# Oswal Specimen Papers ISC Class 12 Mathematics Solutions (Specimen Paper - 10)

## Section-A

(i) (b) ± 1, 0

Explanation :

From the graph, it is clear that f(x) is not differentiable at x = –1, 0 and 1.

(ii)

Explanation :

f : A → B is function or not it can be checked by a graph of the relation. If it is possible to draw a vertical line which cuts the given curve at more than one point then the given relation is not a function and when this vertical line means line parallel to Y-axis cuts the curve at only point then it is a function.

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

Explanation :

$$\text{We have,}\\\text{sin}\begin{bmatrix}\frac{\pi}{2}-\text{sin}^{\normalsize-1}\bigg(\frac{-\sqrt{3}}{2}\bigg)\end{bmatrix}\\\text{Let\space}\text{sin}^{-1}\bigg(\frac{-\sqrt{3}}{2}\bigg) = x\\\Rarr\space \text{sin x} =\frac{-\sqrt{3}}{2}\\= -\text{sin}\frac{\pi}{3} = \text{sin}\bigg(-\frac{\pi}{3}\bigg)\\\Rarr\space x =-\frac{\pi}{3}\\\Rarr\space \text{sin}^{\normalsize-1}\bigg(\frac{-\sqrt{3}}{2}\bigg) =-\frac{\pi}{3}\\ \therefore\space \text{sin}\begin{bmatrix}\frac{\pi}{2} -\bigg(-\frac{\pi}{3}\bigg)\end{bmatrix}$$

$$= \text{sin}\bigg(\frac{\pi}{2} + \frac{\pi}{3}\bigg)\\=\text{cos}\frac{\pi}{3} =\frac{1}{2}$$

(iv) (c) Determinant is a number associated to a square matrix

$$\text{(v)\space (b)}\space\begin{bmatrix}1 &0\\0 &1\end{bmatrix}$$

Explanation :

$$\text{Given,\space}\text{A} = \begin{bmatrix}0 &1\\ 1 &0\end{bmatrix}\\\therefore\space \text{A}^{2} =\begin{bmatrix}0 &1\\1 &0\end{bmatrix}\begin{bmatrix}0 &1\\1 &0\end{bmatrix}\\ = \begin{bmatrix}1 &0\\0 &1\end{bmatrix}$$

(vi) (b) 3

Explanation :

Let

$$\text{A} =\begin{bmatrix}1 &2\\k &6\end{bmatrix}\\\therefore\space \text{A}^{\normalsize-1}\space\text{does not exist if}\\|\text{A}| = 0\\\therefore\space \begin{vmatrix}1 &2\\k &6\end{vmatrix} = 6\\\Rarr\space\text{6 – 2k = 0}\\\Rarr\space\text{2k = 6}\\\Rarr\space\text{k = 3.}$$

$$\text{(vii)\space(b)\space}\frac{5}{2}$$

Explanation :

Given curve   x2 + 3y + y2 = 5

$$\Rarr\space 2x + 3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\\\Rarr\space (3 + 2y)\frac{dy}{dx} = -2x\\\Rarr\space \frac{dy}{dx} =\frac{-2x}{3 + 2y}\\\therefore\space\text{Slope of the normal at (1, 1)=}\\-\frac{1}{\frac{dy}{dx}}\\= -\frac{1}{\frac{-2x}{3 + 2y}} =\frac{3 + 2y}{2x}\\=\frac{3 + 2×1}{2×1} =\frac{5}{2}.$$

(viii) (c) x + y = 0

Explanation :

Given : Curve y = sin x

$$\therefore\space \frac{dy}{dx} =\text{cos x}\\\Rarr\space \frac{dy}{dx}|_{(0,0)} = 1\\\Rarr\space \text{Slope of normal is – 1.}$$

$$\therefore\space \text{Equation is,}\space \\\text{y-0} = -1(x-0)\\\Rarr\space (x + y) = 0.$$

(ix) (d) Statement (1) is false and statement (2) is true.

Explanation :

From the graph, it is clear that statement 1 is false, and statement 2 is true.

(x) (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Explanation :

Reason is true now

$$\text{P(E}_{1}) = ^{4}\text{C}_{3}p^{3}q \\= 4×\bigg(\frac{1}{2}\bigg)^{3}×\frac{1}{2}\\\bigg(\because\space \text{P} = q =\frac{1}{2}\bigg)\\=\frac{4}{16} =\frac{1}{4}\\\Rarr\space \text{P(}\text{E}_{2}) = ^{8}\text{C}_{5}\bigg(\frac{1}{2}\bigg)^{5}×\bigg(\frac{1}{2}\bigg)^{3}\\=\frac{56}{2^{8}} =\frac{7}{32}\\\text{Therefore,}\space\frac{1}{4}\gt\frac{7}{32}$$

P(E1) > P(E2)

Assertion is true.

Reason is correct explanation of Assertion.

(xi) Consider,

$$\text{L.H.S. = }\text{sin}^{\normalsize-1}\bigg(\frac{\sqrt{3}}{2}\bigg) +\\ 2\text{tan}^{\normalsize-1}\bigg(\frac{1}{\sqrt{3}}\bigg)\\=\frac{\pi}{3} + 2×\frac{\pi}{6} \\= \frac{\pi}{3} + \frac{\pi}{3}\\=\frac{2\pi}{3}\space =\text{R.H.S.}\\\textbf{Hence Proved.}$$

(xii) Given | A | = 4

∴ | – 2A | = (– 2)3 | A |

= – 8 × 4 = – 32. Ans.

(xiii) (b) 1 and 2

Explanation :

x = 1 and x = 2 is non differentiable.

(xiv) Here,

$$\text{P(A)} =\frac{4}{5},\space \text{P(B)} =\frac{1}{3}\\\text{P(A')} =\frac{1}{5},\space \text{P(\text{B}')} =\frac{2}{3}\\\therefore\space \text{The required probability = }\\\text{P(A').P(B')}=\\\frac{1}{5}×\frac{2}{3} = \frac{2}{15}\\\textbf{Ans.}$$

(xv) Multiples of 5 are 5, 10, 15 and 20

Multiples of 7 are 7, 14

Total favourable events = 4 + 2 = 6

Total number of possible outcomes = 20

∴ Probability that the ball drawn is marked with a number multiple of 5 or 7

$$= \frac{6}{20} =\frac{3}{10}.\space\textbf{Ans.}$$

(i) Here, R = {(a, b) : b = a + 1}

∴ R = {(a, a + 1) : a, a + 1 ∈ (1, 2, 3, 4, 5, 6)}

$$\Rarr\space \text{R} =\\\lbrace(1,2),(2,3),(3,4),(4,5),(5,6)\rbrace$$

(a) R is not reflexive as (a, a) ∉ R

(b) R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R

(c) R is not transitive as (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R

OR

(ii) Given, f(x) = x2 + 4

Let f(x1) = f(x2)

⇒ x12 + 4 = x22 + 4

⇒ x1 = x2

Thus, f(x) is one-one.

Since, x2 + 4 is a real number. Thus, for every y in the co-domain of f, there exists a number x in R+ such that f(x) = y = x2 + 4.

Thus, we can say that f(x) is onto.

Now, f(x) is one-one and onto. Hence, f(x) is invertible.

Let f(x) = y ⇒ x2 + 4 = y

⇒ x2 = y − 4

$$\text{i.e.}\space x =\sqrt{y-4}\\\text{Also,\space} x =f^{\normalsize-1}(y)\\f^{\normalsize-1}(y) =\sqrt{y-4}.\\\textbf{Hence Proved.}$$

Given : Equation of curve is,

y2 = px3 + q

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

$$2y\frac{dy}{dx} = 3px^{2}\\\Rarr\space \frac{dy}{dx} =\frac{3px^{2}}{2y}\\\therefore\space \bigg(\frac{dy}{dx}\bigg)_{(2,3)} =\frac{3p×2^{2}}{2×3}\\\text{or\space} m =\bigg(\frac{dy}{dx}\bigg)_{(2,3)} \\=\frac{12p}{6} = 2p$$

Since, y = 4x – 7 is the tangent to the curve at point (2, 3).

So, on comparing with y = mx + c, we get

m = 4

Now, 2p = 4

$$\Rarr\space p = \frac{4}{2} = 2$$

Since, point (2, 3) lies on the curve,

∴ 32 = p × 23 + q

⇒ 9 = 2 × 8 + q

⇒ 9 – 16 = q

⇒ q = –7

Hence, p = 2 and q = – 7 Ans.

We know that

sin–1(sin x) = x

$$\text{So,\space}\begin{bmatrix}\text{sin}^{\normalsize-1}\bigg(\frac{-17\pi}{8}\bigg)\end{bmatrix}\\= \text{sin}^{\normalsize-1}\begin{bmatrix}-\text{sin}\frac{17\pi}{8}\end{bmatrix}\\=\text{sin}^{\normalsize-1}\begin{bmatrix}-\text{sin}\bigg(2\pi + \frac{\pi}{8}\bigg)\end{bmatrix}\\ =\text{sin}^{\normalsize-1}\bigg(-\text{sin}\frac{\pi}{8}\bigg)\\=\text{sin}^{\normalsize-1}\begin{bmatrix}\text{sin}\bigg(-\frac{\pi}{8}\bigg)\end{bmatrix}\\ =-\frac{\pi}{8}\space\textbf{Ans.}$$

(i) Given,

$$y =\frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}\\=\frac{e^{4x} +1}{e^{4x} - 1}$$

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

$$\frac{dy}{dx} = \\\frac{4e^{4x}(e^{4x} -1)- 4e^{4x}(e^{4x} +1)}{(e^{4x}-1)^{2}}\\ =\frac{-8e^{4x}}{(e^{4x}-1)^{2}}\space\textbf{Ans.}$$

OR

(ii) Given,

x = a sin3 t

$$\therefore\space \frac{dx}{dt} = 3a\text{sin}^{2}t.\text{cos t}\\\text{Also,\space y = a cos}^{3}t\\\therefore\space\frac{dy}{dt} = 3a\text{cos}^{2}t(-\text{sin t})\\\therefore\space\frac{dy}{dx} =\frac{-3a\text{cos}^{2}t. \text{sin t}}{\text{3a sin}^{2}t. \text{cos t}}\\\begin{bmatrix}\because\space \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{bmatrix}\\\Rarr\space \frac{dy}{dx} =-\text{cot t}.\space\textbf{Ans.}$$

Given, y = 2 cos (log x) + 3 sin (log x)

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

$$\frac{dy}{dx} = -2\space\text{sin(log x)}.\frac{1}{x} +\\ 3\space\text{cos(log x)}.\frac{1}{x}\\\Rarr\space x\frac{dy}{dx} =-2\text{sin}(\text{log x}) + 3\space\text{cos(log x)}$$

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

$$x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} \\=-2\text{cos(log x)}.\frac{1}{x} + 3\text{cos(log x)} .\frac{1}{x}\\\Rarr\space x^{2}\frac{d^{2}y}{dx^{2}} + x.\frac{dy}{dx}\\= -(2 \text{cos(log x)}) + 3\space\text{sin(log x)}\\\Rarr\space x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} =-y\\\Rarr\space x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + y = 0\\\textbf{Hence Proved.}$$

Given function is

$$\text{f(x)} = \frac{x^{4}}{4}-x^{3}-5x^{2} + 24x + 12\\\Rarr\space\\f'(x)=\frac{4x^{3}}{4}-3x^{2}-10x + 24$$

For critical points, put f ′(x) = 0

∴ x3 – 3x2 – 10x + 24 = 0

(x – 2) (x2 – x – 12) = 0

(x – 2) (x – 4) (x + 3) = 0

⇒ x = 2, 4, – 3

Therefore, we have the intervals (– ∞, – 3), (– 3, 2), (2, 4) and (4, ∞).

Since f’(x) > 0 in (– 3, 2) ∪ (4, ∞).

∴ f(x) is increasing in interval (– 3, 2) ∪ (4, ∞).

And f′(x)< 0 in (– ∞, – 3) ∪ (2, 4)

∴ f(x) is decreasing in (– ∞, – 3) ∪ (2, 4).

Ans.

Sample space n(S) = {(H, H), (H, T), (T, H), (T, T)}

n(S) = 4

(i) When A throws two heads

n(A) = (H, H)

n(A) = 1

$$\text{P(A)} = \frac{n(\text{A})}{n(\text{S})} =\frac{1}{4}\\\textbf{Ans.}\\\text{(ii) P(B does not throw two heads) =}\\1 -\frac{1}{4} =\frac{3}{4}.$$

(iii) If A starts the game, then A can win in either 1st or 3rd or 5th trial and so on.

$$\text{P(2 heads)}=\frac{1}{4},\\\text{P(not 2 heads)} =\frac{3}{4}\\\therefore\space \text{P(A wins the game) =}\\\frac{1}{4} + \bigg(\frac{3}{4}\bigg)^{2}×\frac{1}{4} +\\\bigg(\frac{3}{4}\bigg)^{4}×\frac{1}{4}+.....\\=\frac{1}{4}\begin{bmatrix}1 + \bigg(\frac{3}{4}\bigg)^{2} + \bigg(\frac{3}{4}\bigg)^{4}+ .....\end{bmatrix}\\=\frac{1}{4}\begin{bmatrix}\frac{1}{1 - \frac{9}{16}}\end{bmatrix}\\\begin{bmatrix}\because\space\text{For infinite G.P.S. = }\frac{a}{1-r}\end{bmatrix}\\= \frac{1}{4}×\frac{16}{7} =\frac{4}{7}\space\textbf{Ans.}$$

(iv) If B starts the game and A wins, then A can win in either 2nd or 4th or 6th trial and so on.

∴ P(A wins the game) =

$$\frac{3}{4}×\frac{1}{4} + \bigg(\frac{3}{4}\bigg)^{3}×\frac{1}{4} +\\ \bigg(\frac{3}{4}\bigg)^{5}×\frac{1}{4}+.......\\=\frac{3}{16} + \frac{3}{16}×\bigg(\frac{3}{4}\bigg)^{2} +\\\frac{3}{16}×\bigg(\frac{3}{14}\bigg)^{4} +......\\=\frac{3}{16}\begin{bmatrix}1 + \bigg(\frac{3}{4}\bigg)^{2} + \bigg(\frac{3}{4}\bigg)^{4}+......\end{bmatrix}\\ = \frac{3}{16}×\frac{1}{1 -\frac{9}{16 }}$$

$$= \frac{3}{16}×\frac{16}{7}\\ = \frac{3}{7}\space\textbf{Ans.}$$

(i) Let

$$\text{I} = \frac{1}{2}\int(2 \text{cos} 2x \text{cos} 4x)\\\text{cos 6x}\space dx\\\Rarr\space\text{I} =\frac{1}{2}\int(\text{cos 6x + cos 2x})\\\text{cos 6x}dx\\\lbrack\because\space \text{2 cos A cos B}=\\\text{cos}(\text{A + B}) + \text{cos}(\text{A - B})\rbrack\\ =\frac{1}{2}\int\text{cos 6x cos 6xdx} +\\\frac{1}{2}\int\text{cos 2x cos 6x dx}\\=\frac{1}{4}\int 2\text{cos}6x\space \text{cos} 6x dx +\\\frac{1}{4} \int 2\text{cos}2x\text{cos} 6x dx$$

$$= \frac{1}{4}\int(\text{cos} 12 x + \text{cos 0})dx +\\ \frac{1}{4}\int\text{cos} 8x + \text{cos}\space\text{4x}dx\\= \frac{1}{4}\begin{bmatrix}\int(\text{cos 12x + 1 + cos 8x + cos 4x})dx\end{bmatrix}\\=\frac{1}{4}\begin{bmatrix}\frac{\text{sin 12 x}}{12} + \frac{\text{sin 8x}}{8}+ \frac{sin 4x}{4} +x\end{bmatrix}\\+\text{C}\space\textbf{Ans.}$$

OR

$$\text{(ii)\space}\int\frac{\text{cos}2x - cos 2\alpha}{\text{cos x - cos}\alpha}dx =\\\int\frac{2\text{cos}^{2}x -1-2\text{cos}^{2}\alpha+1}{\text{cos} x - cos\space\alpha}dx\\\lbrack\because\space \text{cos 2}\theta = 2\text{cos}^{2}\theta -1\rbrack\\=\int\frac{2\space\text{cos}^{2}x - 2\text{cos}^{2}\alpha}{\text{cos}x - cos\alpha}dx\\ =\int\frac{2(cos x - cos\alpha)(cos x + cos \alpha)}{(\text{cos x - cos}\alpha)}dx\\ = 2\int(\text{cos x + cos}\alpha)dx\\ = 2\begin{bmatrix}\int\text{cos x dx} + \int\text{cos}\alpha dx\end{bmatrix}$$

= 2 sin x + 2x cos α + C Ans.

(i) Given differential equation is

$$\frac{dy}{dx} = 1 + x^{2} + y^{2} + x^{2}y^{2}\\ =(1 + x^{2}) +y^{2}(1 + x^{2})\\\Rarr\space \frac{dy}{dx} = (1 + x^{2})(1 + y^{2})\\\Rarr\space \frac{dy}{1 +y^{2}} =(1 + x^{2})dx$$

On integrating both sides, we have

$$\int\frac{dy}{1 + y^{2}} =\int(1 + x^{2})dx\\\text{tan}^{\normalsize-1}y = x + \frac{x^{3}}{3} + \text{C}\space\text{...(i)}$$

put y = 1 and x = 0 in equation (i),

tan–1 1 = 0 + 0 + C

$$\text{C} =\frac{\pi}{4}\\\text{Equation (i) becomes:}\\\text{tan}^{-1}y = x + \frac{x^{3}}{3} +\frac{\pi}{4}\\\Rarr\space y =\text{tan}\bigg(x + \frac{x^{3}}{3}+ \frac{\pi}{4}\bigg)$$

is the required particular solution of given equation. Ans.

OR

(ii) The given differential equation is

(1 – y2 ) (1 + log x) dx + 2xy dy = 0

$$\frac{(1 + \text{log}\space x)}{x}dx =\frac{-2y}{(1 - y^{2})}dy$$

On integrating both sides, we have

$$\int\frac{1 + \text{log}\space x}{x}dx \\=\int\frac{-2y}{(1-y^{2})}dy$$

In first integral,

put 1 + log x = t

$$\Rarr\space \frac{1}{x}\text{dx} = dt$$

Also in second integral,

put 1 – y2 = u

⇒ – 2y dy = du

$$\therefore\space \int t.dt = \int\frac{1}{u}du\\\Rarr\space \frac{t^{2}}{2}-\text{log|u|} =\text{C}\\\text{or}\\\frac{1}{2}(1 + \text{log x})^{2} -\\\text{log}|1 - y^{2}| =\text{C}$$

It is given that y = 0 when x = 1

$$\text{So,}\space \frac{1}{2}(1 + \text{log}\space 1)^{2} -\\\text{log}|1 - 0^{2}| =\text{C}\\\Rarr\space \text{C} =\frac{1}{2}\\\therefore\space \frac{(1 + \text{log}\space x)^{2}}{2} \\-\text{log}|1- y^{2}| =\frac{1}{2}\\\text{or\space} (1 + \text{log}\space x)^{2} -\\ 2\text{log}|1 - y^{2}| = 1$$

It is the required particular solution. Ans.

Let the number of children be x and amount distributed by Ishan to each child be ₹y.

As per the given information

(x – 8) (y + 10) = xy

⇒ xy + 10x – 8y – 80 = xy

⇒ 5x – 4y = 40 ...(i)

⇒ (x + 16) (y – 10) = xy

⇒ –10x + 16y + xy – 160 = xy

⇒ –5x + 8y = 80 ...(ii)

Now, we can express equation (i) and equation (ii) in matrix form as follows:

$$\text{A} =\begin{bmatrix}5 &\normalsize-4\\\normalsize-5 &8\end{bmatrix},\space\\ \text{X} =\begin{bmatrix}x\\y\end{bmatrix}\space\text{and B} =\begin{bmatrix}40\\ 80\end{bmatrix}\\\Rarr\space \text{X} = \text{A}^{\normalsize-1}\space\text{B}\\\Rarr\space x =\frac{|\text{adj A}|}{|\text{A}|}\text{B}$$

A11 = 8, A12 = 4

A21 = 5, A22 = 5

$$\therefore\space \text{adj A} =\begin{bmatrix}8 &4\\ 5 &5\end{bmatrix}\\\text{and\space} |\text{A}| = 40-20 = 20\\\Rarr\space \text{X} =\frac{1}{20}\begin{bmatrix}8 &4\\5 &5\end{bmatrix}\begin{bmatrix}40\\ 80\end{bmatrix}\\\Rarr\space \text{X} =\frac{1}{20}\begin{bmatrix}640\\ 600\end{bmatrix}\\\Rarr\space \begin{bmatrix}x\\y\end{bmatrix}\begin{bmatrix}32\\30\end{bmatrix}\\\Rarr\space x = 32, y = 30$$

Number of children = 32 Ans.

Amount donate by Ishan = ₹(32 × 30) = ₹960 Ans.

(i) Let ex3 = t

∴ 3x2 ex3 dx = dt

$$\therefore\space \int x^{2}(e^{x^{3}})\text{cos}(2 e^{x^{3}})dx \\=\int\frac{1}{3}\text{cos}\space 2t dt\\ =\frac{1}{3}.\frac{1}{2}\text{sin 2t} + \text{C}\\= \frac{1}{6}\text{sin}(2e^{x^{3}}) +\text{C}\\\textbf{Ans.}$$

OR

$$\text{(ii)\space}\int x(\text{tan}^{\normalsize-1}x)^{2}dx\\\text{Let tan}^{\normalsize-1}x = t\\\Rarr\space x =\text{tan t}\\ dx =\text{sec}^{2}t dt\\\int x(\text{tan}^{\normalsize-1}x)^{2}dx =\\\int t^{2}.\text{tan t sec}^{2}\text{t dt}\\\lbrack\text{Integrating by parts}\rbrack\\ = t^{2}.\frac{\text{tan}^{2}t}{2}-\int\frac{2t \space \text{tan}^{2}t}{2}dt$$

$$=\frac{1}{2}t^{2}\text{tan}^{2}t -\\\int t(\text{sec}^{2}t -1)dt\\ =\frac{1}{2}t^{2}\text{tan}^{2}t - \int t\text{sec}^{2}tdt\\ + \int tdt\\ =\frac{1}{2}t^{2}\text{tan}^{2}t + \frac{t^{2}}{2}-\\\int t\text{sec}^{2}tdt\\=\frac{1}{2}t^{2}(\text{tan}^{2}t + 1) -\\\begin{Bmatrix}t \text{tan t} -\int\text{tan t}dt\end{Bmatrix}$$

$$=\frac{1}{2}t^{2}(\text{tan}^{2}t+1) -\\\text{t tan t + log}|\text{sec t}| + \text{C}\\=\frac{1}{2}(1 +x^{2})(\text{tan}^{\normalsize-1})^{2} -\\ x\text{tan}^{\normalsize-1}x + \text{log}(\sqrt{1 + x^{2}})+\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2} -\\ x\space\text{tan}^{\normalsize-1}x + \\\text{log}(\sqrt{1 + x^{2}})+\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2} - x\space\text{tan}^{\normalsize-1}x +\\\frac{1}{2}\text{log}(1 + x^{2}) + \text{C}\\\textbf{Ans.}$$

(i) Given,

$$(1 + x^{2})\frac{dy}{dx} = (e^{\text{m tan}^{\normalsize-1}x}-y)\\\frac{dy}{dx} =\\ \frac{e^{\text{m tan}^{\normalsize-1}}}{1 + x^{2}}-\frac{y}{1 + x^{2}}\\\Rarr\space \frac{dy}{dx} + \frac{1}{1 + x^{2}}y\\=\frac{e^{m tan^{\normalsize-1}x}}{1 + x^{2}}\\\text{This equation is of the form}\\\frac{dy}{dx} +\text{Py} =\text{Q(x)}\\\text{where\space}\text{P} =\frac{1}{1 + x^{2}}\text{and Q(x)} \\=\frac{e^{m tan^{-1}}x}{1 + x^{2}}$$

$$\text{I.F.} = e\int{\text{Pdx}} \\=\int e^{\bigg(\frac{1}{1 + x^{2}}\bigg)dx} = e^{\text{tan}{\normalsize-1}x}$$

Hence, solution of linear differential equation is given by

y × I.F. = × ∫I.F Q(x) dx

$$\Rarr\space y×e^{\text{tan}^{\normalsize-1}x} =\\\int e^{\text{tan}^{-1}x}.\frac{e^{m \text{tan}^{\normalsize-1}x}}{1 + x^{2}}dx\\\Rarr\space y×e^{\text{tan}^{\normalsize-1}x}=\\\int\frac{e^{(1 + m)\text{tan}^{\normalsize-1}x}}{1 + x^{2}}dx\\\text{Let \space} e^{(1 + m)\text{tan}^{\normalsize-1}x} = t\\\Rarr\space (1 + m)\frac{e^{(1 + m)\text{tan}^{\normalsize-1}x}}{1 + x^{2}}\\dx = dt\\\therefore\space y×e^{\text{tan}^{-1}x} \\ =\int\frac{1}{1 + m}dt$$

$$\Rarr\space y×e^{\text{tan}^{\normalsize-1}x}=\\\frac{1}{(1 + m)}t + \text{C}\\=\frac{1}{1 + m}e^{(1+m)\text{tan}^{\normalsize-1}x} + \text{C}$$

Now, when x = 0, then y = 1,

$$\therefore\space 1×e^{\text{tan}^{\normalsize-1}0} \\= \frac{1}{1 +m}e^{(1 + m)\text{tan}^{\normalsize-1}}0 +\text{C}\\\Rarr\space 1× e^{0} =\\\frac{1}{1 + m}e^{0} + \text{C}\\\Rarr\space 1 -\frac{1}{1 +m} =\text{C}\\\therefore\space \text{C} =\frac{m}{m+1}$$

Hence, particular solution of given differential equation is

$$y×e^{\text{tan}^{-1}x} =\frac{1}{m+1}e^{(1 + m)}\text{tan}^{\normalsize-1}x+\\\frac{m}{m+1}\space\\\lbrack\text{where m}\neq -1\rbrack\space\textbf{Ans.}$$

If m = – 1 then particular solution of the given differential equation is

$$y×e^{\text{tan}^{\normalsize-1}x} =\\\int e^{\text{tan}^{\normalsize-1}x}.\frac{e^{-\text{tan}^{-1}x}}{1 + x^{2}}dx\\\Rarr\space y×e^{\text{tan}^{\normalsize-1}x}=\\\frac{e^{0}}{1+x^{2}}dx =\int\frac{dx}{1 +x^{2}}\\\Rarr\space y×e^{\text{tan}^{\normalsize-1}x} \\= \text{tan}^{\normalsize-1}x+ \text{C}$$

When x = 0, then y = 1,

$$\therefore\space 1×e^{\text{tan}^{\normalsize-1}0} \\= \text{tan}^{\normalsize-1} 0 + \text{C}\\\Rarr\space 1×e^{0} = 0 + \text{C}\\\Rarr\space \text{C = 1}$$

Hence, particular solution of differential equation is

$$y×e^{\text{tan}^{\normalsize-1}x} \\=\text{tan}^{\normalsize-1}x +1.\\\textbf{Ans.}$$

OR

(ii) Let ABC be an isosceles triangle with AB = AC and a circle of radius r unit with centre I is inscribed in ∆ABC.

∴ BD = DC

Let AE = AF = x units,

[Tangents drawn from an external point are equal in length]

CE = CD = BD = y units

∴ BD = BF = y units

Perimeter of triangle = AB + BC + AC

= x + y + 2y + x + y

P = 2x + 4y ...(i)

In ∆AIE, ∠AEI = 90°

$$\therefore\space \frac{\text{AE}}{\text{IE}} = \text{cot\space}\theta$$

AE = r cot θ = x ...(ii)

$$\therefore\space \frac{\text{DC}}{\text{AC}} = \text{sin}\space\theta\\\frac{y}{\text{x + y}} =\text{sin}\space\theta$$

⇒ x sin θ + y sin θ = y

⇒ r cot θ sin θ = y(1 – sin θ)

[Using (ii)]

$$\Rarr\space \frac{\text{r cos}\space\theta}{\text{1 - sin}\space \theta} = y\space\text{...(iii)}$$

From equations (i), (ii) and (iii),

P = 2x + 4y

$$\text{P} = 2r\space\text{cot}\space\theta + \frac{4r\space\text{cos}\space\theta}{\text{1 - sin}\space\theta}\\\therefore\space \frac{\text{dP}}{d\theta} =\frac{d}{d\theta}\text{2r cot}\theta +\\\frac{d}{d\theta}\bigg(\frac{4r\space \text{cos}\space\theta}{\text{1 - sin}\theta}\bigg)\\ = -2r\space\text{cosec}^{2}\theta + \\4r\begin{bmatrix}\frac{(1 - sin\space\theta)(-\text{sin}\theta) -\text{cos}\theta(-\text{cos}\theta)}{(1 - sin\space\theta)^{2}}\end{bmatrix}\\=- 2r\text{cosec}^{2}\theta + \\4r\begin{bmatrix}\frac{-\text{sin}\theta + \text{sin}^{2}\theta + \text{cos}^{2}\theta}{(1 - sin\space \theta)^{2}}\end{bmatrix}\\ = -2r\space\text{cosec}^{2}\theta +\\ 4r\begin{bmatrix}\frac{1 - \text{sin}\theta}{(1 -\text{sin}\theta)^{2}}\end{bmatrix}$$

$$= - 2r\space \text{cosec}^{2}\theta +\\ 4r\begin{bmatrix}\frac{\text{1 - sin}\theta}{(1 - sin\space\theta)^{2}}\end{bmatrix}\\= -2r\space\text{cosec}^{2}\theta + \frac{4r}{\text{1 - sin}\space\theta}\\ = 2r\begin{bmatrix}\frac{\normalsize-1}{\text{sin}^{2}\theta} + \frac{2}{\text{1 - sin}\space\theta}\end{bmatrix}\\ =2r\begin{bmatrix}\frac{-1 + \text{sin}\space \theta + 2\text{sin}^{2}\theta}{\text{sin}^{2}\theta(1 - sin\space\theta)}\end{bmatrix}\\\therefore\space\frac{\text{dP}}{\text{d}\theta} =\frac{2r(2\text{sin}\theta -1)(\text{sin}\space \theta +1)}{\text{sin}^{2}\theta (1 - sin\space \theta)}\\\text{For maxima and minima,}\\\text{Put,\space} \frac{\text{dP}}{\text{d}\theta} = 0\\\therefore\space \frac{2r(2\text{sin}\theta -1)(\text{sin}\theta +1)}{\text{sin}^{2}\theta(1 - \text{sin}\theta)} = 0$$

$$\text{2r}(2\text{sin}\space\theta-1)(\text{sin}\space\theta +1) = 0$$

∵ r ≠ 0

∴ 2 sin θ – 1 = 0

$$\Rarr\space \text{sin}\space\theta =\frac{1}{2}\\\Rarr\space \theta = 30\degree \text{or}\frac{\pi}{6},$$

or sin θ + 1 = 0

$$\Rarr\space \text{sin}\space\theta =-1\\\Rarr\space \theta =-\frac{\pi}{2}$$

$$\because\space \theta =-\frac{\pi}{2}\space\text{is not possible}\\\therefore\space \theta = 30\degree\space\text{or}\space\frac{\pi}{6}\\\text{Now,\space}\frac{d^{2}\text{P}}{d\theta^{2}} =\\\frac{d}{d\theta}\begin{bmatrix}-2r\space\text{cosec}^{2}\theta + \frac{4r}{1-sin\theta}\end{bmatrix}\\= -2r(2\text{cos}\space\theta)(-\text{cosec}\space\theta \text{cot}\space\theta)-4r.\\\frac{1}{(1 - sin\space\theta)^{2}}( -\text{cos}\space\theta)\\ = 4r\begin{bmatrix}\text{cosec}^{2}\theta\text{cot}\theta +\frac{\text{cos}\space\theta}{(1 - sin\space\theta)^{2}}\end{bmatrix}\\= 4r\begin{bmatrix}\text{cosec}^{2}\frac{\pi}{6}\text{cot}\frac{\pi}{6} + \frac{\text{cos}\frac{\pi}{6}}{\bigg(1 -\text{sin}\frac{\pi}{6}\bigg)^{2}}\end{bmatrix}$$

$$= 4r(4.\sqrt{3} + 2\sqrt{3})\\\therefore\space \bigg(\frac{d^{2}\text{P}}{\text{d}\theta^{2}}\bigg)_{\theta =\frac{\pi}{6}}\gt 0$$

Hence, perimeter is minimum at π/6.

Now, Perimeter of ∆ABC = 2x + 4y

$$= 2r\text{cot\space}\theta +\frac{4r \text{cos}\space\theta}{\text{1 - sin}\space\theta}\\= 2r\text{cot}\frac{\pi}{6} + \frac{4r\text{cos}\frac{\pi}{6}}{\text{1 - sin}\frac{\pi}{6}}\\ = 2r.\sqrt{3} + \frac{4r.\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}\\= r(2\sqrt{3} + 4\sqrt{3}) = 6\sqrt{3}r\\\therefore\space \text{Least perimeter of ΔABC is}\\6\sqrt{3r}\space\text{units.}$$

Hence Proved.

(i) n(W|A) = 2

n(S) = 5

$$\text{P(W|A)} =\frac{2}{5}\\\text{Similarly}\\\text{P(W|B)} =\frac{1}{5}, \text{P(W|C)} =\frac{4}{5}\\\text{(ii)\space \text{Given\space}\text{P(A)}} =\frac{2}{5},\text{P(B)} =\frac{2}{5},\\\text{P(C)} =\frac{1}{5}\\\text{P(Probability that the drawn ball is white)} =\\\text{P(A). P(W|A)} + \text{P(B).P(W|B)}\\+ \text{P(C).\text{P(W|C)}}\\(\because\space \text{Theorem of total probability})\\=\frac{2}{5}×\frac{2}{5} + \frac{2}{5}×\frac{1}{5} + \frac{1}{5}×\frac{4}{5}$$

$$= \frac{4}{25} +\frac{2}{25} +\frac{4}{25}\\= \frac{10}{25} = \frac{2}{5}\space\textbf{Ans.}$$

(iii) Let E denote that drawn ball is black then

P(E) = P(A).P(E|A) + P(B).P(E|B) + P(C).P(E|C)

$$\text{P(E)} = \frac{2}{5}×\frac{3}{5} + \frac{2}{5}×\frac{4}{5} + \frac{1}{5}×\frac{1}{5}\\\text{P(E)} =\frac{6}{25} +\frac{8}{25} +\frac{1}{25}\\=\frac{15}{25} = \frac{3}{5}\space\textbf{Ans.}\\\text{(iv) By Baye’s theorem, we have}\\\text{P(C|W)} =\\\frac{\text{P(C)}\text{P(W|C)}}{\text{P(A)}\text{P(W|A)} + \text{P(B)}\text{P}(\text{W|B}) + \text{P(C)}\text{P(W|C)}}\\ =\frac{\frac{1}{5}×\frac{4}{5}}{\frac{2}{5}×\frac{2}{5} + \frac{2}{5}×\frac{1}{5} + \frac{1}{5}×\frac{4}{5}}\\=\frac{4}{10} = \frac{2}{5}\space\textbf{Ans.}$$

## Section-B

$$\text{(i)\space (c)\space}\vec{b},\vec{c}\space\text{and}\space\vec{d}\\\text{Vector}\space\vec{b},\vec{c}\space\text{and}\space\vec{d}\\\text{start from same point.}$$

(ii) (b) 60°

Angle between two planes a1 x + b1 y + c1 = 0 and a2x + b2y + c2 = 0 is given by,

∴ cos θ =

$$\begin{vmatrix}\frac{a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}\end{vmatrix}\\\text{Here a}_1 = 2 , b_{1} = -1, c_{1} = 1\\\text{and\space} a_{2} = 1, b_{2}=1, c_{2}= 2\\\therefore\space \text{cos\space}\theta =\\\begin{vmatrix}\frac{2(1) + (\normalsize-1)(1) + (1)(2)}{\sqrt{2^{2} + 1^{2} + 1^{2}}\sqrt{1^{2}+ 1^{2} + 2^{2}}}\end{vmatrix}\\\text{cos}\space\theta =\begin{vmatrix}\frac{2-1+2}{\sqrt{6}\sqrt{6}}\end{vmatrix}\\=\frac{1}{2}\\\Rarr\space \theta = 60\degree =\frac{\pi}{3}\\\text{(iii)\space \text{For}\space}\bigg(\frac{1}{\sqrt{2}},\frac{1}{2},k\bigg)\space\\\text{to represent direction cosines,}$$

we should have

$$\bigg(\frac{1}{\sqrt{2}}\bigg)^{2} + \bigg(\frac{1}{2}\bigg)^{2} + k^{2} = 1\\\text{or\space}\frac{1}{2} +\frac{1}{4} + k^{2}=1\\\Rarr\space k =\pm\frac{1}{2}.$$

$$\text{(iv)\space}\vec{a} = 2\hat{i}-\hat{j} + \hat{k},\vec{b} = \hat{i} + 2\hat{j}-3\hat{k}\\\text{and}\space\vec{c} = 3\hat{i} + x\hat{j} +5\hat{k}\\\text{If}\space \vec{a},\vec{b}\space\text{and}\space\vec{c}\space\text{are coplanar, then}\\\begin{vmatrix}2 &\normalsize-1 &1\\1 &2 &\normalsize-3\\3 &x &5\end{vmatrix} = 0\\\Rarr\space 2\begin{vmatrix}2 &-3\\x &5\end{vmatrix} +1\begin{vmatrix}1 &\normalsize-3\\3 &5\end{vmatrix} +\\ 1\begin{vmatrix}1 &2\\3 &x\end{vmatrix} = 0 \\\Rarr\space 2(10 + 3x) + (5 + 9) + \\(x-6) = 0\\\Rarr\space 20 + 6x +14 +x-6 =0\\\Rarr\space 7x = 28$$

∴ x = 4. Ans.

$$\text{(v)\space}\vec{\text{OP}} × \vec{\text{OQ}}=\\\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\ 1& 2 &1\\2 &1 &3\end{vmatrix}\\ = 5\hat{i} -\hat{j}-3\hat{k}\\\vec{\text{PQ}}×\vec{\text{PR}} =\\\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\ 1 &\normalsize-1 &2\\\normalsize-2 &\normalsize-1 &1\end{vmatrix}\\ = \hat{i} - 5\hat{j}-3\hat{k}\\\text{Required angle cos}\space\theta =\\\frac{5 +5+9}{\sqrt{25+1+9}\sqrt{1 + 25 + 9}}\\\text{cos}\space\theta =\frac{19}{35}$$

$$\theta = \text{cos}^{\normalsize-1}\bigg(\frac{19}{35}\bigg)$$

Ans.

(i) The given lines are

$$\frac{x+1}{\normalsize-3} = \frac{y-3}{2} =\frac{z+2}{1}\\\text{and\space}\frac{x}{1} = \frac{y-7}{\normalsize-3} =\frac{z+7}{2}$$

Direction ratios of the given lines are (– 3, 2, 1) and (1, – 3, 2) and these are not proportional.

∴ Given lines are not parallel.

Hence, these lines are intersecting.

Hence Proved.

If lines are coplanar then

$$\begin{vmatrix}x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1}\\a_{1} &b_{1} &c_{1}\\a_{2} &b_{2} &c_{2}\end{vmatrix} = 0\\\begin{vmatrix}0-(\normalsize-1) &7-3 &-7-(\normalsize-2)\\\normalsize-3 &2 &1\\ 1 &\normalsize-3 &2\end{vmatrix} = 0\\\begin{vmatrix}1 &4 &\normalsize-5\\\normalsize-3 &2 &1\\ 1 &\normalsize-3 &2\end{vmatrix} = 0\\\Rarr\space 1\begin{vmatrix}2 &1\\\normalsize-3 &2\end{vmatrix}-4\begin{vmatrix}\normalsize-3 &1\\1 &2\end{vmatrix}+\\(\normalsize-5)\begin{vmatrix}\normalsize-3 &2\\1 &\normalsize-3\end{vmatrix} = 0\\\Rarr\space (4+3)-4(-6-1)-5(9-2) = 0\\\Rarr\space 7 + 28 -35 = 0$$

Hence, the given lines are coplanar.

Hence Proved.

Equation of the plane containing these lines is

$$\begin{vmatrix}x-x_{1} &y-y_{1} &z-z_{1}\\x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1}\\a_{1} &b_{1} &c_{1}\end{vmatrix}=0\\\begin{vmatrix}x-(\normalsize-1) &y-3 &z+2\\1 &4 &\normalsize-5\\\normalsize-3 &2 &1\end{vmatrix} = 0\\\begin{vmatrix}x+1 &y-3 &z+2\\1 &4 &\normalsize-5\\-3 &2 &1\end{vmatrix} = 0\\\Rarr\space (x+1)\begin{vmatrix}4 &\normalsize-5\\2 &1\end{vmatrix}-\\(y-3)\begin{vmatrix}1 &\normalsize-5\\\normalsize-3 &1\end{vmatrix} + \\(z+2)\begin{vmatrix}1 &4\\\normalsize-3 &2\end{vmatrix} = 0$$

⇒ (x + 1) (4 + 10) – (y – 3) (1 – 15) + (z + 2) (2 + 12) = 0

⇒ 14(x + 1) – (– 14) (y – 3) + 14(z + 2) = 0

⇒ 14x + 14 + 14y – 42 + 14z + 28 = 0

⇒ 14x + 14y + 14z = 0

⇒ x + y + z = 0

Hence, equation of the plane is x + y + z = 0. Ans.

OR

(ii) Given line is

5x – 25 = 14 – 7y = 35z

⇒ 5(x – 5) = – 7(y – 2) = 35z

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

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

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

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

(i) Given : Coordinates of vertices of ΔABC are A(1, 2, 3), B (2, – 1, 4) and C(4, 5, – 1)

$$\therefore\space \vec{\text{AB}} = \hat{i}- 3\hat{j} +\hat{k}\\\text{and}\space \vec{\text{AC}} = 3\hat{i} +3\hat{j}-4\hat{k}\\\text{We know that,}\\\text{ar}(\Delta\text{ABC}) =\frac{1}{2}\bigg|\vec{\text{AB}}×\vec{\text{AC}}\bigg|\\\text{Now,\space} \vec{\text{AB}} ×\vec{\text{AC}} =\\\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\1 &\normalsize-3 &1\\3 &3 &\normalsize-4\end{vmatrix}\\= \hat{i}(12-3)- \hat{j}(-4-3) + \hat{k}(3+ 9)\\ = 9\hat{i} + 7\hat{j} + 12 \hat{k}$$

$$\therefore\space |\vec{\text{AB}}×\vec{\text{AC}}| =\\\sqrt{9^{2} + 7^{2} + (12)^{2}} =\sqrt{274}\\\text{So,\space \text{ar}(}\Delta\space\text{ABC}) =\\\frac{1}{2}\sqrt{274}\space\text{sq.units}\\\textbf{OR}\\\text{(ii)\space \text{Here},\space} \vec{a} = (\hat{i} + \hat{j} + \hat{k}).\hat{n}\\\text{is unit vector}\\\vec{b} = 2\hat{i} + 4\hat{j}-5\hat{k}\\\vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k}\\\vec{b} +\vec{c} = \\(2+\lambda)\hat{i} + 6\hat{j}- 2\hat{k}\\\text{Then,\space}\hat{n} = \frac{(2 + \lambda)\hat{i} + 6\hat{j}-2\hat{k}}{\sqrt{(2 + \lambda)^{2} + 36 + 4}}$$

$$\text{Given,\space} \vec{a}.\hat{n} = 1\\(\hat{i}+ \hat{j} + \hat{k}).\bigg(\frac{(2 + \lambda)\hat{i} + 6\hat{j}- 2\hat{k}}{\sqrt{(2 + \lambda)^{2} + 40} }\bigg)\\=1$$

$$\Rarr\space (2 +\lambda)+ 6-2 =\\\sqrt{(2 + \lambda)^{2} + 40}\\\lbrack\because\space \hat{i}.\hat{i} =1, \hat{j}.\hat{j} =1, \hat{k}.\hat{k} =1\rbrack\\\Rarr\space (2 + \lambda) +4 =\\\sqrt{(2 + \lambda)^{2} + 40}\\\Rarr\space \lambda +6 =\\\sqrt{(2 + \lambda)^{2} + 40}$$

On squaring both sides, we get

(6 + λ)2 = (2 + l)2 + 40

⇒ 36 + λ2 + 12λ = 4 + λ2 + 4λ + 40

⇒ 36 + 12 λ − 4 − 4λ − 40 = 0

⇒ 8λ − 8 = 0

⇒ λ = 1

$$\text{Then, \space} \vec{b} + \vec{c} =\\ (2 + \lambda)\hat{i} + 6\hat{j}-2 \hat{k}\\ = 3\hat{i} + 6\hat{j}-2 \hat{k}\\\text{unit vector along\space} (\vec{b} + \vec{c})=\\\frac{3\hat{i} + 6\hat{j}-2\hat{k}}{\sqrt{9 + 36 + 4}}\\=\frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{9 + 36 + 4}}\\ = \frac{3 \hat{i} + 6\hat{j}-2 \hat{k}}{\sqrt{49}}\\=\frac{3\hat{i} + 6\hat{j}-2\hat{k}}{7}\\\textbf{Ans.}$$

Given, {(x, y) : 0 ≤ y ≤ x2 , 0 ≤ y ≤ x, 0 ≤ x ≤ 2}

⇒ y ≤ x2 ...(i)

y ≤ x ...(ii)

x ≤ 2 ...(iii)

x ≥ 0 ...(iv)

y ≥ 0 ...(v)

Considering inequalities as equation :

y = x2, y = x, x = 2, x = 0, y = 0

Solving y = x2 and y = x

⇒ x2 = x

⇒ x(x – 1) = 0

⇒ x = 0, x = 1

∴ y = 0, y = 1

∴ Points of intersection of curve (i) and the (ii) are (0, 0) and (1, 1).

$$\therefore\space\text{Required area =}\\\int^{1}_{0}(\text{y of the parabola})dx + \\\int^{2}_{1}(\text{y of the line})dx \\\therefore\space \text{Required area =}\\\int^{1}_{0}x^{2}dx + \int^{2}_{1}x dx\\=\begin{bmatrix}\frac{x^{3}}{3}\end{bmatrix}^{1}_{0} + \begin{bmatrix}\frac{x^{2}}{2}\end{bmatrix}^{2}_{1}\\=\bigg(\frac{1}{3}-0\bigg) + \bigg(\frac{4}{2}-\frac{1}{2}\bigg)\\=\frac{1}{3} + \bigg(2 -\frac{1}{2}\bigg)\\ =\frac{1}{3} +\frac{3}{2} =\frac{2 + 9}{6}\\=\frac{11}{6}\space\text{square units.}$$

## Section-C

(i) (b) ₹30.015

C(x) = 0.005x3 – 0.02x2 + 30x + 5000

$$\text{M.C.} =\frac{d}{dx}\lbrace\text{(C(x))}\rbrace \\=\frac{d}{dx}(0.005 x^{3} -0.02x^{2} + 30x + 5000)$$

= 0.015x2 – 0.04x + 30 + 0

∴ M.C. = 0.015x2 – 0.04x + 30

(M.C.)x = 3 = 0.015(3)2 – 0.04(3) + 30

= 0.135 – 0.120 + 30

= 30.015

∴ (M.C.)x = 3 = ₹30.015. Ans.

(iii) C(x) = 0.007x3 – 0.003x2 + 15x + 4000

$$\text{M.C.}=\frac{d}{dx}\lbrace\text{C(x)}\rbrace =\\\frac{d}{dx}(0.007 x^{3} - 0.003x^{2} + 15x + 4000)$$

M.C. = 0.021x2 – 0.006x + 15 + 0

MC at x = 17 M.C. = 0.021(17)2 – 0.006(17) + 15

M.C. = 0.021(289) – 0.006(17) + 15

M.C. = 6.069 – 0.102 + 15

M.C. = ₹20.967. Ans.

(iv) (c) Regresion

(v) Given, X = 0.85Y and Y = 0.89X

∴ bxy = 0.85 and byx = 0.89

Coefficient of correlation is given as,

$$r =\pm\sqrt{b_{xy}×b_{yx}}\\=\pm\sqrt{0.85×0.89}\\=\pm\sqrt{0.7565}\\=\pm 0.87\\\because\space b_{xy}.b_{yx}\gt 0\\\therefore\space r =0.87\\\textbf{Ans.}$$

(i) Given, Cost price of x items =

$$₹\bigg(\frac{x}{5} + 500\bigg)\\\text{Selling price of x items =}\\₹\bigg(5 -\frac{x}{100}\bigg)x\\= ₹\bigg(5x-\frac{x^{2}}{100}\bigg)\\\therefore\text{Profit = S.P. – C.P.}\\ = ₹\bigg(5x -\frac{x^{2}}{100}-\frac{x}{5}-500\bigg)\\\text{P(x)} =\frac{24x}{5}-\frac{x^{2}}{100}-500\\$$

Differentiating w.r.t. x, we get

$$\frac{\text{dP}}{\text{dx}} =\frac{d}{dx}\bigg(\frac{24x}{5}\bigg) -\\\frac{d}{dx}\bigg(\frac{x^{2}}{100}\bigg)-\frac{d}{dx}(500)\\\frac{\text{dP}}{\text{dx}} =\frac{24}{5}-\frac{x}{50}-0\\\text{For P to be maximum,}\\\frac{\text{dP}}{\text{dx}} =0\\\therefore\space \frac{24}{5}-\frac{x}{50} = 0\\\Rarr\space \frac{x}{50} = \frac{24}{5}\\\Rarr\space x =\frac{24×50}{5} = 240$$

$$\Rarr\space \frac{\text{d}^{2}\text{P}}{dx^{2}} =\frac{d}{dx}\bigg(\frac{24}{5}\bigg)-\\\frac{d}{dx}\bigg(\frac{x}{50}\bigg)\\\Rarr\space \frac{\text{d}^{2}\text{P}}{\text{dx}^{2}} =\frac{\normalsize-1}{50}\\\because\space \frac{\text{d}^{2}\text{P}}{\text{dx}^{2}}\lt 0$$

∴ Profit is maximum when 240 items are sold.

Maximum profit at x = 240,

$$\text{P(240)}= \frac{24}{5}×240-\\\frac{240 × 240}{100}-500$$

= 24 × 48 – 576 – 500

= 1152 – 1076

= ₹76

Hence, maximum profit is ₹76. Ans.

OR

(ii) Let the annual subscription be increased by ₹ x.

∴ Charges per subscriber = ₹ (300 + x)

and Number of subscribers = (500 – x)

Annual income (I) = ₹ (500 – x)(300 + x)

= ₹ (150000 + 200x – x2)

I = 150000 + 200x – x2

Differentiating w.r.t. x, we get

$$\frac{\text{dI}}{\text{dx}} =\frac{d}{dx}(150000)+ \\\frac{d}{dx}(200x) -\frac{d}{dx}(x^{2})\\\Rarr\space \frac{\text{dI}}{\text{dx}} = 0 + 200 - 2x\\\text{For income to be maximum,}\\\frac{\text{dI}}{dx} =0\\\Rarr\space 200-2x = 0\\\Rarr\space x =100\\\Rarr\space \frac{d^{2}\text{I}}{dx^{2}} =\frac{d}{dx}(200 - 2x)\\\Rarr\space\frac{d^{2}\text{I}}{dx^{2}} = 0-2 =-2$$

$$\Rarr\space \frac{d^{2}\text{I}}{dx^{2}}\lt 0$$

Hence, annual income is maximum at increment of ₹100.

New annual income = ₹400 × 400 = ₹160000

Old annual income = ₹500 × 300 = ₹150000

Increase in annual income = ₹10000.

Ans.

 Marks in Physics x Marks in Chemistry y dx = x – 40 dy = y – 36 (dx)2 (dy)2 dxdy 46 40 6 4 36 16 24 42 38 2 2 4 4 4 44 36 4 0 16 0 0 40 35 0 -1 0 1 0 43 39 3 3 9 9 9 41 37 1 1 1 1 1 45 41 5 5 25 25 25 Σdx = 21 Σdy = 14 Σ(dx)2 = 14 Σ(dy)2 = 56 Σdxdy = 63

$$\text{Here, n = 7}\\ b_{xy} =\frac{n\Sigma dx.dy - \Sigma dx\Sigma dy}{n\Sigma (dy)^{2} - (\Sigma dy)^{2}}\\ b_{xy} =\frac{7×63 - 21×14}{7×56-(14)^{2}}\\ b_{xy} =\frac{441 - 294}{392 - 196}\\\therefore\space b_{xy} = \frac{147}{196} = 0.75\\ b_{yx} = \frac{n\Sigma dx.dy - \Sigma dx\Sigma dy}{n\Sigma(dx)^{2} -(\Sigma dx)^{2}}\\ b_{yx} =\frac{7×63 - 21×14}{7×91-(21)^{2}}\\=\frac{441 - 294}{637 - 441} = \frac{147}{196} = 0.75$$

$$\bar{x} = \text{A} +\frac{\Sigma dx}{n}, \\\bar{y} =\text{A} + \frac{\Sigma dy}{n}\\\bar{x} = 40 + \frac{21}{7} = 43,\\\bar{y} = 36 +\frac{14}{7} = 38\\\text{Equation of x on y}\\ x -\bar{x} = b_{xy}(y-\bar{y})$$

x – 43 = 0.75(y – 38)

x = 0.75y – 28.5 + 43

x = 0.75y + 14.5

Equation of y on x

$$y -\bar{y} = b_{yx}(x- \bar{x})\\ y- 38 = 0.75(x- 43)$$

y = 0.75x – 32.25 + 38

y = 0.75x + 5.75

Hence, the regression coefficients are 0.75 and 0.75 and the regression equations are x = 0.75y + 14.5 and y = 0.75x + 5.75.

Ans.

(i) Let the number of type A cake made be x and the number of type B cake made be y.

To maximise the number of cakes.

Z = x + y

Subject to constraint :

200x + 100y ≤ 5000

⇒ 2x + y ≤ 50 ...(i)

Also 25x + 50y ≤ 1000

⇒ x + 2y ≤ 40 ...(ii)

and x ≥ 0, y ≥ 0

Consider, 2x + y = 50

 x 0 25 10 y 50 0 30

x + 2y = 40

 x 0 40 10 y 20 0 15

The feasible region is the shaded region.

 Corner Points Z = x + y A (25, 0), Z = 25 + 0 = 25 B (20, 10), Z = 20 + 10 = 30 C (0, 20), Z = 0 + 20 = 20

Hence, maximum number of cakes = 20 + 10 = 30.

Ans.

OR

(ii) We have,

Maximise Z = 34x + 45y

Subject to the constraints:

x + y ≤ 300

2x + 3y ≤ 70

x ≥ 0, y ≥ 0

Converting the given inequalities into equations, we obtain the following equations:

x + y = 300

2x + 3y = 70

Then, x + y = 300 and

 x 0 300 y 300 0

2x + 3y = 70

 x 0 35 y $$\frac{70}{3}$$ 0

Plotting these points on the graph, we get the shaded feasible region i.e., OCDO.

 Corner point Value of Z = 34x + 45y O (0, 0) 34(0) + 45(0) = 0 C (35, 0) 34(35) + 45(0) = 1190 D (0, 70/3) 34(0) + 45(70/3) = 1050

Clearly, the maximum value of Z is 1190 at (35, 0).