac{4+v^2}{2v}\)

⇒ \(\frac{2v}{4+v^2}\,dv=\frac{1}{x}dx\)

⇒\(\int\frac{2v}{4+v^2}\,dv=\int\, \frac{1}{x}\,dx\)

⇒ log | 4 + v² | = logx + log c

⇒ log | 4 + v² | = log cx

⇒ 4+v² = cx

⇒ 4 + \((\frac{y}{x})^2=cx\)      \(∵v=\frac{y}{x}\)

⇒ 4x² + y² = cx³  

which is solution of given differential equation.

y²dx - x²dy = x dy - ydx

⇒ (x² + x)dy = (y² + y)dx

⇒ \(\frac{dy}{y(y+1)}=\frac{dx}{x(x+1)}\)

⇒ dy(1/y - 1/(y + 1)) = dx(1/x - 1/(x + 1))

⇒ log y - log (y + 1) = log x - log (x + 1) + log C

⇒ log(y/y+1) = log(x/x+1) + log C

⇒ log(y/(y+1) x (x+1)/x) = log C

⇒ \(\frac{(x+1)y}{x(y+1)}=C\)

⇒ (x + 1)y = Cx(y + 1)

which is solution of given differential equation.

Correct option is (a) 1.02m/s²

Given \(\vec V=2\hat i+t^2\hat j-9\hat k\) 

Acceleration a = \(\frac{d\vec v}{dt}\) 

a = \(\frac d{dt}\)(2 + t² - 9)

a = 2t

\(\because\) t = 0.51 sec

a = 2 x 0.51

a = 1.02 m/s²

It is given that \(\vec V = 2\hat i + t^2\hat j-9\hat k\)

Velocity in the x direction = 2 m/s ⇒ not changing with respect to time

Therefore, acceleration in x direction is zero. 

Similarly, the acceleration in z direction is also zero. 

In y direction, the velocity is t² ⇒ acceleration,

 \(a = \dfrac{dv}{dt} = \dfrac{d}{dt}{(t^2)} = 2t = 2\times 0.51 = 1.02~\text{m/s}^2\)

The resultant acceleration will be approximately 1 m/s² since the acceleration in other directions is zero. 

Therefore, option A is correct. 

x²y³ + xy = 1

Differentiating both sides w.r.t. x

3 x²y²\(\frac{dy}{dx}+2xy^3+x\frac{dy}{dx}+y=0\) 

⇒ \(\frac{dy}{dx}(3x^2y^2+x)=-(y+2xy^3)\)

⇒ \(\frac{dy}{dx} = \frac{-y+2xy^3}{x+3x^2y^2}\)

order – 3 degree – 2

\(\frac{d}{dx}(x^3cosec x)\)

 = x³\(\frac{d}{dx}cosec x+(\frac d{dx}x^3)cosec x\) 

 = - x³ cosec x cot x + 3x² cosec x

 = x² cosec x(3 - x cot x)

P(\(\frac{B}{A}\)) = \(\frac{P(B ∩ A)}{P(A)}\)

⇒ 0.4 x 0.8 = P(B∩A)

⇒ P(B∩A) = 0.32

P.I. = \(\frac1{D^3+3D+2}[2cos(2x+3)+2e^x+x^2]\) 

 = \(\frac{2cos(2x+3)}{-4D+3D+2}+\frac{2e^x}{1+3+2}\) + \(\frac12(1+\frac{3D+D^3}2)^{-1}x^2\) 

 = \(\frac{2(2+D)}{4-D^2}\)cos(2x + 3) + \(\frac{e^x}3\) + \(\frac12(1-(\frac{3D+D^3}2)+(\frac{3D+D^3}2)^2)x^2\)

= \(\frac{2(2+D)}{4-(-4)}\)cos(2x + 3) + \(\frac{e^x}3\) + \(\frac12(x^2-\frac{6x}2+\frac94\times2)\) 

= \(\frac14\) (2cos(2x + 3) - 2sin(2x + 3)) +  \(\frac{e^x}3\) + \(\frac{x^2}2-\frac{3x}2+\frac94\) 

= \(\frac14\) (cos (2x + 3) - sin (2x + 3)) + \(\frac{e^x}3\) + \(\frac{x^2}2-\frac{3x}2+\frac94\)

In Mathematics, a differential equation is an equation with one or more derivatives of a function. The derivative of the function is given by dy/dx. In other words, it is defined as the equation that contains derivatives of one or more dependent variables with respect to one or more independent variables.

Example, Consider the equation y′=3x2, which is an example of a differential equation because it includes a derivative. There is a relationship between the variables x and y:y is an unknown function of x. Furthermore, the left-hand side of the equation is the derivative of y.

\(\frac{d^2y}{dx^2}+y = cotx\)

\(\therefore\) It's complementary equation is

m² + m = 0

⇒ m(m + 1) = 0

⇒ m = 0, -1

C. F.  = Ge^0x + C₂e^-x = C₁ + C₂e^-x

Let y₁  = 1, y₂ = e^-x

w(y_1,  y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}1&e^{-x}\\0&-e^{-x}\end{vmatrix}\) = -cot x e^-x

Q(x) = cot x

w₁ = \(\begin{vmatrix}0&y_2\\Q(x)&y'_2\end{vmatrix}=\) \(\begin{vmatrix}0&e^{-x}\\cot x&-e^{-x}\end{vmatrix}=-cot x e^{-x}\)

and w₂ = \(\begin{vmatrix}y_1&0\\y'_1&Q(x)\end{vmatrix}\)\(=\begin{vmatrix}1&0\\0&cotx\end{vmatrix}\) = cot x

Now, u₁ = \(\int\frac{w_1}{w}dx=\int\frac{-cotxe^{-x}}{-e^{-x}}dx\)

 = \(\int cot xdx=log sin\,x\)

and u₂ = \(\int\frac{w_2}{w}dx = \int\frac{cotx}{-e^{-x}}dx\) 

\(=-\int e^xcot\,xdx\) 

\(\therefore\) P.I. = u₁y₁ + u₂y₂

 = log sin x - e^-x\(\int e^xcot\,x dx\) 

\(\therefore\) Complete solution is

y =  C.F. + P. I.

 = C₁ + C₂ e^-x + log sinx - e^-x\(\int e^xcot\,xdx\)

∫sinx. sin(cosx).dx 

Take cos x = t 

Diff w.r.t x, we have 

-sinx = dt/dx

-sin x.dx = dt 

sin x.dx = -dt 

⇒ ∫sin(cosx).sinx.dx 

= ∫sin(t)λ - dt = - ∫ sin t.dt 

= -x – cos t. + c =cost + c = cos(cosx) + c

B) Inductive reasoning

Manual car washers and Automatic car washers.

Correct option is  B) Rhombus

Correct option is  C) (6, 8, 10)

B) Contradiction

A) Inductive reasoning

Necleopolyhydroviruses are used for the biological control of insect pesto.

Degrees and radians are ways of measuring angles. A radian is equal to the amount an angle would have to be open to capture an arc of the circle's circumference of equal length to the circle's radius. 360° (360 degrees) is equal to 2π radians.

Rosie was the name of the first transgenic cow. Gene for human alpha-lactalbumin was introduced in its gene, which made the milk nutritionally richer

Alleles of a particular gene differ from each other on the basis of certain
changes (i.e., mutations) in the genetic material (segment of DNA or RNA).
Different alleles of a gene increases the variability or variation among the
organisms.

A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0. A proportion is an equation in which two ratios are set equal to each other. For example, if there is 1 boy and 3 girls you could write the ratio as: 1 : 3 (for every one boy there are 3 girls).

Correct option:

(B) 45º 

Given that no two persons sitting adjacent, have hats of same colour. So hats of all colours must be used. Also hats of different colours cannot be used in 1 + 1 + 3 combination because any three hats cannot be of same colour. 

Therefore only combination left is 2 + 2 + 1. 

There are a total of 3 cases of selecting hats which are 

2R + 2B + 1G or 2B + 2G + 1R or 2G + 2R + 1B

To distribute these hats first we select a person in ⁵C₁ ways and distribute that hat which is one of it's colour. Then for remaining four persons there are two ways of distributing hats of alternate colours. So total ways will be equal to 3 × ⁵C₁ × 2 = 30

Option : (A).Courier  Tertiary Sector

\(\begin{vmatrix}1& a & a^2 \\[0.3em]1 & b &b^2 \\[0.3em]1 & c & c^2\end{vmatrix}\) 

= \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & b-a &b^2-c^2 \\[0.3em]0 & c-a & c^2-a^2\end{vmatrix}\)

(By applying R₂→R₂-R₁ and R₃→R₃-R₁)

= \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & b-a &(b-a)(b+a) \\[0.3em]0 & c-a & (c-a)(c+a)\end{vmatrix}\)

= (b-a) (c-a) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 1 &c+a\end{vmatrix}\)

(By applying R₂→\(\frac{R_2}{b-a}\) and R₃→\(\frac{R_3}{c-a}\))

 = (b-a) (c-a) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 0 &c-b\end{vmatrix}\)

(By applying R₃→R₃-R₂)

= (b-a) (c-a) (c-b) \(\begin{vmatrix}1& a & a^2 \\[0.3em]0 & 1 &b+a \\[0.3em]0 & 0 &1\end{vmatrix}\)

(By applying R₃→\(\frac{R_3}{c-b}\))

= - (a-b) (c-a) x -(b-c) x 1

(By expanding determinant)

= (a-b) (b-c) (c-a).

Let a = -1/2, b = 1/2

Equivalent number to -1/2 & 1/2, while multiplying both numerator and denominator by 3,

a = -1/2  = -1/2 x 3/3 = -3/6

b = 1/2 = 1/2 x 3/3 = 3/6

Four rational numbers between -3/6 and 3/6

are -2/6, -1/6, 1/6, 2/6 or -1/3, -1/6, 1/6, 1/3

Hence, required four rational numbers are -1/3, -1/6, 1/6 and 1/3

\(SP=CP\times \frac{100+9}{100}\)

\(CP=\frac{SP\times 100}{100+g}\)

\(=\frac{3250\times100}{130}=\frac{32500}{13}=2500\)

\(\frac{10587}{250}=\frac{10587}{250}\times \frac{4}4\) = \(\frac{42348}{1000}\) = 42.348

\(\lim\limits_{x \to 0} (1+5x^2)^{(\frac{1}{x^2-3x})}\) 

(1^∞ type)

= Exp{\(\lim\limits_{x \to 0} (1+5x^2-1)\times{\frac{1}{x^2-3x}}\)}

= Exp{\(\lim\limits_{x \to 0} \frac{5x^2}{x^2-3x}\)}

= Exp{\(\lim\limits_{x \to 0} \frac{5x}{x-3}\)}

= Exp{\(\frac{5\times0}{0-3}\)}  (By taking limit)

= e⁰ = 1

∴ \(\lim\limits_{x \to 0} (1+5x^2)^{\frac{1}{x^2-3x}}\) = 1

i. 75 = 3 × 25 

= 3 × 5 × 5 

135 = 3 × 45 

= 3 × 3 × 15 

= 3 × 3 × 3 × 5 

∴ HCF of 75 and 135 = 3 × 5 

= 15 

LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3 

= 675 

ii. 114 = 2 × 57 

= 2 × 3 × 19 

76 = 2 × 38 

= 2 × 2 × 19 

∴ HCF of 114 and 76 = 2 × 19

= 38 

LCM of 114 and 76 = 2 × 19 × 3 × 2 

= 228 

iii. 153 = 3 × 51 

= 3 × 3 × 17 

187 = 11 × 17 

∴ HCF of 153 and 187 = 17 

LCM of 153 and 187 = 17 × 3 × 3 × 11 

= 1683

iv. 32 = 2 × 16 

= 2 × 2 × 8 

= 2 × 2 × 2 × 4 

= 2 × 2 × 2 × 2 × 2 

24 = 2 × 12

= 2 × 2 × 6 

= 2 × 2 × 2 × 3 

48 = 2 × 24 

= 2 × 2 × 12 

= 2 × 2 × 2 × 6 

= 2 × 2 × 2 × 2 × 3 

∴ HCF of 32, 24 and 48 = 2 × 2 × 2

= 8 

LCM of 32, 24 and 48 = 2 × 2 × 2 × 2 × 2 × 3 

= 96

Correct option:

(A) 1 

i. 5/12 + 7/16

20/48 + 21/48

= (20 + 21)/48

= 41/48

ii. 3(2/5) - 2(1/4)

= 17 - 9/4

= 68/20 - 45/20 

= (68 - 45)/20

= 23/20

iii. 12 x (-10)/3

= 4 x (-2)

= -8

iv 4(3/8) ÷ 25/18

= 7/4 x 9/5

= 63/20

i. 45 ÷ 5 + 120 x 4 - 12

= 9 + 80 -12

= 89 - 12

= 77

ii. (38 - 8) x 2 ÷ 5 + 13

= 30 x 2 ÷ 5 + 13

= 12 + 13

= 25

iii. 5/3 + 4/7 ÷ 32/21

= 5/3 + 4/7 x 21/32

= 5/3 + 4/7 x 21/32

= 5/3 + 3/8 

= 40/24 + 9/24

= 49/24

iv. 3 x {4[85 + 5 - (15 - 3)] + 2}

= 3 x {4[90 - 5] + 2}

= 3 x {4 x 85 + 2}

= 3 x (340 + 2)

= 3 x 342

= 1026

Correct option:

(A) 10 

Correct option (D) 1/3 πr²h  

Explanation:

Volume of cone =1/3 πr²h

Equation of line joining points (2, 4, 5) and (3, 5, -4) is

\(\frac{x-2}{3-2}=\frac{y-4}{5-4}=\frac{z-5}{-4-5}\)----(1)

⇒ \(\frac{x-2}1=\frac{y-4}1 = \frac{z-5}{-9}(Let)\)

\(\therefore\) Co-ordinates of arbitrary points of line are (\(\lambda+2,\lambda+4, -9\lambda+5\))----(2)

Required point is intersection point of line (1) and xz - plane.

\(\therefore\) \(\lambda+4=0\)

⇒ \(\lambda=-4\) (\(\because\) y - co-ordinate of xz-plane is 0)

Therefore co-ordinates of required point are (-4 + 2, -4 + 4, -9 x -4 + 5) (From (2))

or (-2, 0, 41).

Hence, co-ordinates of required point are (-2, 0, 41)