Oswal Specimen Papers ISC Class 12 Mathematics Solutions (Specimen Paper - 8)

Section-A

(i) (b) symmetric

Explanation :

L denotes the set of all straight lines in a plane and R be a relation defined on L. Then, by the rule l ⊥ m.
Then, m ⊥ l.

∴ R is symmetric.

(ii) (b) (2, 4)

Explanation :

$$\frac{d^{2}y}{dx^{2}} +\bigg(\frac{dy}{dx}\bigg)^{1/4} = -x^{1/5}\\\Rarr\space \bigg(\frac{dy}{dx}\bigg)^{1/4} =-\bigg(x^{1/5} + \frac{d^{2}y}{dx^{2}}\bigg)\\\Rarr\space \bigg[\bigg(\frac{dy}{ddx}\bigg)^{1/4}\bigg]^{4} =\begin{bmatrix}-\bigg(x^{1/5} + \frac{d^{2}y}{dx^{2}}\bigg)\end{bmatrix}^{4}\\\Rarr\space \bigg(\frac{dy}{dx}\bigg) =\bigg(x^{1/5} + \frac{d^{2}y}{dx^{2}}\bigg)^{4}\\\text{Here, highest order of derivative is}\space\frac{d^{2}y}{dx^{2}}\space\\\text{and its degree is 4.}$$

∴ Order = 2

Degree = 4

(iii) (a) – 1 < k < 1

Explanation :

Given, f(x) = x³ – 9kx² + 27x + 30

f ¢(x) = 3x² – 18kx + 27

= 3(x² – 6kx + 9)

for increasing function,

f '(x) > 0

∴ 3(x²– 6kx + 9) > 0

x² – 6kx + 9 > 0

We know ax²+ bx + c = 0, if a < 0 and b² – 4ac < 0

∴ 36k² – 36 < 0

k² – 1 < 0

(k + 1) (k – 1) < 0

∴ – 1 < k < 1.

(iv) (b) Minor of an element can never be equal to cofactor of element

Explanation :

∵ C_ij = (– 1)^{i + j} M_ij

When i + j = even number, then

M_ij = C_ij.

$$\text{(v)\space (a)\space}\frac{12^{x}}{\text{log 12}} + \text{C}$$

Explanation :

$$\text{Let\space}\text{I =}\int 4^{x}3^{x}dx\\\Rarr\space \text{I} =\int 12^{x}dx\\\Rarr\space \text{I}=\frac{12^{x}}{\text{log 12}} + \text{C}$$

(vi) (d) 512

Explanation :

The number of all possible matrices of order 3 × 3 is 2⁹ = 512.

$$\text{(vii)\space (a)}\frac{1}{2}$$

Explanation :

Given curves,

y = 2e^x and y = ae^–x

Let the point of intersection of both curves be (x₁, y₁).

$$m_{1} =\bigg(\frac{dy}{dx}\bigg)_{x_{1}, y_{1}} \\= 2\frac{d}{dx}e^{x} = 2e^{x_{1}}\\\because m_{2} =\bigg(\frac{dy}{dx}\bigg)_{(x_{1}, y_{1})}\\= a\frac{d}{dx}e^{\normalsize-x} =-ae^{-x_{1}}$$

Curves intersect orthogonally,

∴ m₁m₂ = – 1

$$2e^{x}_{1}.(-ae^{-x_{1}}) =-1\\ - 2a =-1\\\Rarr\space a =\frac{1}{2}.$$

(viii) (a) Both the statements are true.

Explanation :

Using properties

$$\text{tan}^{\normalsize-1}x + \text{cot}^{\normalsize-1}x =\frac{\pi}{2}\\\Rarr\space \frac{a}{x} = \frac{x}{b}\\\Rarr\space x =\sqrt{\text{ab}}$$

Statement-(1) is true.

$$\text{tan}^{\normalsize-1}\frac{m}{n} +\text{tan}^{\normalsize-1}\frac{n\bigg(1 -\frac{m}{n}\bigg)}{n\bigg(1 + \frac{m}{n}\bigg)}\\\Rarr\space \text{tan}^{\normalsize-1}\frac{m}{n} + \text{tan}^{\normalsize-1}1-\text{tan}^{\normalsize-1}\frac{m}{n}\\\Rarr\space \text{tan}^{\normalsize-1} 1 =\text{tan}^{\normalsize-1}\text{tan}\frac{\pi}{4} =\frac{\pi}{4}$$

Statement (2) is true.

(ix) (a) 0.39

Explanation :

Given, P(A) = 0.25, P(B) = 0.50 and P(A ∩B) = 0.14

We know, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.25 + 0.50 – 0.14

= 0.61

$$\text{Required probability =}\text{P(}\bar{\text{A}} ∩ \bar{\text{B}})\\ 1 - \text{P(A ∪ B)}\\ = 1 - 0.61 = 0.39.$$

(x) (a) Both assertion and reason are correct and reason is the correct explanation of assertion.

Explanation :

$$\text{Since,\space}\\\text{ maximum value of sin}^{\normalsize-1}x_{i}\space\text{is}\frac{\pi}{2}$$

$$\therefore\space \displaystyle\sum_{i=1}^{2n}\text{sin}^{\normalsize-1}\space x_{i} =n\pi\\\text{is possible, if}\\x_{1} = x_{2} = x_{3} =... x_{2n} = 1\\\therefore\space \displaystyle\sum_{i=1}^{n}\space x_{i} = 1 + 1 + 1 + ....\\\text{upto n times = n}\\\therefore\space \displaystyle\sum_{i =1}^{n}\space x_{i}^{2} = 1^{2} + 1^{2} +1^{2} +1^{2}+...\\\text{upto n times = n}\\\text{and}\space \displaystyle\sum^{n}_{i=1}\space x_{i}^{3} = 1^{3} + 1^{3} + 1^{3} +...\\\text{upto n times = n}$$

$$\text{Hence,}\space\displaystyle\sum^{n}_{i=1}x_{i} = \displaystyle\sum^{n}_{i=1}x_{i}^{2} \\=\displaystyle\sum^{n}_{i=1}x_{i}^{3} =n$$

(xi) Given,

$$\text{y = log}_{e}(x + \sqrt{x^{2} + k^{2}})$$

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

$$\frac{dy}{dx} =\frac{1}{x +\sqrt{x^{2} + k^{2}}}\\\bigg(1 + \frac{2x}{2\sqrt{x^{2} + k^{2}}}\bigg)\\=\frac{1}{x + \sqrt{x^{2} + k^{2}}}\bigg(\frac{\sqrt{x^{2} + k^{2}} + x}{\sqrt{x^{2} + k^{2}}}\bigg)\\=\frac{1}{\sqrt{x^{2} + k^{2}}}\space\textbf{Ans.}$$

(xii) Given,

$$\text{f}(\theta) =\begin{bmatrix}\text{cos}\space\theta &i\text{sin}\space\theta\\i\text{sin}\theta &\text{cos}\theta\end{bmatrix}\\\therefore\space f(2\theta) =\begin{bmatrix}\text{cos}\space2\theta &i\text{sin}2\theta\\i\text{sin}\space2\theta &\text{cos}2\theta\end{bmatrix}\\\text{Now}\space f(\theta).f(\theta)=\begin{bmatrix}\text{cos}\theta &i\text{sin\space}\theta\\i\text{sin}\theta &\text{cos}\space \theta\end{bmatrix}\\\begin{bmatrix}\text{cos}\theta &\text{i sin}\theta\\i\text{sin}\theta &\text{cos}\theta\end{bmatrix}$$

$$=\\\begin{bmatrix}\text{cos}^{2}\theta + i^{2}\text{sin}\theta &i\text{cos}\theta\text{sin}\theta + i\text{sin}\theta\text{cos}\theta\\i\text{sin}\theta\text{cos}\theta + i\text{sin}\theta\text{cos}\theta &i^{2}\text{sin}^{2}\theta +\text{cos}^{2}\theta\end{bmatrix}\\=\\\begin{bmatrix}\text{cos}^{2}\theta -\text{sin}^{2}\theta &2i\text{sin}\theta\text{cos}\theta\\ 2i\text{sin}\theta\text{cos}\theta &\text{cos}^{2}\theta -\text{sin}^{2}\theta\end{bmatrix}\\\lbrack\because i^{2}=-1\rbrack\\=\begin{bmatrix}\text{cos}2\theta &i\text{sin}2\theta\\i\text{sin}2\theta &\text{cos}2\theta\end{bmatrix}$$

Hence, f(θ).f(θ) = f(2θ).

Hence Proved.

(xiii) Given,

x = a sin³ t

$$\therefore\space \frac{dx}{dt} = 3a\text{sin}^{2}t.\text{cos t}$$

Also, y = a cos³ t

$$\therefore\space\frac{dy}{dt}=\text{3a\space\text{cos}}^{2}t(-\text{sin}t)\\\therefore\space \frac{dy}{dx} =\frac{-3a\text{cos}^{2}t.\text{sin}t}{\text{3a}\text{sin}^{2}t.\text{cos}t}\\\bigg[\because\space\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\bigg]\\\Rarr\space\frac{dy}{dx} =-\text{cot t}.$$

$$\text{{(xiv) Given,}\space}\text{P(A)}=\frac{4}{5}\\\text{and \text{P(B)}}=\frac{2}{3}\\\therefore\space\text{P(}\bar{\text{A}}) = 1-\frac{4}{5} =\frac{1}{5},\\\text{P(}\bar{\text{B}}) = 1-\frac{2}{3}=\frac{1}{3}.$$

Probability that only A hits the target

$$=\text{P(A)}.\text{P(}\bar{\text{B}})\\=\frac{4}{5}×\frac{1}{3}\\=\frac{4}{15}.\space\textbf{Ans.}$$

(xv) ΣP(x) = 1

0 + k + 3k + 4k + k + 2k² + 5k² + 3k² = 1

⇒ 10k² + 9k – 1 = 0

⇒ 10k² + 10k – k – 1 = 0

⇒ 10k(k + 1) – 1(k + 1) = 0

⇒ (k + 1)(10k – 1) = 0

If k + 1 = 0

⇒ k = – 1

Probability of an event can not be less than 0.

∴ k = – 1 is not possible

If 10k – 1 = 0

$$\therefore\space k = \frac{1}{10}\space\textbf{Ans.}$$

Now, P(x > 6) = P(x = 7) = 3k²

$$\Rarr\space \text{P(x}\gt 6) = 3\bigg(\frac{1}{10}\bigg)^{2}\\=\frac{3}{100}\\\therefore\space \text{P(x}\gt 6) = 0.03\space\textbf{Ans.}$$

Solution 2.

(i) Given :

$$\text{y =}\sqrt{\text{sin x + y}}$$

On squaring both sides, we get

⇒ y² = sin x + y

⇒ y² – y = sin x

$$\Rarr\space \frac{d}{dx}y^{2}-\frac{d}{dx}y =\frac{d}{dx}\text{sin x}\\\Rarr\space 2y\frac{dy}{dx}-\frac{dy}{dx}=\text{cos x}\\\Rarr\space \frac{dy}{dx} =\frac{\text{cos x}}{\text{2y-1}}.\\\textbf{\text{Hence Proved.}}$$

(ii) We have

$$y =\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n}\space\text{...(i)}$$

On differentiating both sides w.r.t. x, we get

$$\frac{dy}{dx} = \frac{d}{dx}\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n}\\= n\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n-1}\\\frac{d}{dx}\bigg(x + \sqrt{x^{2} + a^{2}}\bigg)\\=n\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n-1}\\\begin{bmatrix}1 + \frac{1}{2x\sqrt{x^{2} + a^{2}}}\frac{d}{dx}(x^{2} + a^{2})\end{bmatrix}$$

$$=n\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n-1}\\\begin{bmatrix}1 + \frac{1}{2\sqrt{x^{2} + a^{2}}}.2x\end{bmatrix}\\= n\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n-1}\\\begin{bmatrix}\frac{\sqrt{x^{2} + a^{2}}+ x}{\sqrt{x^{2} + a^{2}}}\end{bmatrix}\\=\frac{n\bigg[x + \sqrt{x^{2} + a^{2}}\bigg]^{n}}{\sqrt{x^{2} + a^{2}}}\\\frac{dy}{dx}=\frac{ny}{\sqrt{x^{2} + a^{2}}}\\\lbrack\text{from (i)}\rbrack\space\\\textbf{Hence Proved.}$$

Solution 3.

We observe that function f(x) is a greatest integer function denotes f(x) = [x].

f(x) = 0 for all x ∈ [0, 1)

So, f : R → R is not one-one.

Also, f : R → R does not attain non-integral values, therefore, non-integer points in R(co-domain) do not have their pre-images in the domain.

So, f : R → R is not onto.

Hence, f : R → R is neither one-one nor onto.

Solution 4.

Let A = [a_ij] be a given matrix

Since, it is skew-symmetric

∴ A’ = – A

∴ a_ji = – a_ij For all i, j

⇒ a_ii = – a_ii For all values of i

⇒ 2a_ii = 0 For all values of i

⇒ a_ii = 0 For all values of i

⇒ a₁₁ = a₂₂ = a₃₃ = ......... = a_nn = 0

Hence, all the diagonal elements of a skew-symmetric matrix are zero (as diagonal elements are : a₁₁, a₂₂, .......a_nn).

Hence Proved.

Solution 5.

$$\text{(i)\space}\int\frac{\text{sec}^{2}x}{\text{cosec}^{2}x} =\\\int\frac{1}{\text{cos}^{2}x}×\frac{\text{sin}^{2}x}{1}\text{dx}$$

$$=\int\text{tan}^{2}x\space dx\\=\int(sec^{2}x -1)dx\\=\int\text{sec}^{2}xdx -\int 1 dx$$

= tan x – x + C Ans.

$$\text{(ii)\space} \int^{5}_{4}|x-5|dx =\int^{5}_{4}-(x-5)dx\\=\int^{5}_{4}(5-x)dx\\=\bigg[5x -\frac{x^{2}}{2}\bigg]^{5}_{4}\\=\bigg[5×5-\frac{5×5}{2}\bigg]-\\\bigg[5×4-\frac{4×4}{2}\bigg]\\=\bigg[25-\frac{25}{2}\bigg] -\bigg[20-\frac{16}{2}\bigg]$$

$$=\frac{50-25}{2}-12\\=\frac{25}{2}-12\\=\frac{25-24}{2} =\frac{1}{2}\space\textbf{Ans.}$$

Solution 6.

(3xy + y² ) dx + (x² + xy) dy = 0

$$\frac{dy}{dx}=-\frac{3xy + y^{2}}{x^{2} + xy}\\\text{Putting y = vx}\\\therefore\space \frac{dy}{dx} =v +x\frac{dv}{dx}\\\Rarr\space v+x\frac{dv}{dx} =-\frac{3vx^{2} + v^{2}x^{2}}{x^{2} + vx^{2}}\\=-\frac{3v + v^{2}}{1 + v}\\\Rarr\space x\frac{dv}{dx} =-\frac{3v + v^{2}}{1 + v}-v\\=\frac{-3v-v^{2}-v -v^{2}}{1 + v}\\\Rarr\space \frac{1 + v}{2v^{2} + 4v}dv =\frac{-dx}{x}$$

$$\Rarr\space \frac{2v +2}{v^{2} + 2v}dv =\frac{-4dx}{x}\\\Rarr\space\int\frac{(2v+ 2)}{v^{2} + 2v}dv =-4\int\frac{dx}{x}\\\text{(On integrating both sides)}\\\Rarr\space \text{log}|v^{2} + 2v|=\\-4\text{log}|x| + \text{log C}\\\Rarr\space \text{log}|v^{2} + 2v| =\text{log}\frac{\text{C}}{x^{4}}\\\therefore\space v^{2} + 2v =\frac{\text{C}}{x^{4}}\\\Rarr\space \frac{y^{2}}{x^{2}} + \frac{2y}{x} =\frac{\text{C}}{X^{4}}\\\lbrack\because\space \text{y = vx}\rbrack\\\Rarr\space \frac{y^{2}}{x^{2}} + \frac{2y}{x} =\frac{\text{C}}{x^{4}}$$

[∵ y = vx]

x² y² + 2x³ y = C

When x = 1, y = 1

1 + 2 = C

∴ C = 3

Hence, particular solution is x²y² + 2x³y = 3 Ans.

Solution 7.

To prove :

$$\text{tan}^{\normalsize-1}\frac{1}{2} =\frac{\pi}{4}-\frac{1}{2}\text{cos}^{\normalsize-1}\frac{4}{5}\\\Rarr\space \frac{\pi}{4}-\text{tan}^{\normalsize-1}\frac{1}{2}\\=\frac{1}{2}\text{cos}^{\normalsize-1}\frac{4}{5}\\\Rarr\space 2\bigg(\frac{\pi}{4} -\text{tan}^{\normalsize-1}\frac{1}{2}\bigg)\\=\text{cos}^{\normalsize-1}\frac{4}{5}\\\therefore\space \text{L.H.S.} = 2\bigg(\text{tan}^{\normalsize-1}-\text{tan}^{\normalsize-1}\frac{1}{2}\bigg)\\= 2\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{1 -\frac{1}{2}}{1 + \frac{1}{2}}\end{pmatrix}\\= 2\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{1}{2}}{\frac{3}{2}}\end{pmatrix}$$

$$= 2\text{tan}^{-1}\frac{1}{3} \\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{2×\frac{1}{3}}{1 -\bigg(\frac{1}{3}\bigg)^{2}}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{2}{3}}{\frac{8}{9}}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\frac{3}{4}\\\text{Let\space} \text{tan}^{-1}\frac{3}{4} =\theta \\= \text{tan}\space\theta =\frac{3}{4}\\\therefore\space \text{cos}\theta =\frac{4}{5}$$

$$\Rarr\space \theta =\text{cos}^{\normalsize-1}\frac{4}{5}\\\Rarr\space \text{tan}^{\normalsize-1}\frac{3}{4} =\text{cos}^{\normalsize-1}\frac{4}{5}$$

∴ L.H.S. = R.H.S. Hence Proved.

Solution 8.

$$\text{Given,}\\ y =\text{sin}^{\normalsize-1}(6x\sqrt{1 - 9x^{2}})\\\text{Put,} x =\frac{\text{sin}\theta}{3}\\\Rarr\space\theta =\text{sin}^{\normalsize-1} 3x.\\ y =\text{sin}^{\normalsize-1}\\\begin{pmatrix}6.\frac{\text{sin}\theta}{3}\sqrt{1 -\frac{ 9\text{sin}^{2}\theta}{9}}\end{pmatrix}\\\Rarr\space y =\text{sin}^{\normalsize-1}(2\text{sin}\theta\sqrt{1 - \text{sin}^{2}\theta})\\\Rarr\space y =\text{sin}^{\normalsize-1}(2\text{sin}\theta\sqrt{\text{cos}^{2}\theta})\\\Rarr\space y =\text{sin}^{\normalsize-1}(2\text{sin}\theta\text{cos}\theta)\\\Rarr\space y =\text{sin}^{\normalsize-1}(\text{sin}2\theta \text{cos}\theta)\\\Rarr\space y =\text{sin}^{\normalsize-1}(\text{sin}2\theta)\\= 2\theta = 2\text{sin}^{\normalsize-1}3x.$$

$$\Rarr\space \frac{dy}{dx} =\frac{2}{\sqrt{1 -9x^{2}}}×3 \\=\frac{6}{\sqrt{1 - 9x^{2}}}\space\textbf{Ans.}$$

Solution 9.

(i) Let,

$$\text{I} =\int\frac{x^{4} + x^{2} +1}{x^{2}+ x +1}dx\\=\int\frac{x^{4} + 2x^{2}+ 1-x^{2}}{x^{2} +x +1}dx\\=\int\frac{(x^{2}+1)^{2} -(x)^{2}}{x^{2} + x + 1}dx\\=\int\frac{(x^{2} + x +1)(x^{2} +1 -x)}{(x^{2} +x + 1)}dx\\=\int(x^{2} + 1-x)dx\\=\int x^{2}dx + \int 1dx-\int x dx\\=\frac{x^{3}}{3} + \frac{x^{2}}{2} + c.\space\textbf{Ans.}$$

$$\text{(ii)\space\text{Let},\space}\text{I} =\int^{\frac{\pi}{2}}_{0}\text{log tan x dx}\\\text{...(i)}\\=\int^{\frac{\pi}{2}}_{0}\text{log tan}\bigg(\frac{\pi}{2}-x\bigg)dx\\=\int^{\frac{\pi}{2}}_{0}\text{log cot x dx}\\\text{...(ii)}$$

On adding equations (i) and (ii), we get,

$$\text{2I} = \int^{\frac{\pi}{2}}_{0}\text{log tan x dx} +\\\int^{\frac{\pi}{2}}_{0}\text{log cot x dx}\\\Rarr\space 2\text{I} =\int^{\frac{\pi}{2}}_{0}\text{log\space\text{tan x}. cot x dx}\\=\int^{\frac{\pi}{2}}_{0}\text{log 1 dx}\\=\int^{\frac{\pi}{2}}_{0} 0\space\text{dx}$$

∴ I = 0 Ans.

Solution 10.

(i) Given probabilities of solving a problem by A, B and C are

$$\text{P(A)} =\frac{1}{2},\text{P(B)} =\frac{1}{3}\\\text{and\space P(C)}=\frac{1}{4}\\\Rarr\space \text{P(}\bar{\text{A}}) = 1-\frac{1}{2},\\\text{P(}\bar{\text{B}}) = 1-\frac{1}{3}=\frac{2}{3}\\\text{and}\space \text{P(}\bar{\text{C}}) = 1 -\frac{1}{4}=\frac{3}{4}$$

(a) Probability of solving the problem by exactly two students

$$= \text{P(A)P(B)}\text{P(}\bar{\text{C}}) + \text{P(A)P(}\bar{\text{B}})\text{P(C)}+\\\text{P(}\bar{\text{A}})\text{P(B)}\text{P(C)}\\=\frac{1}{2}×\frac{1}{3}×\frac{3}{4} +\\\frac{1}{2}×\frac{2}{3}×\frac{1}{4} +\\\frac{1}{2}×\frac{1}{3}×\frac{1}{4}\\=\frac{3 + 2+1}{24} =\frac{6}{24} =\frac{1}{4}\space\textbf{Ans.}$$

(b) Probability of solving the problem by at least two of them

$$= \text{P(A)}\text{P(B)}\text{P(}\bar{\text{C}}) + \text{P(A)}\text{P(}\bar{\text{B}})\text{P(C)}+\\\text{P(}\bar{\text{A}})\text{P(B)}\text{P(C)} + \text{P(A)}\text{P(B)}\text{P(C)}\\=\begin{pmatrix}\frac{1}{2}×\frac{1}{3}×\frac{3}{4}\end{pmatrix} + \begin{pmatrix}\frac{1}{2}×\frac{2}{3}×\frac{1}{4}\end{pmatrix}+\\\begin{pmatrix}\frac{1}{2}×\frac{1}{3}×\frac{1}{4}\end{pmatrix}+ \begin{pmatrix}\frac{1}{2}×\frac{1}{3}×\frac{1}{4}\end{pmatrix}\\=\frac{3 + 2 +1}{24}=\frac{7}{24}.\space\textbf{Ans.}$$

(ii) Given Bag A : 4W and 3B, Total = 7

Bag B : 3W and 5B, Total = 8

Probabilities : 2W and 1W or 1W and 1B and 1W or 2B and 1W.

$$\therefore\space\text{The required probability =}\\\space\text{P(2 W and 1 W)} + \\\text{P(1W and 1B and 1 W)}\\ +\space\text{P}(2\text{B} \space\text{and} \space1\text{W})\\=\frac{^4\text{C}_{2}}{^7\text{C}_{2}}×\frac{^5\text{C}_{1}}{^{10}\text{C}_{1}} \\+\frac{^4\text{C}_{1}× ^3\text{C}_{1}}{^7\text{C}_{2}}×\frac{^4\text{C}_{1}}{^{10}\text{C}_{1}}+\\\frac{^3\text{C}_{2}}{^7\text{C}_{2}}×\frac{^3\text{C}_{1}}{^{10}\text{C}_{1}}\\=\frac{1}{7} + \frac{8}{35} + \frac{3}{70} =\frac{29}{70}.\space\textbf{Ans.}$$

Solution 11.

$$\text{L.H.S} =\\\begin{vmatrix}(b + c)^{2} &a^{2} &a^{2}\\b^{2} &(c + a)^{2} &b^{2}\\ c^{2} &c^{2} &(a + b)^{2}\end{vmatrix}$$

$$\text{Applying C}_{1}\xrightarrow{}\text{C}_{1} -\text{C}_{2}\\\text{and\space} \text{C}_{2}\xrightarrow{}\text{C}_{2} - \text{C}_{3},$$

$$=\\\begin{vmatrix}(b + c +a)(b + c-a) &0 &a^{2}\\(b + c + a)(b -c -a) &(c + a+ b)(c + a-b) &b^{2}\\0 &(c +a + b)(c-a-b) &(a +b)^{2}\end{vmatrix}$$

Taking (a + b + c) common from C₁ and C₂,

$$= (a +b + c)^{2}\begin{vmatrix}b +c-a &0 &a^{2}\\b-c-a &c +a-b &b^{2}\\0 &c-a-b &(a+b)^{2}\end{vmatrix}\\\text{Applying\space C}_{1}\xrightarrow{}\text{C}_{1}+\text{C}_{2},\\\text{we get}\\= (a + b +c)^{2}\begin{vmatrix}b+c-a &0 &a^{2}\\0 &c+a-b &b^{2}\\c-a-b &c-a-b &(a+b)^{2}\end{vmatrix}\\\text{Applying R}_{3}\xrightarrow{}\text{R}_{3}-(\text{R}_{1} + \text{R}_{2}),\\=(a + b + c)^{2}\begin{vmatrix}b+c-a &0 &a^{2}\\ 0 &c+a-b &b^{2}\\-2b &-2a &2ab\end{vmatrix}\\\text{Applying\space C}_{1}\xrightarrow{}a\text{C}_{1}\space\text{and}\\\text{C}_{2}\xrightarrow{}b\text{C}_{2},\space\text{we get} $$

$$=\frac{(a +b+c)^{2}}{ab}\\\begin{vmatrix}a(b+c)-a^{2} &0 &a^{2}\\0 &b(c+a)-b^{2} &b^{2}\\- 2ab &-2ab &2ab\end{vmatrix}\\\text{Applying C}_{1}\xrightarrow{}\text{C}_{1} +\text{C}_{3}\space\text{and}\\\text{C}_{2}\xrightarrow{}\text{C}_{2}+\text{C}_{3},\\\text{we get}\\=\frac{(a + b + c)^{2}}{ab}\\\begin{vmatrix}a(b+c) &a^{2} &a^{2}\\b^{2} &b(c+a) &b^{2}\\0 &0 &2ab\end{vmatrix} $$

Expanding along R₃,

$$=\frac{(a + b+c)^{2}}{ab}\\\text{2ab}\lbrack ab(b+c)(c+a)-a^{2}b^{2}\rbrack$$

= 2(a + b + c)² .ab(bc + ab + c² + ac – ab)

= 2ab (a + b + c)² (bc + c² + ac)

= 2abc (a + b + c)³

= R.H.S. Hence Proved.

Solution 12.

$$\text{(i)}\space \frac{3x-1}{(x-1)(x-2)(x-3)}\\=\frac{\text{A}}{x-1} +\frac{\text{B}}{x-2} +\frac{\text{C}}{x-3}$$

3x – 1 = A(x – 2) (x – 3) + B(x – 1) (x – 3) + C(x –1) (x –2)

If x = 1, then

2 = A(– 1) (– 2)

⇒ 2A = 2

⇒ A = 1

If x = 2, then

5 = B(1) (– 1)

⇒ B = – 5

If x = 3, then

8 = C(2) (1)

⇒ C = 4

$$\therefore\space \int\frac{3x-1}{(x-1)(x-2)(x-3)}dx =\\\int\frac{dx}{x-1} + \int\frac{\normalsize-5}{x-2}dx +\int\frac{4}{x-3}dx\\\Rarr\space\int\frac{3x-1}{(x-1)(x-2)(x-3)}dx\\=\text{log}|x-1|-5\text{log}|x-2| +\\ 4\text{log}|x-3|+ \text{C}.\space\textbf{Ans.}$$

$$\text{(ii)\space Let\space}\\\text{I} =\int\frac{e^{x}}{(e^{x}-1)^{2}(e^{x} +2)}dx\\\text{Putting\space} e^{x} = t\space\text{and}\\\text{e}^{x} dx = dt,\space\text{we get}\\\text{I}=\int\frac{dt}{(t-1)^{2}(t+2)}\\\text{Let,\space} \frac{1}{(t-1)^{2}(t+2)}\\=\frac{\text{A}}{t-1} +\frac{\text{B}}{(t-1)^{2}} + \frac{\text{C}}{t+2}$$

⇒ 1 = A(t – 1) (t + 2) + B(t + 2) + C(t – 1)²

Putting, t = 1

⇒ 1 = 3B

⇒ B = 1/3

Putting, t = – 2

⇒ 1 = 9 C

⇒ C = 1/9

Putting, t = 0

⇒ 1 = –2A + 2B + C

$$\Rarr\space 1 = -2\text{A} + \frac{2}{3} + \frac{1}{9}\\\Rarr\space \text{A} = -\frac{1}{9}\\\therefore\space \text{I} =\frac{\normalsize-1}{9}\int\frac{1}{(t-1)}dt +\frac{1}{3}\int\frac{1}{(t-1)^{2}}dt+\\\frac{1}{9}\int\frac{1}{(x+2)}dt\\\text{I} =\frac{\normalsize-1}{9}\text{log}|t-1|-\frac{1}{3.(t-1)} +\\\frac{1}{9}\text{log}|x+2| +\text{C}\\\therefore\space \text{I} =\frac{1}{9}\text{log}\begin{vmatrix}\frac{e^{x}+2}{e^{x}-1}\end{vmatrix}-\frac{1}{3(e^{x}-1)} +\text{C}\\\textbf{Ans.}$$

Solution 13.

(i) Given,

$$x\sqrt{1 + y}+y\sqrt{1 + x} = 0\\\Rarr\space x\sqrt{1 + y} =-y\sqrt{1 + x}$$

On squaring both sides, we get

⇒ x² (1 + y) = y² (1 + x)

⇒ x² + x²y = y² + y² x

⇒ x² + x² y - y² - y² x = 0

⇒ x² - y² + x² y - y² x = 0

⇒ (x - y) (x + y) + xy (x - y) = 0

⇒ (x - y) [x + y + xy] = 0

⇒ x + y + xy = 0 [∵ x ≠ y]

$$\Rarr\space y= -\frac{x}{1 +x}\space\text{...(i)}$$

On differentiating equation (i), w.r.t. x, we get

$$\frac{dy}{dx} =\\\frac{(1+ x)\frac{d(\normalsize-x)}{dx} -(-x)\frac{d}{dx}(1 + x)}{(1 + x^{2})}\\=\frac{(1 + x)(\normalsize-1) + x(1)}{(1 + x)^{2}}\\=\frac{-1-x + x}{(1 + x^{2})}\\=\frac{\normalsize-1}{(1 + x)^{2}}$$

∴ L.H.S = R.H.S. Hence Proved.

(ii) Given,

x = 3 cos t - cos³ t

$$\Rarr\space \frac{dx}{dt} = -3\text{sin t} + 3\text{cos}^{2}t\text{sin t}\\\text{and}\\\text{y = 3 sint - sin}^{3}t\\\Rarr\space \frac{dy}{dt} =\text{3 cos t - 3 sin}^{2}t \text{cos}\space t$$

Slope of the tangent,

$$\frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3\text{cos}t - 3\text{sin}^{2}t\text{cos t}}{-3\text{sin t} + 3\text{cos}^{2}t\text{sin t}}\\=\frac{3\text{cos t}[\text{cos}^{2}t]}{-3\text{sin t}[\text{sin}^{2}t]}\\\Rarr\space \frac{dy}{dx}=\frac{-\text{cos}^{3}t}{\text{sin}^{2}t}\\\Rarr\space\frac{dy}{dx}=\frac{-\text{cos}^{3}t}{\text{sin}^{3}t}\\\therefore\space \text{Slope of the normal =}\\\frac{-dx}{dy}=\frac{\text{sin}^{3}t}{\text{cos}^{3}t}$$

The equation of the normal is given by

$$\frac{y-(3\text{sin t - sin}^{3}t)}{x-(3\text{cos t - cos}^{3}t)}=\frac{\text{sin}^{3}t}{\text{cos}^{3}t}\\\Rarr\space \text{y cos}^{3}t- 3\text{sin t cos}^{3}t +\text{sin}^{3}t\text{cos}^{3}t\\= x\space \text{sin}^{3}t - 3\text{cos}\space t\text{sin}^{3}t +\text{sin}^{3}t\text{cos}^{3}t\\\Rarr\space \text{y cos}^{3}t - x\text{sin}^{3}t =\\= 3(\text{sin t cos}^{3}t -\text{cos}\space t\space\text{sin}^{3}t)\\\Rarr\space y\space \text{cos}^{3}t- x\text{sin}^{3}t =\\\text{3 sin t cos t}(\text{cos}^{2}t -\text{sin}^{2}t)\\\Rarr\space y\text{cos}^{3}t - x\text{sin}^{3}t\\=\frac{3×2}{2}\text{sin t cos t cos 2t}\\\Rarr\space \text{y cos}^{3}t - x\text{sin}^{3}t\\=\frac{3}{2}\text{sin 2t cos 2t}$$

$$\Rarr\space \text{y cos}^{3}t - x\text{sin}^{3}t \\=\frac{3×2}{2×2}\text{sin 2t cos 2t}\\\Rarr\space \text{y cos}^{3}t - x\text{sin}^{3}t\\=\frac{3}{4}\text{sin 4t}\\\Rarr\space 4(y \text{cos}^{3}t - x\text{sin}^{3}t) \\= 3\space\text{sin 4t}\\\textbf{Hence Proved.}$$

Solution 14.

$$\text{Given,\space} \text{P(A)} =\frac{4}{5},\text{P(B)}=\frac{3}{4},\\\text{P(C)}=\frac{2}{3}\\\therefore\space \text{P(}\bar{\text{A}}) =\frac{1}{5},\text{P(}\bar{\text{B}})=\frac{1}{4},\\\text{P(}\bar{\text{C}})=\frac{1}{3}$$

(i) P (exactly two persons hit the target)

$$=\text{P(A}∩ \text{B} ∩\bar{\text{C}}) + \text{P(A} ∩ \bar{\text{B}}∩\text{C})+\\\text{P(}\bar{\text{A}}∩\text{B}∩\text{C})\\=\text{P(A)}.\text{P(B)}.\text{P(}\bar{\text{C}}) +\text{P(A)}.\text{P(}\bar{\text{B}}).\text{P(C)}+\\\text{P(}\bar{\text{A}}).\text{P(B)}.\text{P(C)}\\=\frac{4}{5}.\frac{3}{4}\bigg(1-\frac{2}{5}\bigg) + \frac{4}{5}\bigg(1 -\frac{3}{4}\bigg).\frac{2}{3}+\\\bigg(1 -\frac{4}{5}\bigg).\frac{3}{4}.\frac{2}{3}\\=\frac{4}{5}.\frac{3}{4}.\frac{1}{3} + \frac{4}{5}.\frac{1}{4}.\frac{2}{3}\\+\frac{1}{5}.\frac{3}{4}.\frac{2}{3}\\=\frac{12 + 8+ 6}{60}=\frac{26}{60}=\frac{13}{30}.\\\textbf{Ans.}$$

(ii) P (at least two persons hit the target)

= P (all three persons hit the target) + P (exactly two persons hit the target)

$$=\text{P(A∩B∩C)} + \frac{13}{30}\\\lbrack\text{Using (i)}\rbrack\\=\text{P(A)P(B)P(C)}+\frac{13}{30}\\=\frac{4}{5}.\frac{3}{4}.\frac{2}{3}+\frac{13}{30}\\=\frac{2}{5} + \frac{13}{30}\\=\frac{12 + 13}{30}=\frac{25}{30}=\frac{5}{6}.\\\textbf{Ans.}$$

(iii) P (None hit the target) =

$$\text{P}(\bar{\text{A}}∩\bar{\text{B}}∩\bar{\text{C}})\\=\bigg(1-\frac{4}{5}\bigg).\bigg(1-\frac{3}{4}\bigg)\\.\bigg(1-\frac{2}{3}\bigg)\\=\frac{1}{5}.\frac{1}{4}.\frac{1}{3}=\frac{1}{60}\space\textbf{Ans.}$$

Section-B

Solution 15.

(i) (c) 14

$$\text{Let \space}\vec{a} = 3\hat{i} +\lambda\hat{j} +\hat{k}\\\vec{b} = 2\hat{i}-\hat{j} + 8\hat{k}\\\text{Given,}\\\vec{a}\perp\vec{b}\\\therefore \space\vec{a}.\vec{b} = 0$$

$$(3\hat{i} +\lambda\hat{j} + \hat{k}).(2\hat{i}-\hat{j} + 8\hat{k}) = 0\\6-\lambda +8 -0\\\lambda = 14.$$

(ii) (b) cos^–1 (1/3)

Take one vertex, say O, of the cube as origin and the three mutually perpendicular edges OA, OB and OC as coordinate axes.

Length of the side of the cube = a unit (by figure)

The cube has four diagonals OP, MC, AN and BL.

The direction numbers of OP are < a – 0, a – 0, a – 0 > i.e., < a, a, a >

and hence direction cosines are

$$\begin{pmatrix}\frac{a}{\sqrt{a^{2} + a^{2} +a^{2}}},\frac{a}{\sqrt{a^{2} + a^{2} + a^{2}}},\\\frac{a}{\sqrt{a^{2} + a^{2} + a^{2}}}\end{pmatrix}\\\text{i.e.,}\space\begin{pmatrix}\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\end{pmatrix}$$

Direction numbers of MC are a, a, – a and hence its directions cosines are

$$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{\normalsize-1}{\sqrt{3}}$$

Let θ be the (acute) angle between OP and MC.

Then

$$\text{cos}\space\theta =\begin{vmatrix}\frac{1}{\sqrt{3}}×\frac{1}{\sqrt{3}} +\\\frac{1}{\sqrt{3}}×\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}×\frac{\normalsize-1}{\sqrt{3}}\end{vmatrix}\\\text{cos}\space \theta=\frac{1}{3}\\\theta=\text{cos}^{\normalsize-1}\bigg(\frac{1}{3}\bigg)\space\textbf{Ans.}$$

$$\text{(iii)\space (d)} 5\sqrt{2}$$

$$\text{A(0,0,0)},\text{B}(10\space \hat{i}),\text{H}(h\hat{j} +10 \hat{k}),\\\text{G}(10\hat{i} + h\hat{j} +10\hat{k})\\\vec{\text{AG}} = 10\hat{i} + h\hat{j} + 10\hat{k},\\\vec{\text{BH}} =-10\hat{i} + h\hat{j} +10 \hat{k}\\\text{cos}\theta=\frac{\vec{\text{AG}}\vec{\text{BH}}}{|\vec{\text{AG}}||\vec{\text{BH}}|}\\\Rarr\space \frac{1}{5} =\frac{h^{2}}{h^{2} + 200}\\\Rarr\space 5h^{2} =h^{2} + 200\\\Rarr\space 4h^{2} = 200\\\Rarr\space h =5\sqrt{2}.$$

(iv) Given, 2x = 3y = – z

and 6x = – y = – 4z

$$\Rarr\space \frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\normalsize-1},\\\frac{x}{\frac{1}{6}}=\frac{y}{\normalsize-1}=\frac{z}{\frac{\normalsize-1}{4}}\\\therefore\space a_{1}=\frac{1}{2}, b_{1}=\frac{1}{3}, c_{1} = -1\\ a_{2}=\frac{1}{6}, b_{2}=-1, c_{2}=\frac{1}{4}\\\because\space a_{1}a_{2} +b_{1}b_{2} +c_{1}c_{2}\\=\frac{1}{2}.\frac{1}{6} +\frac{1}{3}.(\normalsize-1) + (\normalsize-1)\bigg(\frac{\normalsize-1}{4}\bigg)\\=\frac{1}{12}-\frac{1}{3}+\frac{1}{4}\\=\frac{1 - 4 +3}{12} = 0$$

∴ Lines are perpendicular to each other.

Hence, the angle between the given lines is 90°. Ans.

$$\text{(v) Sum of the vectors}\\\vec{a},\vec{b},\vec{c}\space \text{is}\\\vec{a} + \vec{b} + \vec{c} = (\hat{i}- 2\hat{j}+ \hat{k})\\ + (2\hat{i}- 4\hat{j}+ 5\hat{k}) +(\hat{i}-6\hat{j}-7\hat{k})\\= 4\hat{i}-12\hat{j}-\hat{k}\space\textbf{Ans.}$$

Solution 16.

(i) Given :

$$\vec{a} = 2\hat{i}-\hat{j} +\hat{k},\vec{b} =\hat{i} +\hat{j}- 2\hat{k}\\\text{and}\space\vec{c} =\hat{i} + 3\hat{j}-5\hat{k}\\\text{Also given,}\\\vec{a}\perp(\lambda\vec{b} + \vec{c})\\\lambda\vec{b} +\vec{c} =\lambda(\hat{i} + \hat{j}-2 \hat{k}) +\\(\hat{i} + 3\hat{j}-5 \hat{k})\\=(\lambda +1)\hat{i} + (\lambda + 3)\hat{j}- (2\lambda + 5)\hat{k}\\\vec{a}.(\lambda \vec{b} +\vec{c}) =0\\\lbrack\because \vec{a}\perp(\lambda\vec{b} + \vec{c})\rbrack$$

$$(2\hat{i}-\hat{j} +\hat{k}).\lbrace(\lambda +1)\hat{i} + (\lambda + 3)\hat{j}-\\(2\lambda +5)\hat{k}\rbrace =0\\\Rarr\space 2(\lambda +1)\\-1(\lambda +3)-1(2\lambda + 5) = 0\\\Rarr\space 2\lambda + 2-\lambda- 3-2\lambda-5 =0\\\Rarr\space -\lambda -6=0\\\Rarr\space \lambda = -6\space\textbf{Ans.}$$

$$\text{(ii)\space}\because\space\hat{a}\space\text{is a unit vector.}\\|\hat{a}| = 1\\\text{Given}\space (2\vec{x} - 3\hat{a}).(2\vec{x} + 3\hat{a}) = 91\\\Rarr\space 4\vec{x}.\vec{x}+ 6\vec{x}\hat{a}-6\hat{a}.\vec{x} -9\hat{a}.\hat{a} = 91\\\Rarr\space 4|\vec{x}|^{2} -9|\hat{a}|^{2} = 91\\\lbrace\because\vec{x}.\hat{a} = \hat{a}.\vec{x}\rbrace\\\Rarr\space 4|\vec{x}|^{2} - 9×1 =91\\\lbrack\therefore\space |\vec{a}|=1\rbrack\\ 4|\vec{x}|^{2} =91 +9 =100\\|\vec{x}|^{2} =\frac{100}{4} = 25\\\therefore\space |\vec{x}| =\sqrt{25} =5\space\textbf{Ans.}$$

Solution 17.

(i) The equation of the given lines are,

$$\frac{x-1}{\normalsize-3} =\frac{y-2}{2\lambda}=\frac{z-3}{2}\text{and}\\\frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{\normalsize-5}$$

If these lines are perpendicular, then

a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒ – 3 × 3l + 2l × 2 + 2 × (– 5) = 0

⇒ – 9λ + 4λ – 10 = 0

⇒ – 5λ = 10

⇒ λ = – 2

Now, the lines are

$$\frac{x-1}{-3}=\frac{y-2}{\normalsize-4}=\frac{z-3}{2}\space\text{and}\\\frac{x-1}{6}=\frac{y-1}{2}=\frac{z-6}{\normalsize-5}$$

The coordinates of any point on first line are given by,

$$\frac{x-1}{\normalsize-3}=\frac{y-2}{\normalsize-4}=\frac{z-3}{2}\space=\alpha$$

or x – 1 = – 3 α ⇒ x = – 3 α + 1

y – 2 = – 4 α ⇒ y = – 4 α + 2

z – 3 = 2 α ⇒ z = 2 α + 3

So, coordinates of any point on this line are,

(– 3 α + 1, – 4 α + 2, 2 α + 3).

The coordinates of any point on second line are given by,

$$\frac{x-1}{-6}=\frac{y-1}{2}=\frac{z-6}{-5}=\beta$$

or x – 1 = – 6β ⇒ x = – 6β + 1

y – 1 = 2 β ⇒ y = 2β + 1

z – 6 = – 5β ⇒ z = – 5β + 6

So, coordinates of any point on second line are

(– 6β + 1, 2β + 1, – 5β + 6).

If lines intersect then they have a common point. So, for some value of a and b, we have

– 3α + 1 = – 6β + 1

⇒ – 3α = – 6β

⇒ α = 2b

and – 4α + 2 = 2β + 1

⇒ – 4α + 1 = 2β

On solving, we have

$$\alpha =\frac{1}{5}\space\text{and}\space\beta =\frac{1}{10}$$

The values of a and b do not satisfy the third equation. Hence, lines do not intersect each other. The values of a and b do not satisfy the third equation. Hence, lines do not intersect each other. Ans.

(ii) Given line is 5x – 25 = 14 – 7y = 35z

⇒ 5 (x – 5) = – 7 (y – 2) = 35z

$$\Rarr\space \frac{x-15}{1/5}=\frac{y-2}{-1/7}=\frac{z-0}{1/35}\\\Rarr\space \frac{x-5}{7}=\frac{y-2}{\normalsize-5}=\frac{z-0}{1}$$

Direction ratios of this line are 7, – 5, 1.

∴ Vector equation of the line which passes through the point A (1, 2, – 1) and its direction ratio are proportional to 7, – 5, 1 is

$$\vec{r} =\hat{i} + 2\hat{j}-\hat{k} +\lambda\bigg(7\hat{i}-5\hat{j} +\hat{k}\bigg)$$

Ans.

Solution 18.

Given : x = 4y – y²

⇒ y² – 4y = – x

⇒ y² – 4y + 4 = – x + 4

⇒ (y – 2)² = – (x – 4)

Given curve represents a left parabola with vertex A(4, 2) and intersect the Y-axis at origin and (0, 4).

$$\text{Required area = }\int^{4}_{0}x dy=\\\int^{4}_{0}(4y-y^{2})dy$$

$$=\bigg[\frac{4y^{2}}{2}-\frac{y^{3}}{3}\bigg]^{4}_{0}\\=\bigg[2y^{2}-\frac{y^{3}}{3}\bigg]^{4}_{0}\\= 32-\frac{64}{3}\\\text{Required area =}\frac{32}{3}\space\text{sq.unit}$$

Ans.

Section-C

Solution 19.

(i) (c) 50

We have Demand function, x = 100 – 8p

$$\Rarr\space p =\frac{100-x}{8}$$

We know, Revenue function = Price × Quantity

$$=\bigg(\frac{100-x}{8}\bigg)(x)\\\text{R(x)} =\frac{10x-x^{2}}{8}\\\text{M.R.} =\frac{d}{dx}\bigg(\frac{100x - x^{2}}{8}\bigg)\\\text{M.R}=\frac{100-2x}{8}\\\frac{100-2x}{8} = 0$$

(Given MR = 0)

⇒ x = 50.

(ii) (a) ₹2,17,500

Advertising budget of two branches differ by ₹30,000 then difference in their sales

= ₹7.25 × 30,000

= ₹2,17,500.

(iii) We have,

P(x) = 0.005x³ + 0·02x²+ 30x

$$\frac{d}{dx}\text{P(x)} =\frac{d}{dx}(0.005x^{3}) +\\\frac{d}{dx}(0.02x^{2})+ \frac{d}{dx}(30 x)$$

= 0.015x2 + 0.04x + 30

Marginal increase in pollution when x = 3

= 0.015(3)² + (0.04) (3) + 30

= 0.135 + 0.12 + 30

= 30.255 units. Ans.

(iv) Given, MR = 50 – 2x + 5x²

∫ MR dx = ∫ (50 – 2x + 5x²) dx

R(x) = ∫ 50 dx – 2 ∫ x dx + 5 ∫ x² dx

$$\text{R(x)} = 50x -\frac{2x^{2}}{2} +\frac{5x^{3}}{3} + c\\\text{R(x) = 50x - x}^{2} +\frac{5}{3}x^{3} + c$$

When x = 0, R(x) = 0

∴ c = 0

$$\therefore\space \text{R(x)} = 50x-x^{2} +\frac{5x^{3}}{3}$$

Ans.

(v) Intersection point of two regression line is called the means of x and y variates respectively.

Therefore (x, y) = (6, 8).

Solution 20.

(i) Given :

$$\text{C(x)}=\frac{x^{3}}{3}-7x^{2} +111 x + 50$$

and x = 100 – p

⇒ p = 100 – x

(a) Revenue function, R(x) = Quantity × Price

= x (100 – x)

∴ R(x) = 100x – x² Ans.

(b) Profit function, P(x) = R(x) – C(x)

$$ = 100x-x^{2}-\frac{x^{3}}{3} +\\ 7x^{2} - 111x - 50\\\text{P(x)} =-\frac{x^{3}}{3}+6x^{2}-11x-50 $$

Differentiating w.r.t. x, we get

$$\frac{d}{dx}(\text{P(x)}) =\\\frac{d}{dx}\bigg(-\frac{x^{3}}{3} + 6x^{2}-11x -50\bigg)\\=-\frac{3x^{2}}{3} + 6.2x-11.1-0\\\Rarr\space \frac{d}{dx}\text{P(x)} = -x^{2} +12 x -11$$

Again, differentiating w.r.t. x, we get

$$\Rarr\space \frac{d^{2}\text{P(x)}}{dx^{2}} = -2x + 12$$

For maximum and minimum value,

$$\text{Put}\space \frac{d}{dx}\text{P(x)} = 0$$

⇒ – x² + 12x – 11 = 0

∴ x² – 12x + 11 = 0

⇒ (x – 11)(x – 1) = 0

⇒ x = 11 and 1

$$\bigg(\frac{d^{2}\text{P}}{dx^{2}}\bigg)_{at x=1}\\ -2 + 12 = 10\\\because \space\bigg(\frac{d^{2}\text{P(x)}}{dx^{2}}\bigg)_{x=1}\gt 0$$

∴ Profit is minimum at x = 1

$$\bigg(\frac{d^{2}\text{P}}{dx^{2}}\bigg)_{x=11} \\= -2×11 + 12 =-10\\\because\space \bigg(\frac{d^{2}\text{P}}{dx^{2}}\bigg)_{x=1}\lt 0$$

Hence, profit is maximum at x = 11. Ans.

$$\text{(ii)\space}\text{p = 2}\bigg(100-\frac{x}{4}\bigg)\\= 200-\frac{x}{2}$$

Revenue function R(x) = Price × Quantity

= p × x

$$=\bigg(200-\frac{x}{2}\bigg)×x\\\text{R(x)} = 200x-\frac{x^{2}}{2}\\\text{Marginal revenue (M.R.) = }\\\frac{d}{dx}\bigg(200 x-\frac{x^{2}}{2}\bigg)$$

= 200 – x

$$\text{Cost function =} 120x +\frac{x^{2}}{2}\\\text{Marginal cost (M.C.) =}\\\frac{d}{dx}\bigg(120x + \frac{x^{2}}{2}\bigg)$$

M.C. = 120 + x

∵ M.R. = M.C. [Given]

⇒ 200 – x = 120 + x

⇒ 2x = 200 – 120

⇒ x = 40

Hence, number of radios = 40. Ans.

Solution 21.

Number of obseravtion n = 5

Construct the following table

x	y	x²	y²	xy
3	6	9	36	18
4	5	16	25	20
5	4	25	16	20
6	3	36	9	18
7	2	49	4	14
Total : 25	20	135	90	90

$$\bar{x}=\frac{\Sigma x}{n}=\frac{25}{5} =5,\\\bar{y}=\frac{\Sigma y}{n}=\frac{20}{5} = 4\\ b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x \Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}\\=\frac{90-\frac{1}{5}×25×20}{135 -\frac{1}{5}×(25)^{2}}\\=\frac{90-100}{135-125}=-1\\ b_{xy}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma y^{2}-\frac{1}{n}(\Sigma y)^{2}}\\=\frac{90-\frac{1}{5}×25×20}{90-\frac{1}{5}×(20)^{2}}$$

$$=\frac{90-100}{90-80}=-1$$

Regression line of y on x axis

$$y-\bar{y} =b_{yx}(x -\bar{x})\\\text{or}\\ y-4 =(\normalsize-1)(x-5)\\\Rarr\space y = -x+9\\\Rarr\space y + x = 9$$

Regression line of x on y is

$$x -\bar{x} = b_{xy}(y -\bar{y})\\\text{or}\space x-5 =(\normalsize-1)(y-4)\\\Rarr\space x =-y + 9.$$

Ans.

Solution 22.

(i) Let manufacturer makes x number of toys of type A and y number of toys of type B.

∴ Maximise Z = 7.50x + 5y

Subject to constraints :

12x + 6y ≤ 360 or 2x + y ≤ 60 ...(i)

18x + 0y ≤ 360 or x ≤ 20 ...(ii)

6x + 9y ≤ 360 or 2x + 3y ≤ 120 ...(iii)

x ≥ 0, y ≥ 0

Consider,

2x + y = 60

x	0	30	20
y	60	0	20

2x + 3y = 120

x	0	60	30
y	40	0	20

OABCD is the required feasible region

Coordinates	Maximise Z = 7.50x + 5y
A (0, 40)	Z = 7.5 × 0 + 5 × 40 = ₹200.00
B(15, 30)	Z = 7.5 × 15 + 5 × 30 = ₹262.50
C(20, 20)	Z = 7.5 × 20 + 5 × 20 = ₹250.00
D(20, 0)	Z = 7.5 × 20 + 5 × 0 = ₹150.00

Hence, profit is maximum when 15 toys of type A and 30 toys of type B are manufactured in a day. Ans.

(ii) Let the Refinery A run x days and Refinery B run y days.

Refinery	High Grade	Medium Grade	Low Grade	Cost per day
A	100	300	200	₹400
B	200	400	100	₹300
Requirement	12,000	20,000	15,000

Object : To minimise cost

Z = 400x + 300y

Subject to constraints,

100x + 200y ≥ 12,000

or x + 2y ≥ 120 …(i)

300x + 400y ≥ 20,000

or 3x + 4y ≥ 200 …(ii)

200x + 100y ≥ 15,000

or 2x + y ≥ 150 …(iii)

and x ≥ 0, y ≥ 0

Consider,

x + 2y = 120

x	0	120	60
y	60	0	30

3x + 4y = 200

x	0	20	40
y	50	35	20

2x + y = 150

x	0	75	50
y	150	0	50

ABC is the required open feasible region.

Coordinates	Minimise cost Z = 400x + 300y
A(0, 150)	Z = 400 × 0 + 300 × 150 = ₹45,000
B(60, 30)	Z = 400 × 60 + 300 × 30 = ₹33,000
C(120, 0)	Z = 400 × 120 + 300 × 0 = ₹48,000