# Oswal 36 Sample Question Papers ISC Class 12 Maths Solutions

## 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) (b)

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.

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

Explanation :

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

$$\therefore\space\text{sin}\begin{bmatrix}\frac{\pi}{2}-\bigg(-\frac{\pi}{3}\bigg)\end{bmatrix}\\=\text{sin}\bigg(\frac{\pi}{2} + \frac{\pi}{3}\bigg)\\\text{cos}\frac{\pi}{3} =\frac{1}{2}$$

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

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

Explanation :

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

(vi) (b) 3

Explanation :

$$\text{Let\space A =}\begin{bmatrix}1 &2\\k &6\end{bmatrix}$$

∴  A– 1 does not exist if | A | = 0

$$\therefore\qquad\begin{vmatrix}1 &2\\k &6\end{vmatrix} = 0$$

$$\Rarr\space\text{6 – 2k = 0}\\\Rarr\space\text{2k = 6}\\\Rarr\space\text{k = 3.}$$

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

Explanation :

Given curve x2 + 3y + y2 = 5

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

(viii) (c) x + y = 0

Explanation :

Given : Curve y = sin x

$$\therefore\space\frac{dy}{dx} =\text{cos x}$$

$$\Rarr\space\frac{dy}{dx}\biggm\vert_{(0,0)} = 1\\\Rarr\space\text{Slope of normal is – 1.}\\\therefore\space\text{Equation is,}\\\text{y – 0 = – 1(x – 0).}\\\Rarr\space\text{x + y = 0.}$$

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

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

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

P(E1) > P(E2)

Assertion is true.

Reason is correct explanation of Assertion.

(xi) Consider,

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

= R.H.S. Hence Proved.

(xii) Given | A | = 4

$$\therefore\space\text{|-2A|} =(-2)^{3}|\text{A}|$$

= – 8 × 4 = – 32. Ans.

(xiii) (b) 1 and 2

Explanation :

x = 1 and x = 2 is non differentiable.

(xiv) Here,

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

(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)}

⇒ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

(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)

⇒ x21 + 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

$$\Rarr\space x^{2}= y-4\\\text{i.e\space} x =\sqrt{y-4}\\\text{Also,\space}x = f^{\normalsize-1}(y)\\\text{f}^{\normalsize-1}(y) =\sqrt{y-4}\space\\\textbf{Hence Proved.}$$

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

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

(i) Given,

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

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 sin3t

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

Also, y = a cos3t

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

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

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

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

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

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

Hence Proved.

Given function is

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

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

$$\Rarr\space x=2,4,\normalsize-3$$

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

$$\text{P(A)} =\frac{n(A)}{n(S)} =\frac{1}{4}\\\textbf{Ans.}$$

(ii) P(B does not throw two heads) =

$$1 -\frac{1}{4} =\frac{3}{4}.\space\textbf{Ans.}$$

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

$$\text{P(A wins the game)}=\\\frac{1}{4} + \bigg(\frac{3}{4}\bigg)^{2}×\frac{1}{4} +\bigg(\frac{3}{4}\bigg)^{4}×\frac{1}{4} +.....\\=\frac{1}{4}\begin{bmatrix}1 + \bigg(\frac{3}{4}\bigg)^{2} + \bigg(\frac{3}{4}\bigg)^{4} + ......\end{bmatrix}\\=\frac{1}{4}\begin{bmatrix}\frac{1}{1 -\frac{9}{16}}\end{bmatrix}\\\begin{bmatrix}\because\space\text{For infinite G.P.S.} =\frac{a}{1-r}\end{bmatrix}\\=\frac{1}{4}×\frac{16}{7} =\frac{4}{7}\qquad\textbf{Ans.}$$

(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) =

$$\frac{3}{4}×\frac{1}{4} +\bigg(\frac{3}{4}\bigg)^{3}×\frac{1}{4} +\\\bigg(\frac{3}{4}\bigg)^{5}×\frac{1}{4} + .......\\=\frac{3}{16} +\frac{3}{16}×\bigg(\frac{3}{4}\bigg)^{2}+\\\frac{3}{16}×\bigg(\frac{3}{4}\bigg)^{4} +........\\=\frac{3}{16}\begin{bmatrix}1 +\bigg(\frac{3}{4}\bigg)^{2} + \bigg(\frac{3}{4}\bigg)^{4} +.....\end{bmatrix}\\=\frac{3}{16}×\frac{1}{1-\frac{9}{16}} =\frac{3}{16}×\frac{16}{7}\\=\frac{3}{7}\qquad\textbf{Ans.}$$

(i) Let

$$\text{I =}\frac{1}{2}\int(\text{2 cos 2x cos 4x})\\\space\text{cos 6x}\space dx\\\Rarr\space\text{I =}\frac{1}{2}\int\text{(cos 6x + cos 2x)}\\\text{cos 6x dx}$$

[ 2 cos A cos B = cos (A + B) + cos (A – B)]

$$=\frac{1}{2}\int\text{(cos 6x cos6x dx)} +\\\frac{1}{2}\int\text{cos 2x cos6x dx}\\=\frac{1}{4}\int\text{2 cos 6x cos 6x dx} +\\\frac{1}{4}\int\text{2 cos 2x cos6x dx}\\=\frac{1}{4}\int(\text{cos 12 x + cos 0})dx +\\\frac{1}{4}\int\text{cos8x cos 4x dx}\\=\frac{1}{4}\\\begin{bmatrix}\int(\text{cos 12 x +1 +cos 8x +cos 4x})dx\end{bmatrix}$$

$$=\frac{1}{4}\begin{bmatrix}\frac{\text{sin 12 x}}{12} + \frac{\text{sin 8 x}}{8} + \frac{\text{sin 4 x}}{4}\end{bmatrix}\\+\space\text{C}\space\textbf{Ans.}$$

OR

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

= 2 sin x + 2x cosα + C Ans.

(i) Given differential equation is

$$\frac{dy}{dx} = 1+x^{2}+ y^{2} +x^{2}y^{2}$$

= (1 + x2) + y2 (1 + x2)

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

On integrating both sides, we have

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

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

tan–1 1 = 0 + 0 + C

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

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

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

On integrating both sides, we have

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

In first integral,

put 1 + log x = t

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

Also in second integral,

put 1 – y2 = u

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

It is given that y = 0 when x = 1

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

or (1 + log x)2 – 2 log|1 – y2| = 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:

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

A11 = 8, A12 = 4

A21 = 5, A22 = 5

$$\therefore\space\text{adjA =}\begin{bmatrix}8 &4\\5 &5\end{bmatrix}$$

and | A | = 40 – 20 = 20

$$\Rarr\space X =\frac{1}{20}\begin{bmatrix}8 &4\\5 &5\end{bmatrix}\begin{bmatrix}40\\80\end{bmatrix}$$

$$\Rarr\space\text{X} =\frac{1}{20}\begin{bmatrix}640\\600\end{bmatrix}\\\Rarr\space\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}32\\30\end{bmatrix}$$

$$\Rarr\space\text{x = 32, y = 30}$$

Number of children = 32 Ans.

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

(i) Let ex3 = t

∴ 3x2 ex3 dx = dt

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

OR

$$\text{(ii)\space}\int x(\text{tan}^{\normalsize-1}x)^{2}dx\\\text{Let}\space\text{tan}^{\normalsize-1}x = t\\\Rarr\space x =\text{tan t}$$

dx = sec2 t dt

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

$$=\frac{1}{2}t^{2}(\text{tan}^{2}t +1) -\\\begin{Bmatrix}\text{t tan t} -\\\int\text{tan t} dt\end{Bmatrix}\\=\frac{1}{2}t^{2}(\text{tan}^{2} + 1) -\\\text{t tan t + log}|\text{sec t}| +\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2} -\\x\text{tan}^{\normalsize-1}x + \text{log}(\sqrt{1 + x}^{2}) +\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2} - x\space\text{tan}^{\normalsize-1}x+\\\text{log}(\sqrt{1 + x^{2}})+\text{C}\\=\frac{1}{2}(1 + x^{2})(\text{tan}^{\normalsize-1}x)^{2}-$$

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

(i) Given,

$$(1 + x^{2})\frac{dy}{dx} \\=(e^{m} tan^{\normalsize-1}x - y)\\\frac{dy}{dx} =\frac{e^{m tan^{\normalsize-1}x}}{1 +x^{2}}-\frac{y}{1 +x^{2}}\\\Rarr\space\frac{dy}{dx} + \frac{1}{1 + x^{2}}y\\=\frac{e^{m tan^{\normalsize-1}x}}{1 +x^{2}}$$

This equation is of the form

$$\frac{dy}{dx} +\text{Py = Q(x)}\\\text{where P}=\frac{1}{1 +x^{2}}\text{and}\space\\\text{Q(x)} =\frac{e^{m tan^{\normalsize-1}x}}{1 + x^{2}}\\\text{I.F. = e}^{\int\text{P dx}} =\int e^{\bigg(\frac{1}{1+x^{2}}\bigg)dx}\\= e^{\text{tan}\normalsize-1\space x}$$

Hence, solution of linear differential equation is given by

$$y×\text{I.F. =}\int\text{I.F × Q(x)dx}\\\Rarr\space y×e^{\text{tan}^{\normalsize-1}x} =\\\int e^{\text{tan}^{\normalsize-1}x}.\frac{e^{m\space tan^{-1}x}}{1 +x^{2}}dx\\\Rarr\space y× e^{tan^{\normalsize-1}x} \\=\int\frac{e^{(1 + m)tan^{\normalsize-1}x}}{1 +x^{2}}dx$$

Let e(1 + m) tan -1 x = t

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

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

$$\therefore\space\text{1× e}^{\text{tan}^{\normalsize-1}0}\\=\frac{1}{\text{1 + m}}e^{(1+m)\text{tan}^{\normalsize-1}0} +\text{C}\\\Rarr\space 1×e^{0} =\frac{1}{1+m}e^{0} +\text{C}\\\Rarr\space1 -\frac{1}{\text{1 -m}} =\text{C}\\\therefore\space\text{C} = \frac{m}{m+1}$$

Hence, particular solution of given differential equation is

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

[where m ≠ – 1] Ans.

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

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

When x = 0, then y = 1,

∴ 1 × etan-1 0 = tan–1 0 + C

$$\Rarr\space 1×e^{0} = 0 +\text{C}\\\Rarr\space\text{C = 1}$$

Hence, particular solution of differential equation is

y × etan−1x = tan−1 x+1. 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 ΔABC.

∴ 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 ΔAIE, ∠AEI = 90°

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

AE = r cot θ = x ...(ii)

$$\frac{y}{x+y} =\text{sin}\space\theta\\\Rarr\space\text{x sin θ + y sin θ = y}\\\Rarr\space\text{r cot}\space\theta\space\text{sin}\space\theta = y(\text{1 - sin}\space\theta)\\\lbrack\text{Using (ii)}\rbrack\\\Rarr\space\frac{\text{r cos}\theta}{\text{1 - sin}\theta} = y\space\text{...(iii)}$$

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

P = 2x + 4y

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

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

For maxima and minima,

$$\text{Put,\qquad}\frac{\text{dP}}{\text{d}\theta} = 0\\\therefore\qquad\frac{\text{2r(2 sin}\theta-1)(\text{sin}\space\theta +1)}{\text{sin}^{2}\theta(\text{1 - sin}\space\theta)}\\=0$$

2r(2sinθ − 1)(sinθ + 1) = 0

∴ 2 sin θ – 1 = 0

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

or sin θ + 1 = 0

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

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

$$= 4r(4.\sqrt{3} + 2\sqrt{3})\\\therefore\space\bigg(\frac{d^{2}\text{P}}{d\theta}\bigg)_{\theta =\frac{\pi}{6}}\gt0$$

Hence, perimeter is minimum at π/6.

Now, Perimeter of ΔABC = 2x + 4y

$$= \text{2r cot}\theta + \frac{\text{4r cos}\theta}{\text{1 - sin}\theta}\\=\text{2r cot}\frac{\pi}{6} + \frac{\text{4r cos}\frac{\pi}{6}}{\text{1 - sin}\frac{\pi}{6}}\\= \text{2r.}\sqrt{3} +\frac{4r.\frac{\sqrt{3}}{2}}{1 -\frac{1}{2}}\\= r(2\sqrt{3} + 4\sqrt{3}) = 6\sqrt{3r}$$

$$\therefore\space\text{Least perimeter of}\\\Delta\text{ABC is}\space 6\sqrt{3r}\space\text{units.}$$

Hence Proved.

(i) n(W|A) = 2

n(S) = 5

$$\text{P(W|A)} =\frac{2}{5}\\\text{Similarly}\\\text{P(W|B) =}\frac{1}{5},\space\text{P(W|C)} =\frac{4}{5}$$

$$\text{(ii)\space Given P(A) =}\frac{2}{5},\space\text{P(B) =}\frac{2}{5},\\\text{P(C) =}\frac{1}{5}$$

P(Probability that the drawn ball is white) = P(A).P(W|A) + P(B).P (W|B) + P(C).P(W|C)

( Theorem of total probability)

$$= \frac{2}{5}×\frac{2}{5} + \frac{2}{5}×\frac{1}{5} +\frac{1}{5}×\frac{4}{5}\\=\frac{4}{25} +\frac{2}{25} +\frac{4}{25}\\=\frac{10}{25} = \frac{2}{5}\qquad\textbf{Ans.}$$

(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)

$$\text{P(E) =}\frac{2}{5}×\frac{3}{5} + \frac{2}{5}×\frac{4}{5} +\frac{1}{5}×\frac{1}{5}\\\text{P(E) =}\frac{6}{25} +\frac{8}{25} +\frac{1}{25}\\=\frac{15}{25} =\frac{3}{5}\qquad\textbf{Ans.}$$

(iv) By Baye’s theorem, we have

$$\text{P(C|W) =}\\\frac{\text{P(C)P(W|C)}}{\text{P(A)P(W|A)} + \text{P(B)P(W|B) + P(C)P(W|C)}}\\=\\\frac{\frac{1}{5}×\frac{4}{5}}{\frac{2}{5}×\frac{2}{5} + \frac{2}{5}×\frac{1}{5} + \frac{1}{5}×\frac{4}{5}}\\=\frac{4}{10} =\frac{2}{5}\qquad\textbf{Ans.}$$

## 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,

$$\text{cos}\space\theta =\begin{vmatrix}\frac{a_{1}a_{2} +b_{1}b_{2} +c_{1}c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} +c_{1}^{2}}\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}\end{vmatrix}$$

Here a1= 2, b1 = – 1, c1 = 1

and a2 = 1, b2 = 1, c2 = 2

$$\therefore\space\text{cos}\space\theta = \begin{vmatrix}\frac{2(1) + (\normalsize-1)(1) + (1)(2)}{\sqrt{2^{2} + 1^{2} +1^{2}}\sqrt{1^{2} + 1^{2} + 2^{2}}}\end{vmatrix}\\\text{cos}\space\theta =\begin{vmatrix}\frac{2-1+2}{\sqrt{6}\sqrt{6}}\end{vmatrix}=\frac{1}{2}\\\therefore\theta =60\degree =\frac{\pi}{3}$$

$$\text{(iii)\space For}\space\bigg(\frac{1}{\sqrt{2}},\frac{1}{2},k\bigg)\space\text{to represent}$$

direction cosines, we should have

$$\bigg(\frac{1}{\sqrt{2}}\bigg)^{2} + \bigg(\frac{1}{2}\bigg)^{2}+ k^{2}=1\\\text{or}\space\frac{1}{2} + \frac{1}{4}+k^{2} =1\\\Rarr\space k =\pm\frac{1}{2}.\\\text{(iv)\space}\vec{a} =2\hat{i} -\hat{j} + \hat{k}, \vec{b} =\hat{i} + 2\hat{j} - 3\hat{k}\\\text{and}\space\vec{c} = 3\hat{i} + x\hat{j} +5\hat{k}\\\text{If\space}\vec{a},\vec{b}\space\text{and}\space\vec{c}\space\text{are coplanar, then}\\\begin{vmatrix}2 &\normalsize-1 &1\\1 &2 &\normalsize-3\\ 3 &x &5\end{vmatrix} = 0\\\Rarr\space 2\begin{vmatrix}2 &\normalsize-3\\x &5\end{vmatrix} + 1\begin{vmatrix}1 &\normalsize-3\\3 &5\end{vmatrix} +\\ 1\begin{vmatrix}1 &2\\ 3 &x\end{vmatrix} = 0$$

$$\Rarr\space\text{2(10 + 3x) + (5 + 9)}\\\text{+ (x - 6)} = 0\\\Rarr\space\text{20 + 6x + 14 + x – 6 = 0}\\\Rarr\space\text{7x = 28}\\\therefore\space\text{x = 4.}\qquad\textbf{Ans.}\\\text{(v)\space}\vec{\text{OP}×\text{OQ}} =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 &2 &1\\2 &1 &3\end{vmatrix}\\=\space 5\hat{i} -\hat{j} -3\hat{k}\\\vec{\text{PQ}}×\vec{\text{PR}} =\begin{vmatrix}\hat{i} &\hat{j}&\hat{k}\\1 &\normalsize-1 &2\\\normalsize-2 &\normalsize-1 &1\end{vmatrix}\\=\hat{i} -5\hat{j} - 3\hat{k}$$

Required angle cos θ =

$$\frac{5+5+9}{\sqrt{25 +1 +9}\sqrt{1 + 25 +9}}\\\text{cos}\space\theta =\frac{19}{35}\\\theta =\text{cos}^{\normalsize-1}\bigg(\frac{19}{35}\bigg)\space\textbf{Ans.}$$

(i) The given lines are

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

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

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

$$(\normalsize-5)\begin{vmatrix}\normalsize-3 &2\\1 &\normalsize-3\end{vmatrix} = 0\\\Rarr\space\text{(4+3) - 4(-6-1) -5(9-2)}\\= 0\\\Rarr\space\text{7 + 28 -35} = 0$$

Hence, the given lines are coplanar.

Hence Proved.

Equation of the plane containing these lines is

$$\begin{vmatrix}x -x_{1} &y-y_{1} &z-z_{1}\\x_{2} -x_{1} &y_{2}-y_{1} &z_{2}-z_{1}\\a_{1} &b_{1} &c_{1}\end{vmatrix}\\= 0\\\begin{vmatrix}x-(\normalsize-1) &y-3 &z-(\normalsize-2)\\0-(\normalsize-1) &7-3 &-7-(\normalsize-2)\\\normalsize-3 &2 &1\end{vmatrix}\\= 0\\\begin{vmatrix}x+1 &y-3 &z+2\\1 &4 &\normalsize-5\\\normalsize-3 &2 &1\end{vmatrix}\\= 0\\\Rarr\space (x+1)\begin{vmatrix}4 &\normalsize-5\\2 &1\end{vmatrix}-(y-3)\begin{bmatrix}1 &\normalsize-5\\\normalsize-3 &1\end{bmatrix} +$$

$$(z+2)\begin{vmatrix}1 &4\\\normalsize-3 &2\end{vmatrix} = 0$$

$$\Rarr\space (x+1)(4+10)-(y-3)(1- 15) +\\(z+2)(2+12) = 0\\\Rarr\space 14(x+1)-(\normalsize-14)(y-3)\\+ 14(z+2) = 0\\\Rarr\space\text{14x + 14 + 14y – 42 +}\\\text{14z + 28 = 0}\\\Rarr\space\text{14x + 14y +14z = 0}\\\Rarr\space\text{x + y +z} = 0$$

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

OR

(ii) Given line is

5x – 25 = 14 – 7y = 35z

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

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

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

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

$$\therefore\space\vec{\text{AB}} = \hat{i} -3\hat{j} +\hat{k}\\\text{and}\space\vec{\text{AC}} = 3\hat{i} + 3\hat{j}-4\hat{k}$$

We know that,

$$\text{ar}(\Delta\text{ABC}) =\frac{1}{2}\begin{vmatrix}\vec{\text{AB}}×\vec{\text{AC}}\end{vmatrix}\\\text{Now,\qquad}\vec{\text{AB}}×\vec{\text{AC}} =\\\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\ 1 &\normalsize-3 &1\\3 &3 &\normalsize-4\end{vmatrix}\\=\space\hat{i}(12-3) -\hat{j}(-4-3) +\hat{k}(3+9)\\ =9\hat{i} + 7\hat{j}+12\hat{k}\\\therefore\space|\vec{\text{AB}}×\vec{\text{AC}}| =\sqrt{9^{2} + 7^{2} + (12)^{2}}\\=\sqrt{274}\\\text{So,}\qquad ar(\Delta \text{ABC})=\\\frac{1}{2}\sqrt{274}\space\text{sq. units}\qquad\textbf{Ans.}$$

OR

(ii) Here,

$$\vec{a} =(\hat{i} +\hat{j} +\hat{k}).\hat{n}\\\text{is unit vector}\\\vec{b} = 2\hat{i} + 4\hat{j} -5\hat{k}\\\vec{c} =\lambda\hat{i} +2\hat{j} +3\hat{k}\\\vec{b} +\vec{c} =(2 +\lambda)\hat{i} + 6\hat{j}-2\hat{k}\\\text{Then,\space}\hat{n} =\frac{(2+\lambda)\hat{i} + 6\hat{j}-2\hat{k}}{\sqrt{(2 + \lambda)^{2} +36 +4}}\\\text{Given,\space}\vec{a}.\hat{n} = 1\\(\hat{i} + \hat{j} +\hat{k}).\begin{pmatrix}\frac{(2 +\lambda)\hat{i} +6\hat{j}-2\hat{k}}{\sqrt{(2 +\lambda)^{2} +40}}\end{pmatrix}\\= 1\\\Rarr\space(2 +\lambda) +6-2 =\\\sqrt{(2 + \lambda)^{2} +40}$$

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

On squaring both sides, we get

(6 + λ)2 = (2 + λ)2 + 40

$$\Rarr\space 36 +\lambda^{2} + 12\lambda =\\ 4 + \lambda^{2} +4\lambda+40\\\Rarr\space 36 +12\lambda -4-4\lambda -40 = 0\\\Rarr\space 8\lambda -8 = 0\\\Rarr\space\lambda = 1\\\text{Then,\space}\vec{b} + \vec{c} =\\(2 +\lambda)\hat{i} + 6\hat{j} -2 \hat{k}\\ = 3\hat{i} +6\hat{j}-2\hat{k}\\\text{unit vector along}\space(\vec{b} +\vec{c})\\=\frac{3\hat{i} +6\hat{j}-2 \hat{k}}{\sqrt{49}}\\=\frac{3\hat{i} +6\hat{j}-2\hat{k}}{7}\qquad\textbf{Ans.}$$

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).

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

$$=\frac{1}{3} +\frac{3}{2} =\frac{2 +9}{6}\\=\frac{11}{6}\space\text{square units.}$$

## Section-C

(i) (b) ₹30.015

C(x) = 0.005x3 – 0.02x2 + 30x + 5000

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

= 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. (b)

(ii) (a)

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

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

MC at x = 17

M.C. = 0.021x2 – 0.006x + 15 + 0

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,

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

(i) Given, Cost price of x items =

$$₹\bigg(\frac{x}{5} +500\bigg)\\\text{Selling price of x items =}\\₹\bigg(5 -\frac{x}{100}\bigg)x\\= ₹\bigg(5x -\frac{x^{2}}{100}\bigg)$$

∴ Profit = S.P. – C.P.

$$= ₹\bigg(5x -\frac{x^{2}}{100}-\frac{x}{5}-500\bigg)\\\text{P(x) =}\frac{24 x}{5} -\frac{x^{2}}{100}-500$$

Differentiating w.r.t. x, we get

$$\frac{dP}{dx} =\frac{d}{dx}\bigg(\frac{24x}{5}\bigg) -\frac{d}{dx}\bigg(\frac{x^{2}}{100}\bigg)-\\\frac{d}{dx}(500)\\\frac{\text{dP}}{\text{dx}} =\frac{24}{5} -\frac{x}{50}-0$$

For P to be maximum,

$$\frac{dP}{dx} =0\\\therefore\space\frac{24}{5}-\frac{x}{50} = 0\\\Rarr\space\frac{x}{50} = \frac{24}{5}\\\Rarr\space x =\frac{24×50}{5} = 240\\\Rarr\space\frac{d^{2}\text{P}}{dx^{2}} =\frac{d}{dx}\bigg(\frac{24}{5}\bigg) -\frac{d}{dx}\bigg(\frac{x}{50}\bigg)\\\Rarr\space\frac{d^{2}\text{P}}{dx^{2}} =\frac{\normalsize-1}{50}\\\because\space\frac{d^{2}\text{P}}{\text{dx}^{2}}\lt0$$

∴ Profit is maximum when 240 items are sold.

Maximum profit at x = 240,         Ans.

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

= 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

$$\frac{\text{dI}}{\text{dx}} =\frac{\text{d}}{\text{dx}}(1500000) +\frac{d}{dx}(200 x)-\\\frac{d}{dx}(x^{2})\\\Rarr\space\frac{\text{dI}}{\text{dx}} =0 +200-2x$$

For income to be maximum,

$$\frac{\text{dI}}{\text{dx}} = 0\\\Rarr\space\text{200 - 2x = 0}\\\Rarr\space\space\text{x = 100}\\\Rarr\space\frac{d^{2}\text{I}}{\text{dx}^{2}} =\frac{d}{dx}(200 - 2x)\\\Rarr\space\frac{d^{2}\text{I}}{\text{dx}^{2}} = 0-2= -2\\\Rarr\space\frac{d^{2}\text{I}}{dx^{2}}\lt 0$$

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 = 91 Σ(dy)2 = 56 Σdxdy = 63

Here, n = 7

$$\text{b}_{xy} =\frac{n\Sigma dx.dy - \Sigma dx\Sigma dy}{n\Sigma(dy)^{2} -(\Sigma dy)^{2}}\\ b_{xy} =\frac{7×63-21×4}{7×56-(14)^{2}}\\b_{xy} =\frac{\text{441 - 294}}{\text{392 - 196}}\\\text{b}_{xy} =\frac{147}{196} = 0.75\\ b_{yx} =\frac{n\Sigma dx.dy -\Sigma dx\Sigma dy}{n\Sigma(dx)^{2} -(\Sigma dx)^{2}}\\b_{yx} =\frac{7× 63-21×14}{7×91-(21)^{2}}\\=\frac{441 -294}{637 - 441} =\frac{147}{196}\\= 0.75$$

$$\bar{x} = A +\frac{\Sigma dx}{n},\space\bar y = A +\frac{\Sigma dy}{n}\\\bar {x} = 40 +\frac{21}{7} = 43,\\\bar{y} = 36 +\frac{14}{7} = 38$$

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

$$\Rarr\space\text{2x+y}\le 50\\\text{...(i)}\\\text{Also\space 25x + 50 y}\leq 1000\\\Rarr\space \text{x +2y}\leq 40\space\text{...(ii)}\\\text{and}\space \text{x}\geq0, y\geq0$$

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).

Ans.

#### ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2024

#### ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2024