# Oswal Specimen Papers ISC Class 12 Mathematics Solutions (Specimen Paper - 9)

## Section-A

Solution 1.

(i) (c) aRb ⟺ a < b

Explanation :

aRb ⟺ a < b

a < b then b > a

∴ It is not symmetric.

∴ It is not an equivalence relation.

(ii) (b) Symmetric

Explanation :

We observed in figure

R = {(L1, L2) : L1 is perpendicular to L2}

R is not reflexive, as a line L1 can not be perpendicular to itself.

R is symmetric as (L1, L2) ∈ R

R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3, then L1 can never be perpendicular to L3. In fact L1 is parallel to L3.

$$\text{(iii)\space (a)}\space\sqrt{\frac{1 + x^{2}}{2 + x^{2}}}$$

Explanation :

sin cot–1 (cos tan– 1 x)

Let tan–1 x = y ➾ x = tan y

x2 = tan2 y

x2 = sec2 y – 1

$$\therefore\space \text{sec y} =\sqrt{1 + x^{2}}$$

Now, sin cot–1 (cos tan– 1 x) = sin cot–1 (cos y)

$$=\text{sin cot}^{-1}(\text{cos y})\\=\text{sin cot}^{\normalsize-1}\bigg(\frac{1}{\sqrt{1 + x^{2}}}\bigg)\\\text{Let,\space}\text{cot}^{\normalsize-1}\bigg(\frac{1}{\sqrt{1 + x^{2}}}\bigg) =\theta\\\therefore\space \frac{1}{\sqrt{1 +x^{2}}} =\text{cot}\theta\\=\frac{1}{1 + x^{2}}=\text{cot}^{2}\theta\\=\text{cosec}^{2}\theta-1\\1 + \frac{1}{1 + x^{2}} =\text{cosec}^{2}\theta\\\frac{2 +x^{2}}{1 +x^{2}}=\text{cosec}^{2}\theta$$

$$\Rarr\space \text{sin}\theta =\sqrt{\frac{1 +x^{2}}{2 + x^{2}}}$$

(iv) (d) a11 c11 + a22 c22 + a33 c33

Explanation :

|A| = a11 c11 + a22 c22 + a33 c33

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

Explanation :

$$\text{A} =\begin{bmatrix}i &0\\0 &i\end{bmatrix}\\\text{A}^{2} =\begin{bmatrix}i &0\\0 &i\end{bmatrix}\begin{bmatrix}i &0\\0 &i\end{bmatrix}\\=\begin{bmatrix}i^{2}+0 &0 +0\\0+0 &0+i^{2}\end{bmatrix}\\\text{A}^{2} =\begin{bmatrix}-1 &0\\0 &-1\end{bmatrix}$$

∴ A4 = I

$$\text{Hence\space}(A)^{4n}=\begin{bmatrix}1 &0\\0 &1\end{bmatrix}$$

(vi) (c) x = y

Explanation :

$$\text{A}=\begin{bmatrix}5 &x\\ y &0\end{bmatrix}\\\therefore\space \text{A}^{\text{T}}=\begin{bmatrix}5 &y\\x &0\end{bmatrix}\\\text{A = A}^{\text{T}}\space \Rarr\space \begin{bmatrix}5 &x\\ y &0\end{bmatrix}\\=\begin{bmatrix}5 &y\\ x &0\end{bmatrix}$$

∴ x = y

(vii) (a) AP : PD = 16 : 3

Explanation :

Take A as origin.

$$\text{Let}\space \vec{\text{AB}} =\vec{\text{b}}\\\text{and}\space \vec{\text{AC}} =\vec{\text{c}}$$

Therefore, by hypothesis

$$\vec{\text{AD}} =\frac{\vec{b} + 3\vec{c}}{4}\text{and}\\\vec{\text{AE}}=\frac{4\vec{c}}{5}$$

According to figure,

BP : PE = λ : 1 and AP : PD = μ : 1

$$\text{Then}\space\frac{\vec{b}+ \lambda(4\vec{c}/5)}{\lambda +1} =\frac{\mu(\vec{b} + 3\vec{c})}{4(\mu +1)}$$

Equating the coefficients of

$$\vec{b}\space\text{and}\space\vec{c}\space\text{on both sides, we get}\\\frac{1}{\lambda + 1}=\frac{\mu}{4 (\mu +1)}\\\text{...(i)}\\\text{and}\space \frac{4\lambda}{5(\lambda +1)} =\frac{3\mu}{4(\mu+1)}\\\text{...(ii)}\\\text{by (i) and (ii)}\\\frac{4\lambda}{5(\lambda +1)}=\frac{3}{\lambda +1}\\\Rarr\space 4\lambda = 15\space\text{or}\\\lambda =\frac{15}{4}$$

Hence BP : PE = 15 : 4

$$\text{Substituting the value of}\space\lambda =\frac{15}{4}\\\text{in equation (i), we get}\\\frac{4}{19}=\frac{\mu}{4(\mu +1)}\\\Rarr\space \mu =\frac{16}{3}$$

∴ AP : PD = 16 : 3

$$\text{(viii)\space (b)}\frac{13}{\sqrt{21}}$$

Explanation :

$$\text{Plane\space} \vec{r}.(\hat{i} - 2\hat{j} +4\hat{k}) = 9\\\text{...(i)}\\\text{and Point}\space \vec{a} = 2\hat{i} + \hat{j}-\hat{k}\\\text{Here}\space \vec{n} =\hat{i}- 2\hat{j} +4\hat{k}\\|\vec{n}| =\sqrt{1^{2} + (-2)^{2} + 4^{2}}\\=\sqrt{1 + 4 + 16} =\sqrt{21}\\\text{Then distance of point}\space\vec{a}\\\text{from plane (1)}\\=\frac{|\vec{a}.\vec{n}-9|}{|\vec{n}|}\\=\\\frac{|(2)(1) + (1)(-2)+ (\normalsize-1)(4)-9|}{\sqrt{21}}$$

$$=\frac{|2-2-4-9|}{\sqrt{21}} =\\\frac{13}{\sqrt{21}}\text{units}.$$

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

Explanation :

$$y^{2} = ax^{3} + b\space\text{...(i)}\\ 2y\frac{dy}{dx}= 3ax^{2}\\\Rarr\space \frac{dy}{dx}=\frac{3ax^{2}}{2y}\\\bigg(\frac{dy}{dx}\bigg)_{(2,3)}=\frac{3a×2^{2}}{2×3}\\=\frac{12a}{2×3} = 2a$$

Since the equation of the tangent at the point (2, 3) on the given curve is y = 4x – 5 and its slope = 4

Then 2a = 4, a = 2 …(i)

At the point (2, 3) lies on the given curve.

32 = a × 23 + b

⇒ 9 = 2 × 8 + b

⇒ b = – 7

Hence a = 2 and b = – 7

Hence statement (1) is true.

Statement (2): Curve y = x3 – x + 1

$$\frac{dy}{dx} = 3x^{2}-1\\\bigg(\frac{dy}{dx}\bigg)_{x=2} = 11$$

Statement (2) is true.

(x) (d) Assertion is false and Reason is true.

Explanation :

$$\textbf{Reason:}\space \text{P(A|B)}=\frac{\text{P(A ∩ B)}}{\text{P(B)}}\\\text{(by definition)}\\\text{P(}\bar{\text{B}}) =\text{P}\lbrack(\text{A}∪\bar{\text{A}} ∩\bar{\text{B}})\rbrack\\=\text{P}\lbrack(\text{A} ∩\bar{\text{B}})∪(\bar{\text{A}}∩\bar{\text{B}})\rbrack$$

Reason is true.

$$\textbf{Assertion:}\space \text{P(}\bar{\text{A}}|\bar{\text{B}}) + \text{P(}\vec{\text{A}}|\vec{\text{B}})\\= \frac{\text{P(A}∩\bar{\text{B}})}{\text{P(}\bar{\text{B}})} + \frac{\text{P(}\bar{\text{A}} ∩\bar{\text{B}})}{\text{P(}\bar{\text{B}})}\\=\frac{\text{P(A} ∩ \bar{\text{B}}) + \text{P}(\bar{\text{A}}∩\bar{\text{B}})}{\text{P}(\bar{\text{B}})}\\=\frac{\text{P}(\bar{\text{B}})}{\text{P}(\bar{\text{B}})}$$

= 1

Assertion is false.

(xi) (d) (gof) (e) = g[f(e)] = g(q) = α

Explanation :

(gof) (e) = g[f(e)] = g(s) = β

(xii) Let A = [aij] a matrix which is both symmetric and skew-symmetric matrix.

A = [aij] is a symmetric matrix then aij = aji for all i, j. ...(i)

A = [aij] is a skew-symmetric matrix then aij = – aji for all i, j or aji = – aij for all i, j ...(ii)

From (i) and (ii), we get

aij = – aij for all i, j

⇒ aij + aij = 0

⇒ 2aij = 0

⇒ aij = 0 for all i, j

∴ A = [aij] is a null matrix.

Hence Proved.

(xiii) The general equation of all non-horizontal lines in a plane is ax + by + c = 0, a ≠ 0. Differentiating w.r.t. y, we get

$$a\frac{d}{dy}(x) + b\frac{d}{dy}(y)+ \frac{d}{dy}(c)= 0\\\Rarr\space a\frac{dx}{dy} + b×1 +0 =0\\\Rarr\space a\frac{dx}{dy} +b =0$$

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

$$\frac{d}{dy}\bigg(a\frac{dx}{dy}\bigg) + \frac{d}{dy}(b)=0\\\Rarr\space a\frac{d^{2}x}{dy^{2}} + 0 =0\\\Rarr\space \frac{d^{2}x}{dy^{2}} = 0$$

which is the required differential equation. Ans.

(xiv) Let E and F be the events of selection of A and B, respectively.

We know that selection of A and B is independent.

∴ E and F are independent events.

Let P(F) = x

Probability that exactly one of A and B is selected = 0.6

$$\therefore\space \text{P(E ∩} \bar{\text{F}}) + \text{P}(\bar{\text{E} ∩ F}) = 0.6 \\\text{P(E)}.\text{P(}\bar{\text{F}}) + \text{P(}\bar{\text{E}}).\text{P(F)} = 0.6\\\Rarr\\\space (0.7)(1-x) + (1-0.7)(x) = 0.6\\\Rarr\space 0.7 - 0.7x + x -0.7x = 0.6\\\Rarr\space -0.4x + 0.7 =0.6 \\\Rarr\space 0.4x = 0.7 - 0.6\\\Rarr\space x = \frac{0.1}{0.4}=\frac{1}{4}\\\therefore\space \text{Probability that B is selected }\\= \frac{1}{4}.\space \textbf{Ans.}$$

(xv) The sample space has 216 outcomes.

Now Step I = {1, 1, 4), (1, 2, 4) … (1, 6, 4), (2, 1, 4), (2, 2, 4), … (2, 6, 4)

(3, 1, 4), (3, 2, 4), … (3, 6, 4), (4, 1, 4), (4, 2, 4), … (4, 6, 4)

(5, 1, 4), (5, 2, 4), … (5, 6, 4), (6, 1, 4), (6, 2, 4) … (6, 6, 4)}

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

and I ∩ II = {(6, 5, 4)}

$$\text{Now\space} \text{P(I)}=\frac{6}{216}\space\text{and}\\\text{P(I ∩ II)}=\frac{1}{216}\\\text{Then}\space\text{P(I|II)}=\frac{\text{P(I∩II)}}{\text{P(II)}}\\= \frac{\frac{1}{216}}{\frac{6}{216}} =\frac{1}{6}\space\textbf{Ans.}$$

Solution 2.

$$\text{(i)\space Given,\space}\text{f(x)}=\frac{x^{2}-9}{x-3}\\\Rarr\space \text{f(x)} =\frac{(x-3)(x + 3)}{x-3}\\\Rarr\space \lim_{x\to3}\space\text{f(x)} = \lim_{x\to3}(x+3)\\\lim_{x\to3}(x+3)$$

f (3) = 6

Hence, f(x) to be continuous at x = 3, f (3) = 6. Ans.

OR

(ii) Given : x3 + y3 = 3axy

Differentiating w.r.t. x, we get

$$3x^{2} + 3y^{2}\frac{dy}{dx} =\\3a\bigg(1.y + x\frac{dy}{dx}\bigg)\\\Rarr\space x^{2} + y^{2}\frac{dy}{dx} = \\\text{ay + ax}\frac{dy}{dx}\\\Rarr\space \frac{dy}{dx}(y^{2} - ax) = ay-x^{2}\\\Rarr\space \frac{dy}{dx} =\frac{ay-x^{2}}{y^{2}-ax}\space\textbf{Ans.}$$

Solution 3.

$$\text{Since the function f :}\lbrack-1,1\rbrack\xrightarrow{}\\\text{Range (f ) is invertible, as}\space \\ f:\lbrack-1,1\rbrack\xrightarrow{}\text{R}\space\\\text{is invertible function.}$$

Then, fof–1 (x) = x ∀ x ∈ Range (f)

f (f –1 (x)) = x

$$\frac{f^{\normalsize-1}(x)}{f^{\normalsize-1}(x) + 2}= x$$

f–1 (x) = xf–1 (x) + 2x

f–1 (x) – xf–1 (x) = 2x

f–1 (x) (1 – x) = 2x

$$\therefore\space f^{\normalsize-1}(x)=\frac{2x}{1-x}.\space \textbf{Ans.}$$

Solution 4.

Given differential equation is,

ex tan y dx + (1 – ex) sec2 y dy = 0

⇒ (1 – ex) sec2 y dy = – ex tan y dx

$$\Rarr\space \frac{\text{sec}^{2}y}{\text{tan y}}dy =-\frac{e^{x}}{1 - e^{x}}dx$$

Integrating both sides

log|tan y| = log|1 – ex |+ C

$$\begin{bmatrix}\because\space \int\frac{f'(x)}{f(x)}dx =\text{log|f(x)|+ c}\end{bmatrix}\\\Rarr\space \text{log}|\text{tan y}| -\text{log}|1 - e^{x}| = \text{C}\\\begin{bmatrix}\because\space\int\frac{f'(x)}{f(x)}dx = \text{log}|\text{f(x) + c}|\end{bmatrix}\\\Rarr\space \text{log}|\frac{\text{tan y}}{1 - e^{x}}| =\text{C}\\\Rarr\space \frac{\text{tan y}}{1 - e^{x}} = e^{C}$$

⇒ tan y = eC (1 – ex).

which is the required solution. Ans.

Solution 5.

$$\text{(i)\space Let}\space \text{I} =\int^{\frac{\pi}{4}}_{0}\frac{2 \text{cos} 2x}{\text{1 + sin 2x}}dx$$

Let 1 + sin 2x = t

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

$$\text{cos 2x×2}=\frac{dt}{dx}\\\Rarr\space \text{2 cos 2x. dx} = dt$$

At x = 0, t = 1 + sin 2 × 0 = 1 + 0 = 1

$$\text{At x} =\frac{\pi}{4}, t \\= 1 + \text{sin} 2×0 = 1 + 0 = 1\\\text{At x=}\frac{\pi}{4}, t \\= 1 + \text{sin 2}×\frac{\pi}{4}\\= 1 + 1= 2\\\Rarr\space \text{I} =\int^{2}_{1}\frac{dt}{t}\\=\lbrack\text{log} _e t\rbrack^{2}_{1}$$

= log 2 – log 1

= log 2 Ans.

OR

(ii) Let t = x2 + 4x + 5

Then, dt = (2x + 4) dx

$$\Rarr\space \int\frac{x+3}{\sqrt{x^{2} + 4x + 5}}dx \\=\frac{1}{2}\int\frac{(2x + 4) +2}{\sqrt{x^{2} + 4x + 5}}dx\\=\frac{1}{2}\int\frac{2x+4}{\sqrt{x^{2} + 4x + 5}}dx + \\\frac{1}{2}\int\frac{2}{\sqrt{x^{2} + 4x+ 5}}dx\\=\frac{1}{2}\int\frac{dt}{\sqrt{t}} + \int\frac{2}{\sqrt{(x + 2)^{2} + t^{2}}}\\=\frac{1}{2}\lbrack2\sqrt{t} + 2\text{log}\rbrack\\|\sqrt{1 + (x+2)^{2}} + (x + 2)| + c.\\=\sqrt{x^{2} + 4x + 5} + \\\text{log}|\sqrt{(1 + (x + 2)^{2})} + (x+2)| +c.\\\textbf{Ans.}$$

Solution 6.

$$\text{we have}\space y = xc + \frac{a}{c}\space\text{...(i)}$$

Differentiating with respect to x, we get

$$\frac{dy}{dx} = c\space\text{...(ii)}\\\therefore\space x\frac{dy}{dx} +\frac{a}{\frac{dy}{dx}} = xc +\frac{a}{c}\\\begin{bmatrix}\text{Putting}\frac{dy}{dx} = c\end{bmatrix}\\\Rarr\space \frac{xdy}{dx} +\frac{a}{\frac{dy}{dx}} = y\\\text{Hence,}\space y = cx + \frac{a}{c}\space\text{is a solution of}\\\text{the given differential equation.}\\\textbf{Hence Proved.}$$

Solution 7.

$$\text{Given :}\space\text{cos}^{-1}\frac{x}{2} + \text{cos}^{\normalsize-1}\frac{y}{3} =\theta\\\Rarr\space\text{cos}^{\normalsize-1}\begin{bmatrix}\frac{x}{2}×\frac{y}{3}-\sqrt{1 - \frac{x^{2}}{4}}.\sqrt{1-\frac{y^{2}}{9}}\end{bmatrix}\\=\theta\\\begin{bmatrix}\because\space \text{cos}^{\normalsize-1} a + \text{cos}^{\normalsize-1}b=\\\text{cos}^{\normalsize-1}\bigg(ab -\sqrt{1 - a^{2}}.\sqrt{1- b^{2}}\bigg)\end{bmatrix}$$

$$\Rarr\space \frac{xy}{6}-\sqrt{1-\frac{x^{2}}{4}}.\sqrt{1-\frac{y^{2}}{9}}\\=\text{cos}\space \theta\\\Rarr\space \frac{xy}{6}-\text{cos}\space \theta\\=\sqrt{1-\frac{x^{2}}{4}}.\sqrt{1-\frac{y^{2}}{9}}$$

Squaring both sides, we get

$$\bigg(\frac{xy}{6}-\text{cos}\space \theta\bigg)^{2}\\=\bigg(1-\frac{x^{2}}{4}\bigg)\bigg(1-\frac{y^{2}}{4}\bigg)\\\Rarr\space \frac{x^{2}y^{2}}{36} + \text{cos}^{2}\theta-\frac{2xy}{6}\text{cos}\theta\\= 1-\frac{x^{2}}{4}-\frac{y^{2}}{9}+ \frac{x^{2}y^{2}}{36}\\\Rarr\space \frac{x^{2}}{4} + \frac{y^{2}}{9}-\frac{xy\text{cos}\theta}{3}\\= \text{1 - cos}^{2}\theta\\\Rarr\space\frac{9x^{2} + 4y^{2} - 12xy\text{cos}\theta}{36} = \text{sin}^{2}\theta$$

⇒ 9x2 + 4y2 – 12xy cos θ = 36 sin2 θ

⇒ 9x2– 12xy cos θ + 4y2 = 36 sin2 θ

Hence Proved.

Solution 8.

The given equation of one parameter family of curves is

x2 – y2 = c (x2 + y2)2 …(i)

Differentiating (i) with respect to x, we get

$$2x - 2y\frac{dy}{dx} =\\ 2c(x^{2} + y^{2})\bigg(2x + 2y\frac{dy}{dx}\bigg)\\\Rarr\space \bigg(x - y\frac{dy}{dx}\bigg)\\ = 2c(x^{2} + y^{2})\bigg(x + y\frac{dy}{dx}\bigg)\\\text{...(ii)}$$

On substituting the value of c obtained from (i) in (ii), we get

$$\bigg(x -y\frac{dy}{dx}\bigg)=\\\frac{2(x^{2}-y^{2})(x^{2} + y^{2})}{(x^{2} + y^{2})^{2}}\bigg(x +y\frac{dy}{dx}\bigg)\\\Rarr\space (x^{2} + y^{2})\bigg(x-y\frac{dy}{dx}\bigg)\\= 2(x^{2} - y^{2})\\\bigg(x + y\frac{dy}{dx}\bigg)\\\Rarr\space \lbrace x(x^{2} + y^{2})-2x(x^{2}-y^{2})\rbrace=\\\frac{dy}{dx}\lbrace(2y(x^{2} -y^{2})) + y(x^{2} + y^{2})\rbrace\\\Rarr\space (3xy^{2}-x^{3})=\\\frac{dy}{dx}(3x^{2}y- y^{3})$$

⇒ (x3 – 3xy2) dx = (y3 – 3x2y) dy, which is the given differential equation.

Hence Proved.

Solution 9.

$$\text{(i)\space conisder},\int\frac{\text{cosec x}}{\text{log tan}\bigg(\frac{x}{2}\bigg)}\\\text{Put}\space \text{log tan}\bigg(\frac{x}{2}\bigg) = t\\\text{...(i)}$$

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

$$\Rarr\space \frac{1}{\text{tan}\frac{x}{2}}.\frac{1}{2}\text{sec}^{2}\frac{x}{2}=\frac{dt}{dx}\\\Rarr\space \frac{1}{2}.\frac{\frac{\text{cos x}}{2}}{\frac{\text{sin x}}{\frac{2 \text{cos}^{2}x}{2}}} dx = dt\\\Rarr \space \frac{1}{2\text{sin}\frac{x}{2}\text{cos}\frac{x}{2}}dx = dt\\\Rarr\space \frac{1}{\text{sin x}}dx = dt\\\Rarr\space\text{cosec x dx} = dt\\\text{...(ii)}$$

Now, on dividing both sides by ‘t’ and then integrating

$$\therefore\space \int\frac{\text{cosec x dx}}{\text{log tan}\bigg(\frac{x}{2}\bigg)} =\int\frac{dt}{t}\\\int\frac{\text{cosec x dx}}{\text{log tan}\bigg(\frac{x}{2}\bigg)}\\=\text{log t + c}$$

Put the value of ‘t’

$$\int\frac{\text{cosec x dx}}{\text{log tan}\bigg(\frac{x}{2}\bigg)} =\\\text{log}\begin{bmatrix}\text{log tan}\frac{x}{2}\end{bmatrix} + \text{c}.\\\textbf{Ans.}$$

OR

$$\text{(ii)\space}\text{Consider}\space\int^{\frac{\pi}{2}}_{0}\frac{\text{sin x cos x}}{\text{cos}^{2}x + 3 cos x + 2}dx$$

Let, cos x = t

∴ – sin x dx = dt

When x = 0 ⇒ t = 1 and

$$\text{when}\space x =\frac{\pi}{2}\\\Rarr t = 0\\\therefore\space \text{I} =\int^{0}_{1}\frac{-t.dt}{t^{2} + 3t +2}\\=\int^{1}_{0}\frac{t\space dt}{t^{2} + 3t +2}\\\text{Now,\space}\frac{1}{t^{2} + 3t + 2}\\=\frac{1}{(t+2)(t+1)}$$

Taking partial fractions, we get

$$\frac{t}{t^{2}+ 3t+2} \\=\frac{\text{A}}{\text{t+2}} + \frac{\text{B}}{\text{t+1}}$$

t = At + A + Bt + 2B

Equating coefficients of like terms, we get

A + B = 1, A + 2B = 0

⇒ B = – 1 and A = 2

$$\therefore\space\int^{1}_{0}\bigg(\frac{2}{t +2}-\frac{1}{t+1}\bigg)dt\\=\lbrack\text{2 log}(t+2) - \text{log}(t+1)\rbrack^{1}_{0}$$

= 2 log 3 – log 2 – (2 log 2 – log 1)

= 2 log 3 – 3 log 2

= log 9 – log 8

$$=\text{log}\frac{9}{8}.\space\textbf{Ans.}$$

Solution 10.

(i) (a) Since the sum of all the probabilities in a probability distribution is always one. Therefore

P(X = 0) + P(X = 1) + P(X = 2) + … + P(X = 7) = 1

⇒ 0 + k + 2k + 2k + 3k + k2+ 2k2 + 7k2 + k = 1

⇒ 10k2 + 9k – 1 = 0

⇒ (10k – 1) (k + 1) = 0

⇒ 10k – 1 = 0 [∵ k ≥ 0 ∴ k + 1 ≠ 0]

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

(b) P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

⇒ P(X < 6) = 0 + k + 2k + 2k + 3k + k2

⇒ P(X < 6) = k2 + 8k

$$\Rarr\space \text{P(X}\lt 6) =\bigg(\frac{1}{10}\bigg)^{2} + \frac{8}{10}\\\bigg(\because k =\frac{1}{10}\bigg)\\\Rarr\space \text{P(X}\lt 6)=\frac{81}{100}\space\textbf{Ans.}$$

(c) P(X ≥ 6) = P(X = 6) + P(X = 7)

∴ P(X ≥ 6) = 2k2 + 7k2 + k

⇒ P(X ≥ 6) = 9k2 + k

$$= \frac{9}{100} + \frac{1}{10}\\\bigg(\because\space k =\frac{1}{10}\bigg)\\=\frac{19}{100}\space \textbf{Ans.}$$

(d) P(0 < X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

⇒ P(0 < X < 5) = k + 2k + 2k + 3k

= 8k

$$= 8×\frac{1}{10}\space\bigg(\because\space k =\frac{1}{10}\bigg)\\=\frac{4}{5}\space \textbf{Ans.}$$

OR

(ii) Let E1, E2 and A be the events defined as follows :

E1 = Six appears on throwing a die.

E2 = Six does not appear on throwing a die.

and A = the man reports that it is a six

We have,

$$\text{P(E}_{1}) = \frac{1}{6}\\\text{P(E}_{2}) =\frac{5}{6}$$

Now P(A | E1) = Probability that the man reports that there is a six on the die given that six has occurred on the die = 4/5 (probability that the man speaks truth)

and P(A | E2) = Probability that the man reports that there is a six on the die given that six has not occurred on the die (probability that the man does not speak truth).

$$= 1-\frac{4}{5} =\frac{1}{5}$$

By Bayes’ theorem, we have

$$\text{P(E}_{1}|\text{A}) \\=\frac{\text{P(E}_{1})\text{P(A|\text{E}}_{1})}{\text{P}(\text{E}_{1})\text{P}(\text{A}|\text{E}_{1}) + \text{P}(\text{E}_{2})\text{P}(\text{A}|\text{E}_{2})}\\=\frac{\frac{1}{6}×\frac{4}{5}}{\frac{1}{6}×\frac{4}{5} + \frac{5}{6}×\frac{1}{5}}\\=\frac{4}{4 + 5}=\frac{4}{9}$$

Yes, truthfulness always leads to more respect in the society as truth always wins. Ans.

Solution 11.

(a) Let the numbers are x, y, z be the prize amount per person per adaptability, carefulness and calmness respectively.

As per the given data, we get

2x + 4y + 3z = 29000

5x + 2y + 3z = 30500

x + y + z = 9500

$$\text{A} =\begin{bmatrix}2 &4 &3\\5 &2 &3\\ 1&1 &1\end{bmatrix},\text{X} =\begin{bmatrix}x\\y\\z\end{bmatrix},\\\text{B} =\begin{bmatrix}29000\\ 30500\\ 9500\end{bmatrix}$$

Form of matrix multiplication

AX = B

$$\begin{bmatrix}2 &4 &3\\5 &2 &3\\1 &1 &1\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix} =\begin{bmatrix}29000\\ 30500\\9500\end{bmatrix}$$

(b) | A | = 2(2 – 3) –4(5 –3) + 3(5 – 2)

= 2 × – 1 – 4 × 2 + 3 × 3

= – 2 – 8 + 9

= – 1

Hence, the unique solution given by

X = A–1B

C11 = (– 1)1+1 (2 – 3) = – 1, C12 = (– 1)1+2 (5 – 3) = – 2

C13 = ( –1)1+3 (5 – 2) = 3, C21 = (– 1)2+1 (4 – 3) = – 1

C22 = (– 1)2+2 (2 – 3) = – 1, C23 = (– 1)2+3 (2 – 4) = 2

C31 = (– 1)3+1 (12 – 6) = 6, C32 = (– 1)3+2 (6 – 15) = 9

C33 = (– 1)3+3 (4 – 20) = – 16

$$\text{Adj A} =\begin{bmatrix}\normalsize-1 &\normalsize-1 &6\\\normalsize-2 &\normalsize-1 &9\\3 &2 & \normalsize-16\end{bmatrix}\\\text{X} =\text{A}^{\normalsize-1}\text{B} =\frac{1}{|\text{A}|}\text{(adj A)}\text{B}\\\text{X} =\frac{1}{\normalsize-1}\begin{bmatrix}\normalsize-1 &\normalsize-1 &6\\\normalsize-2 &\normalsize-1 &9\\3 &2 &\normalsize-16\end{bmatrix}\begin{bmatrix}29000\\ 30500\\ 9500\end{bmatrix}\\\text{X} =\begin{bmatrix}1 &1 &\normalsize-6\\2 &1 &\normalsize-9\\\normalsize-3 &\normalsize-2 &16\end{bmatrix}\begin{bmatrix}29000\\ 30500\\9500\end{bmatrix}\\\text{X} =\begin{bmatrix}29000 + 30500 - 57000\\ 58000 + 30500-85500\\-87000 -61000 + 152000\end{bmatrix}$$

$$\text{(c)\space} \begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}2500\\3000\\ 4000\end{bmatrix}$$

Hence, x = ₹2500, y = ₹3000, z = ₹4000.

Ans.

Solution 12.

(i) The given differential equation is

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

Rewriting the given differential equation, we get

$$\frac{dy}{dx} + \frac{1}{1 + x^{2}}.y \\=\frac{e^{\text{tan}^{\normalsize-1}x}}{\text{1+ x}^{2}}$$

which is a linear differential equation of the form

$$\frac{dy}{dx} + \text{P}y =\text{Q}\\\text{Here,\space} \text{P} =\frac{1}{\text{1 + x}^{2}}\\\text{and}\space \text{Q} =\frac{e^{\text{tan}^{\normalsize-1}x}}{\text{1 + x}^{2}}\\\therefore\space \int\text{P dx} =\int\frac{dx}{1 + x^{2}} \\=\text{tan}^{\normalsize-1}x\\\text{Now,\space}\text{I.F.} = e^{\int\text{P dx}}\\ = e^{\text{tan}^{\normalsize-1}x}$$

Hence, the solution is

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

Putting, tan-1 x = t

$$\Rarr\space \frac{1}{\text{1 + x}^{2}}dx = dt\\\Rarr\space ye^{\text{tan}^{\normalsize-1}x} =\int e^{2t}dt + \text{C}\\=\frac{1}{2}e^{2t} + \text{C}\\ y.e^{\text{tan}^{\normalsize-1}x}=\\\frac{1}{2}e^{2\text{tan}^{\normalsize-1}x} + \text{C}\\\therefore\space y =\frac{1}{2}e^{\text{tan}^{\normalsize-1}x} + \text{C}e^{-\text{tan}^{\normalsize-1}x}\\\textbf{Ans.}$$

OR

(ii) Let,

$$\text{I} =\int^{\pi}_{0}\frac{x}{\text{1 + sin x}}dx\\\text{I} =\int^{\pi}_{0}\frac{\pi - x}{1 + \text{sin}(\pi - x)}dx\\\begin{bmatrix}\because\space \int^{a}_{0} f(x)dx =\\\int^{a}_{0} f(a-x)dx\end{bmatrix}\\\text{I} =\int^{\pi}_{0}\frac{\pi- x}{1 + \text{sin x}}dx\\\text{I} =\int^{\pi}_{0}\frac{\pi}{\text{1 + sin x}} -\int^{\pi}_{0}\frac{x}{1 + \text{sin x}}dx\\\text{...(ii)}$$

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

$$2\text{I} =\int^{\pi}_{0}\frac{\pi}{\text{1 + sin x}}dx\\\Rarr\space \text{2 I} =\int^{\frac{\pi}{2}}_{0}\frac{\pi}{\text{1 + sin x}}dx +\\\int^{\frac{\pi}{2}}_{0}\frac{\pi}{\text{1 + sin}(\pi - x)}dx\\\begin{bmatrix}\because\space \int^{2a}_{0} f(x)dx =\\\int^{a}_{0}\lbrace f(x) + f(2a - x)\rbrace dx\end{bmatrix}\\\Rarr\space 2\text{I} =\pi\int^{\frac{\pi}{2}}_{0}\frac{dx}{\text{1 + sin x}} + \\\pi\int^{\frac{\pi}{2}}_{0}\frac{dx}{\text{1 + sin x}}\\\Rarr\space 2\text{I} = 2\pi\int^{\frac{\pi}{2}}_{0}\frac{dx}{\text{1 + sin x}}$$

$$\Rarr\space \text{I} = \pi\int^{\frac{\pi}{2}}_{0}\frac{dx}{\text{1 + sin x}}×\frac{\text{1 - sinx }}{\text{1 - sin x}}dx\\\Rarr\space \pi\begin{bmatrix}\int^{\frac{\pi}{2}}_{0}\frac{1}{\text{cos}^{2}x}dx -\int^{\frac{\pi}{2}}_{0}\frac{\text{sin x}}{\text{cos}^{2}x}dx\end{bmatrix}\\\Rarr\space \pi\begin{bmatrix}\int^{\frac{\pi}{2}}_{0}\text{sec}^{2}dx -\\\int^{\frac{\pi}{2}}_{0}\text{sec x tan x dx}\end{bmatrix}\\\Rarr\space \text{I} = \pi\lbrack\text{tan x - sec x}\rbrack^{\frac{\pi}{2}}_{0}\\=\pi\begin{bmatrix}\frac{\text{sin x - 1}}{\text{cos x}}\end{bmatrix}^{\frac{\pi}{2}}_{0}\\\Rarr\space \text{I} =-\pi\begin{bmatrix}\frac{\text{1 - sin x}}{\text{cos x}}.\frac{\text{1 + sin x}}{\text{1 + sin x}}\end{bmatrix}^{\frac{\pi}{2}}_{0}\\=-\pi\begin{bmatrix}\frac{\text{cos x}}{\text{1 + sin x}}\end{bmatrix}^{\frac{\pi}{2}}_{0}\\\Rarr\space \text{I} =-\pi\begin{bmatrix}\frac{0}{2} -\frac{1}{\text{1 + 0}}\end{bmatrix}\\-\pi(\normalsize-1) =\pi\\\therefore\space \int^{\pi}_{0}\frac{x}{\text{1 + sin x}}dx =\pi.\space\textbf{Ans.}$$

Solution 13.

Let ABCD be a rectangle inscribed in a given circle with centre at O and radius a.

Given AB = 2x and BC = 2y

In ΔOAM by Pythagoras theorem

OA2 = AM2 + OM2

⇒ a2 = x2 + y2

$$\Rarr\space y = \sqrt{a^{2}- x^{2}}\\\Rarr\space \frac{\text{dA}}{\text{dx}}= 4\begin{Bmatrix}\sqrt{a^{2} - x^{2}}-\\\frac{x^{2}}{\sqrt{a^{2} - x^{2}}}\end{Bmatrix}\\ = 4\begin{Bmatrix}\frac{a^{2} - 2x^{2}}{\sqrt{a^{2} - x^{2}}}\end{Bmatrix}\\\text{The critical points of A are given by}\\\frac{\text{dA}}{\text{dx}} = 0\\\therefore\space \frac{\text{dA}}{\text{dx}} = 0\\\Rarr\space 4\begin{Bmatrix}\frac{a^{2} - 2x^{2}}{\sqrt{a^{2} - x^{2}}}\end{Bmatrix} = 0\\\Rarr\space a^{2} - 2x^{2} =0\\\Rarr\space x =\frac{a}{\sqrt{2}}$$

$$\text{Now,\space} \frac{\text{dA}}{\text{dx}} = 4\begin{Bmatrix}\frac{a^{2} - 2x^{2}}{\sqrt{a^{2} - x^{2}}}\end{Bmatrix}\\\Rarr\space \frac{d^{2}\text{A}}{dx^{2}} = 4\frac{d}{dx}\lbrace(a^{2} - 2x^{2})(a^{2} - x^{2})^{-\frac{1}{2}}\rbrace\\\Rarr\space\frac{d^{2}\text{A}}{dx^{2}} = 4\lbrace-4x(a^{2} - x^{2})^{\frac{-1}{2}} + \\(a^{2} - 2x^{2})^{\frac{\normalsize-1}{2}}×(a^{2} - x^{2})^{-\frac{3}{2}}(-2 x)\rbrace\\\Rarr\space \frac{d^{2}\text{A}}{dx^{2}} = 4\begin{Bmatrix}\frac{-4 x}{\sqrt{a^{2} - x^{2}}} + \frac{x(a^{2} - 2x^{2})}{(a^{2} - x^{2})^{\frac{3}{2}}}\end{Bmatrix}\\\therefore\space \bigg(\frac{d^{2}\text{A}}{dx^{2}}\bigg)_{x =\frac{a}{\sqrt{2}}} = -16\lt 0$$

Thus, A is maximum when

$$x =\frac{a}{\sqrt{2}},\text{putting x} =\frac{a}{\sqrt{2}}\\\space\text{in (i), we get}$$

$$y =\frac{a}{\sqrt{2}}\\\text{Therefore,}\space x = y =\frac{a}{\sqrt{2}}\\\Rarr\space 2x = 2y =\sqrt{2a}\\\Rarr\space\text{AB = BC}\\\Rarr\space \text{ABCD is a square.}\\\text{ Hence, area A is maximum }\\\text{when the rectangle is a square.}\\\textbf{Hence Proved.}$$

OR

ABC be a Δ inscribed in a given circle with centre O and radius r.

The area of the triangle will be maximum of its vertex.

A opposite to the base BC is at a maximum distance from the base BC. This is possible only when A lies on the diameter perpendicular to BC.

ΔABC must be an isosceles D. Let OD = x

In ΔODB by Pythagoras theorem

$$\text{OB}^{2} = \text{OD}^{2} + \text{BD}^{2}\\\Rarr\space r^{2} = x^{2} + \text{BD}^{2}\\\Rarr\space \text{BD} =\sqrt{r^{2} - x^{2}}\\\therefore\space \text{BC = 2BD}\\= 2\sqrt{r^{2} - x^{2}}$$

but AD = AO + OD

= r + x

Let be area of ΔABC = A

$$\text{A} =\frac{1}{2}(\text{BC×AD})\\\Rarr\space \text{A} = \frac{1}{2}×2\sqrt{r^{2} - x^{2}}×(r + x)\\\Rarr\space \text{A} =(r +x)\sqrt{r^{2} - x^{2}}\\\Rarr\space \frac{dA}{dx} =\sqrt{r^{2} - x^{2}}-\\\frac{x(r+x)}{\sqrt{r^{2} - x^{2}}}\\\Rarr\space \frac{dA}{dx} =\frac{r^{2} - x^{2} - rx - x^{2}}{\sqrt{r^{2} - x^{2}}}\\\Rarr\space \frac{dA}{dx} =\frac{r^{2} - rx - 2x^{2}}{\sqrt{r^{2} - x^{2}}}\\\therefore\space \frac{dA}{dx} = 0$$

$$\Rarr\space \frac{r^{2} - rx- 2x^{2}}{\sqrt{r^{2} - x^{2}}} = 0$$

⇒ r2 – rx – 2x2 = 0

⇒ r2– 2rx + rx – 2x2 = 0

⇒ (r – 2x) (r + x) = 0

⇒ r – 2x = 0

$$\Rarr\space x =\frac{r}{2}\\(\because\space r + x\neq 0)\\\text{Now,\space }\frac{\text{dA}}{dx} =\frac{r^{2} -rx -2x^{2}}{\sqrt{r^{2} - x^{2}}}\\\Rarr\space \frac{d^{2}\text{A}}{dx^{2}} =\frac{(-r-4x)}{\sqrt{r^{2} - 4x^{2}}} + \\\frac{(x^{2} - rx - 2x^{2})x}{(r^{2} - x^{2})^{\frac{3}{2}}}\\\Rarr\space \bigg(\frac{d^{2}\text{A}}{dx^{2}}\bigg)_{x =\frac{r}{2}} =\\-2\sqrt{3}\lt 0\\\text{Thus, A is maximum when}\space x =\frac{r}{2}$$

$$\therefore\space \text{BD} =\sqrt{r^{2} - x^{2}}\\\Rarr\space \text{BD}=\frac{\sqrt{3}r}{2}\\\text{In}\space \Delta\text{ODB}\\\text{tan}\space\theta =\frac{\text{BD}}{\text{OD}}\\\Rarr\space \text{tan}\space\theta =\frac{\frac{\sqrt{3}r}{2}}{\frac{r}{2}} =\sqrt{3}$$

⇒ θ = 60° = ∠BAC

Hence, A is maximum when Δ ABC is equilateral.

Solution 14.

According to question

A : a black ball is selected

E1 : box I is selected

E2 : box II is selected

E3 : box III is selected

E4 : box IV is selected

(i) Since the boxes are chosen at random.

$$\text{Therefore}\space \text{P(E}_{1}) = \frac{1}{4}\\\text{Similarly,}\\\space \text{P(E}_{2}) =\text{P(\text{E}}_{3}) =\text{P(E}_{4})=\frac{1}{4}$$

n(A) = 3, n(S) = 3 + 4 + 5 + 6 = 18

$$\text{P(A|\text{E}}_{1}) =\frac{n(\text{A})}{n(\text{S})}\\=\frac{3}{18} =\frac{1}{6}\space\text{...(i)}$$

(ii) B = a white ball is selected Ans.

n(B | E2) = white ball is selected in II box

n(B | E2) = 2, n(S) = 8

$$\text{P(B | \text{E}}_{2}) = \frac{2}{8} =\frac{1}{4}\\\text{P}(A | \text{E}_{1}) + \text{P}(\text{B} | \text{E}_{2}) =\\\frac{1}{6} + \frac{1}{4}\\=\frac{2 + 3}{12} =\frac{5}{12}$$

$$\text{(iii)\space}\text{P(A |\text{E}}_{2})=\frac{2}{8},\\\text{P}(\text{A} |\text{E}_{3}) =\frac{1}{7}\\\text{P(A |\text{E}}_{4}) =\frac{4}{13}$$

P(box III is selected, given that the drawn ball is black) = P(E3 | A)

By Baye’s theorem

$$\text{P(E}_{3} |\text{A}) =\\\frac{\text{P(E}_{3}).\text{P}(\text{A}|\text{E}_{3})}{\text{P}(\text{E}_{1}).\text{P}(\text{A}|\text{E}_{1}) + \text{P}(\text{E}_{2}).\text{P}(\text{A}|\text{E}_{2})+ \\\text{P(E}_{3}).\text{P}(A|\text{E}_{3}) + \text{P}(\text{E}_{4}).\text{P(A|\text{E}}_{4})}\\=\\\frac{\frac{1}{4}×\frac{1}{7}}{\frac{1}{4}×\frac{3}{18} + \frac{1}{4}×\frac{2}{8} + \frac{1}{4}×\frac{1}{7} + \frac{1}{4}×\frac{4}{13}}$$

= 0.165

Ans.

(iv) P(E1) + P(E2) + P(E3) + P(E4)

$$=\frac{1}{4} + \frac{1}{4} +\frac{1}{4} +\frac{1}{4} = 1.\\\textbf{Ans.}$$

## Section-B

Solution 15.

(i) (c)

$$\frac{x}{a} +\frac{y}{b} +\frac{z}{c} = 1$$

(ii) (a) 60 [a – b, b – c, c – a] = [(a – b) × (b – c)].(c – a)

= [a × b – a × c – b × b + b × c].(c – a)

= [a × b – a × c + b × c].(c – a)

= (a × b).c – (a × c),c + (b × c).c – (a × b).a + (a × c).a – (b × c).a

= (a × b).c – (b × a)a

= (a × b).c – (a × b).c = 0.

(iii) Given, direction ratios of line are 2, – 1, – 2.

∴ Direction cosines of above line are

$$\frac{2}{\sqrt{(2)^{2} + (\normalsize-1)^{2} + (\normalsize-2)^{2}}},\\\frac{\normalsize-1}{\sqrt{(2)^{2} + (-1)^{2} + (-2)^{2}}},\\\frac{-2}{\sqrt{(2)^{2} + (-1)^{2} + (-2)^{2}}}\\=\frac{2}{\sqrt{4 +1+4}},\frac{\normalsize-1}{\sqrt{4 + 1 +4}},\\\frac{\normalsize-2}{\sqrt{4 + 1 + 4}}\\=\frac{2}{\sqrt{9}},\frac{\normalsize-1}{\sqrt{9}},\frac{-2}{\sqrt{9}}\\=\frac{2}{3},\frac{\normalsize-1}{3},\frac{\normalsize-2}{3}\space\textbf{Ans.}$$

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

$$\Rarr\space 2\begin{vmatrix}2 &-3\\ x &5\end{vmatrix} + 1\begin{vmatrix}1 &-3\\ 3 &5\end{vmatrix}\\ + 1\begin{vmatrix}1 &2\\3 &x\end{vmatrix} = 0$$

⇒ 2(10 + 3x) + (5 + 9) + (x – 6) = 0

⇒ 20 + 6x + 14 + x – 6 = 0

⇒ 7x = 28

∴ x = 4. Ans.

(v) Direction Ratios of line L1 = (3 – 0, 5 – 3, 6 – 2)

= (3, 2, 4)

D.R’s of Line L2 = (2, 5, – 4)

a1 = 3, b1 = 2, c1 = 4

a2 = 2, b2 = 5, c2 = – 4

$$\text{cos}\theta =\\\frac{a_{1}a_{2} + b_{1}b_{2} +c_{1}b_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}\sqrt{a_{2}^{2}+ b_{2}^{2} + c_{2}^{2}}}\\\Rarr\space \text{cos}\space \theta \\=\frac{6 + 10 -6}{\sqrt{3^{3} + 2^{2} + 4^{2}} + \sqrt{2^{2} + 5^{2} + (-4)^{2}}}\\\Rarr\space \text{cos}\space\theta = 0\\\Rarr\space \text{cos}\space\theta =\text{cos}\frac{\pi}{2}\\\Rarr\space \theta =\frac{\pi}{2}\space\textbf{Ans.}$$

Solution 16.

(i) Let x be the side of the square to be cut from each corner.

∴ For the box

l = 18 – 2x cm

b = 18 – 2x cm

h = x cm

∴ Volume of the box

V = lbh = (18 – 2x)2. x

$$\frac{\text{dV}}{\text{dx}} = (18 - 2x)^{2}1 + x.2(18 - 2x)(\normalsize-2)$$

= (18 – 2x)2 – 4x (18 – 2x)

= (18 – 2x) (18 – 2x – 4x)

= (18 – 2x) (18 – 6x)

= 2 (9 – x) 2(9 – 3x)

$$\frac{\text{dV}}{\text{dx}} = 4(9-x)(9-3x)\\\text{...(i)}\\\text{For maxima or minima,}\\\frac{\text{dV}}{\text{dx}} = 0\\\Rarr\space 4(9-x)(9-3x) =0\\\Rarr\space 9-x = 0\text{or}\\\text{9 - 3x} = 0\Rarr\space \text{x = 9 or 3}$$

But x = 9 is not possible

[∵ then l = b = 0]

∴ x = 3

Differentiating, (i) w.r.t. x,

$$\frac{d^{2}\text{V}}{dx^{2}} =\\ 4\lbrack(9-x)(\normalsize-3) + (9-3x)(\normalsize-1)\rbrack\\ = 4\lbrack-27 + 3x-9 +3x\rbrack\\= 4\lbrack6x - 36\rbrack\\= 24\lbrack x-6\rbrack\\\bigg(\frac{d^{2}\text{V}}{dx^{2}}\bigg)_{x =3} = 24(3-6)$$

= 24 (– 3) = – ve < 0

⇒ V is maximum when x = 3.

Hence, volume of the box is maximum when the side of square to be cut from each corner
is 3 cm. Ans.

OR

(ii) Let r be the radius and h be the height of the cylinder.

$$\text{Given,\space}\text{V} =\pi r^{2}h =\frac{539}{2}\\\Rarr\space h =\frac{539}{2\pi r^{2}}$$

Let S be the total surface area, then

S = 2πrh + 2πr2

$$= 2\pi r\bigg(\frac{539}{2\pi r^{2}}\bigg) + 2\pi r^{2}\\=\frac{539}{r} +2\pi r^{2}$$

Differentiating w.r.t. r, we get

$$\frac{dS}{dr} =-\frac{539}{r^{2}} + 4\pi r\\\text{For maxima or minima}\\\frac{dS}{dr} = 0\\\Rarr\space -\frac{539}{r^{2}} + 4\pi r = 0\\\Rarr\space r =\bigg(\frac{539}{4\pi}\bigg)^{\frac{1}{3}}=\frac{7}{2}\text{units}$$

Differentiating equation (i) w.r.t. r, we get

$$\therefore\space \frac{d^{2}\text{S}}{dr^{2}} = \frac{1078}{r^{3}} + 4\pi,\\\text{which is positive.}\\\therefore\space \text{ S is minimum, when r =}\frac{7}{2}\space\text{and}\\ h =\frac{539}{2\pi\bigg(\frac{7}{2}\bigg)^{2}}\\= \frac{539}{7×\frac{7}{2}×\frac{22}{7}} = \text{7 units}\\\textbf{Ans.}$$

Solution 17.

$$\text{(i) Given, vector}\space \vec{r}\\\text{makes acute angles with coordinate axes.}\\\therefore\space \vec{r}.\hat{i}\gt 0, \vec{r}.\hat{j}\gt0\space\text{and}\\\vec{r}.\hat{k}\gt 0\\\text{When}\space \vec{r}.\hat{j}\gt 0\\\lbrace(x^{2}-1)\rbrace\hat{i} + 5\hat{j} -3 (x^{2} - 4)\hat{k}\rbrace\hat{i}\gt0\\\Rarr\space (x^{2}-1)\gt 0\\\Rarr\space (x-1)(x+1)\gt 0\\\therefore\space x =(-\infty,-1) ∪(1, \infty)\space\text{...(i)}\\\text{When}\space \vec{r}.\hat{k}\gt 0\\\lbrace(x^{2}-1)\hat{i} + 5\hat{j} - 3(x^{2} - 4)\hat{k}\rbrace\hat{i}\gt 0\\\Rarr\space (x^{2}-1)\gt 0\\\Rarr\space (x-1)(x+1)\gt 0$$

∴ x = (– ∞, – 1) ∪ (1, ∞) ...(i)

$$\text{When}\space \vec{r}.\hat{k}\gt 0\\\lbrace(x^{2}-1)\hat{i} + 5\hat{j}-3(x^{2}-4)\hat{k}\rbrace\hat{i} \gt 0\\\Rarr\space (x^{2}-1)\gt 0\\\Rarr\space (x-1)(x + 1)\gt 0\\\therefore\space x =(-\infty,-1) ∪(1,\infty)\\\text{...(i)}\\\text{When}\space \vec{r}.\hat{k}\gt0\\\lbrace(x^{2}-1)\hat{i} + 5\hat{j} - 3(x^{2} -4)\hat{k}\rbrace.\hat{k}\gt 0\\\Rarr\space -3(x^{2}-4)\gt 0\\\Rarr\space x^{2}-4\lt 0\\\Rarr (x-2)(x+2)\lt 0$$

x = (– 2, 2) ...(ii)

From equations (i) and (ii),

x = (– 2, – 1) ∪ (1, 2). Ans.

OR

(ii) Let A be origin and position vectors of the points B, C and D be

$$\vec{b},\vec{c}\space\text{and}\vec{d}\space\text{respectively.}\\\therefore\space \vec{\text{AB}} =\vec{b},\vec{\text{BC}} =\vec{c} -\vec{b},\\\vec{\text{BD}} = \vec{d} -\vec{b},\vec{\text{CD} =}\vec{d} -\vec{c},\\\vec{\text{AD}} =\vec{d}\space\text{and}\space\vec{\text{CA}} =-\vec{c}$$

Now,

$$\text{L.H.S} =|\vec{\text{AB}}×\vec{\text{CD}} +\vec{\text{BC}}×\vec{\text{AD}}+\\\vec{\text{CA}}×\vec{\text{BD}}|\\= |\vec{b}×(\vec{b} -\vec{c}) + (\vec{c} -\vec{b})×\vec{d}+\\(\vec{\normalsize-c})×(\vec{d} -\vec{b})|\\ =|\vec{b}×\vec{d} -\vec{b}×\vec{c}+ \vec{c}× \vec{d} \\- \vec{b}×\vec{d} -\vec{c}×\vec{d}-\vec{c}×\vec{b}|\\ =\bigg|-\vec{b}×\vec{c} + \vec{c}×\vec{b}\bigg|\\=\bigg|2×(\vec{c}×\vec{b})\bigg|\\ = 2\bigg|\vec{\text{AC}}×\vec{\text{AB}}\bigg|$$

= 2 × 2 ar (∆ABC)

= 4 ar (∆ABC)

= R.H.S. Hence Proved.

Solution 18.

Equation of the line passing through A(2, – 2) and B(4, 3),

$$y -y_{1} =\frac{y_{2} -y_{1}}{x_{2} -x_{1}}(x-x_{1})\\\Rarr\space y + 2 =\frac{3 + 2}{4-2}(x-2)\\\Rarr\space 2y +4 = 5x -10$$

∴ Equation of line AB,

$$x = \frac{2y + 14}{5}$$

Equation of the line passing through B(4, 3) and C(1, 2),

$$\Rarr\space y-3 =\frac{2-3}{1-4}(x-4)$$

$$\Rarr\space y-3=\frac{\normalsize-1}{\normalsize-3}(x-4)\\\Rarr\space 3y-9 = x-4$$

∴ Equation of line BC,

x = 3y – 5

Equation of the line passing through A(2, – 2) and C(1, 2),

$$y +2=\frac{2+2}{1-2}(x-2)\\\Rarr\space -y-2 = 4x-8\\\therefore\space\text{Equation of line AC,}\\ x =\frac{6-y}{4}$$

ar ∆ABC = ar DABF – ar DACE – ar BFEC

$$=\int^{3}_{\normalsize-2}\frac{2y + 14}{5}dy -\int^{2}_{-2}\frac{6-y}{4}dy-\\\int^{3}_{2}(3y-5)dy\\= \frac{1}{5}\lbrack y^{2} + 14y\rbrack^{3}_{\normalsize-2}-\frac{1}{4}\begin{bmatrix}6y -\frac{y^{2}}{2}\end{bmatrix}^{2}_{\normalsize-2}-\\\begin{bmatrix}\frac{3y^{2}}{2} - 5y\end{bmatrix}^{3}_{2}\\=\frac{1}{5}\lbrack 51 + 24\rbrack-\frac{1}{4}\lbrack10 +14\rbrack-\\\begin{bmatrix}-\frac{3}{2} + 4\end{bmatrix}\\ = 15-6-\frac{5}{2} =\frac{13}{2}$$

Hence, area of the required triangle is

$$\space \frac{13}{2}\space\text{sq. units.}\space\text{Ans.}$$

## Section-C

Solution 19.

(i) (d) 50

$$\text{p = 500 + 20x -}\frac{x^{2}}{3}\\\text{R(x)} = p x\\=\bigg(500 + 20x -\frac{x^{2}}{3}\bigg)(x)\\= 500x + 20 x^{2}-\frac{x^{3}}{3}\\\text{M.R.} =\frac{d}{dx}\text{R}(x) \\=\frac{d}{dx}\bigg(500x + 20 x^{2} -\frac{x^{3}}{3}\bigg)\\\Rarr\space \text{M.R.} = 500 + 40 x-\frac{3x^{2}}{3}\\\Rarr\space \text{M.R.} = 500 + 40 x - x^{2}$$

= (50 – x) (10 + x)

∵ M.R. = 0 [Given]

(50 – x) (10 + x) = 0

If 50 – x = 0, then x = 50 unit.

If 10 + x = 0, then x = – 10 unit (Not possible)

∴ x = 50

(ii) (d) 2

$$\text{C(x)} = 5 +\frac{48}{x} + 3x^{2}\\\frac{d}{dx}\text{C(x)} =0 -\frac{48}{x^{2}} + 6x \\\text{For minimum cost}\\\frac{d}{dx}\text{C(x)} = 0\\-\frac{48}{x^{2}} +6x = 0\Rarr\space x^{3} = 8\\\therefore\space x = 2$$

(iii) R(x) = 13x2 + 26x + 15

$$\text{M.R} = \frac{d}{dx}(\text{R}(x)) \\=\frac{d}{dx}(13 x^{2} + 26x + 15)$$

= 26x + 26 + 0

∴ M.R. = 26x + 26

(M.R.)x = 7 = 26 × 7 + 26

= 182 + 26

M.R. at 7 units = ₹ 208 Ans.

(iv) Here Revenue, R(x) = (Price) × (Demand)

= 55x (Let be Demand is x)

Cost, C(x) = 30x + 250 (given in graph)

∴ Profit function, P(x) = R(x) – C(x)

= 55x – (30x + 250)

= 55x – 30x – 250

= 25x – 250

To determine the break-even point, we have P(x) = 0

⇒ 25x – 250 = 0

⇒ x = 10

Hence, the break-even point is x = 10. At this level of production.

Revenue = cost = ₹550 and profit = 0.

(v) (a) 64 crore

$$\text{Given,}\space\bar{\text{X}} = \text{mean of sales = 40}\\\bar{\text{Y}} =\text{mean of advertisement cost}\\ = 6\\\text{Formula: Regression equation of X on Y}\\\text{X -} \bar{\text{X}} = r\frac{\sigma_{x}}{\sigma_{y}}(y -\bar{y})$$

Calculation

⇒ X – 40 = 0.9 (10/1.5)(Y – 6)

⇒ X – 40 = (9/1.5)(Y – 6)

⇒ X – 40 = 6(Y – 6)

⇒ X – 40 = 6Y – 36

⇒ X = 6Y + 4

⇒ X = 6(10) + 4 = 64

∴ The likely sales for a proposed advertisement expenditure of ₹10 crore is 64 crore.

Solution 20.

$$\text{(i) Given,}\\\text{C(x)} = \frac{x^{3}}{3} - 45x^{2} - 900 x + 36\\\text{M.C.} = \frac{d}{dx}\text{C(x)}\\= \frac{3x^{2}}{3} - 90x -900$$

= x2 – 90x – 900

$$\text{Also,\space}\frac{d}{dx}(\text{MC}) = 2x-90$$

Now, the marginal cost is to be minimised i.e.

$$\frac{d}{dx}(\text{M.C.}) = 2x-90\\\frac{d}{dx}(\text{M.C.}) = 0$$

2x – 90 = 0

⇒ x = 45

$$\frac{d^{2}}{dx^{2}}(\text{M.C.}) = 2\\\therefore\space \frac{d^{2}}{dx^{2}}(\text{M.C.})\gt 0$$

So, marginal cost is minimum.

∴ Number of units produced = 45. Ans.

OR

(ii) We have,

$$\text{MR} = 8 + 9x^{2} -\frac{3}{2}x\\\text{We know that}\\\text{MR} =\frac{\text{dR}}{dx},$$

where R is the total revenue function

dR = MR dx

On integrating both sides, we get

∫ dR = ∫ MR dx + c

where c is a constant of integration

$$\text{R} =\int\bigg(8 + 9x^{2}-\frac{3}{2}x\bigg)dx + c\\\text{R} = 8x + 3x^{2}-\frac{3}{4}x^{2} + c\\\text{Now,\space when x = 0, R = 0}\\\Rarr\space c = 0\\\therefore\space \text{R} = 8x +3x^{2}-\frac{3}{4}x^{2}$$

Hence, demand function is

$$\text{P} =\frac{\text{R}}{\text{x}}\\\Rarr\space \text{P} = 8 + 3x^{2}-\frac{3}{4}x\\= 3x^{2}-\frac{3}{4}x + 8\\\textbf{Ans.}$$

Solution 21.

(i) Consider the given data,

 x y dx = x – 8 dy = y – 8 dx2 dxdy 11 10 3 2 9 6 7 8 -1 0 1 0 9 6 1 -2 1 -2 5 5 -3 -3 9 9 8 9 0 1 0 0 6 7 -2 -1 4 2 10 11 2 3 4 6 ∑x = 56 ∑y = 56 ∑dx2= 28 ∑dxdy = 21

$$\therefore\space \bar{x} =\frac{\Sigma x}{7} =\frac{56}{7} =8;\\\bar{y} =\frac{\Sigma y}{7}=\frac{56}{7} =8\\\text{The regression equation of y on x is}\space\\\text{y -}\bar{y} =\frac{\Sigma d_xd_y}{\Sigma x^{2}}(x - \bar{x})\\\Rarr\space y-8=\frac{21}{28}(x-8)\\\Rarr\space y-8 =\frac{3}{4}(x-8)\\\Rarr\space 4y-32 = 3x - 24\\\Rarr\space 4y = 3x + 8\\\textbf{Ans.}$$

(ii) When x = 13 ⇒ 4y = 39 + 8

$$\Rarr\space y =\frac{47}{4}= 11.75\\\text{Hence, the initial start should be}\\₹ 11.75\space\textbf{Ans.}$$

Solution 22.

(i) Let x be the number of executive class tickets sold and y be the number of economy class tickets sold.

Let Z be the total profit.

Then, Maximize, Z = 500x + 400y

Subject to constraints : x + y ≤ 200

x ≥ 20

y ≥ 4x

x ≥ 0, y ≥ 0

Consider, x + y = 200

 x 200 0 y 0 200

y = 4x

 x 0 20 40 y 0 80 160

The shaded portion is the required feasible region.

 Corner Points Z = 500x + 400y A (20, 80) Z = 500 × 20 + 400 × 80 = ₹42000 B (40, 160) Z = 500 × 40 + 400 × 160 = ₹84000 C (20, 180) Z = 500 × 20 + 400 × 180 = ₹82000

Hence, maximum profit of ₹84,000 is earned by selling 40 executive class and 160 economy class tickets.

Ans.

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

To maximize 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.