# Oswal Specimen Papers ISC Class 12 Mathematics Solutions (Specimen Paper - 7)

## Section-A

Solution 1.

(i) (d) None of these

Explanation :

a ≤ b3 if a = -2 then b =-8

$$\because -2\nleq -8$$

∴ It is not reflexive.

Also, less than or equal to relation is not symmetric and transitive.

(ii) (c) Function is called identity function

(iii) (a) x

Explanation :

sin [cot–1 {tan (cos–1 x)}]

Let cos–1 x = y ⇒ x = cos y ...(i)

Now, sin [cot–1 {tan (cos–1 x)} = sin [cot–1 {tan y}]

$$= sin\bigg[\text{cot}^{\normalsize-1} \text{cot}\bigg(\frac{\pi}{2}-y\bigg)\bigg]\\\begin{bmatrix}\therefore\space \text{tan y} =\text{cot}\bigg(\frac{\pi}{2}-y\bigg)\end{bmatrix}\\=\text{sin}\bigg(\frac{\pi}{2}-y\bigg)\\\begin{bmatrix}\therefore\space \text{cos y} =\text{sin}\bigg(\frac{\pi}{2}-y\bigg)\end{bmatrix}$$

= cos y = x [∵ from (i)]

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

Explanation :

$$\Delta =\begin{vmatrix}1 & 1 & 1\\ 1 & 1+\text{sin}\space\theta &1\\ 1 + \text{cos}\space\theta & 1 & 1\end{vmatrix}\\\text{Applying R}_{1}\xrightarrow{} \text{R}_{1}-\text{R}_{3},\\\Delta=\begin{vmatrix}-\text{cos}\theta &0 &0\\ 1 & 1 +\text{sin}\space\theta &1\\1 +\text{cos}\space\theta &1 &1\end{vmatrix}\\= -\text{cos}\theta\begin{vmatrix}1 + \text{sin}\theta &1\\1 &1\end{vmatrix}$$

= – cos θ (1 + sin θ – 1) ⇒ – sin θ cos θ

= – sin θ cos θ

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

We know

$$\frac{-\pi}{2}\leq\text{sin}\space\theta\leq\frac{\pi}{2}$$

∴ Maximum value of

$$\text{sin 2}\theta = \text{sin}\frac{\pi}{2}$$

$$\therefore\space \theta=\frac{\pi}{4},\frac{3\pi}{4}\\\text{Hence, Maximum value =}\\\begin{vmatrix}\frac{\normalsize-1}{2}\text{sin}2\theta\end{vmatrix}\\=\begin{vmatrix}\frac{-1}{2}×1\end{vmatrix}=\frac{1}{2}$$

(v) (c) I

Explanation :

A2 = A [Given]

Now, (I + A)3– 7A = I3+ A3 + 3I2A + 3IA2– 7A

= I + A3 + 3A + 3A2– 7A [ ∵ I = I2 = I3]

= I + A2.A + 3A + 3A – 7A

[∵ A2 = A]

= I + A.A – A

= I + A2 – A

= I + A – A [∵ A2= A]

= I

$$\text{(vi)\space(c)\space}\frac{7}{8}$$

Explanation :

$$\therefore\space \text{P(A)} =\frac{4}{5}, \text{P(A ∩ B)}=\frac{7}{10}\\\because\space \text{P(B|A)}=\frac{\text{P(A∩B)}}{\text{P(A)}} =\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7}{8}.$$

(vii) (d) – 11

Explanation :

$$\begin{vmatrix}x &2 &x\\ x^{2} &x &6\\ x &x &6\end{vmatrix}$$

= ax4 + bx3 + cx2 + dx + e

$$x\begin{vmatrix}x &6 \\ x &6\end{vmatrix}-2\begin{vmatrix}x^{2} &6 \\ x &6\end{vmatrix}+x\begin{vmatrix}x^{2} &x\\ x &x\end{vmatrix}$$

= ax4 + bx3 + cx2 + dx + e

⇒ 6x2 – 6x2 – 12x2 + 12x + x4 – x3 = ax4
+ bx3 + cx2 + dx + e

⇒ x4 – x3 – 12x2 + 12x = ax4 + bx3 + cx2+ dx + e

∴ a = 1, b = – 1, c = – 12, d = 12, e = 0

∴ 5a + 4b + 3c + 2d + e = 5(1) + 4(– 1) + 3(– 12) + 2(12) + 0

= 5 – 4 – 36 + 24 + 0 = – 11.

(viii) (d) 1

Explanation :

S = {1, 2, 3, 4, 5, 6}

A = {4, 5, 6}

B = {1, 2, 3, 4}

$$\text{P(A)} =\frac{3}{6}\\\text{P(B)} =\frac{4}{6}$$

A ∩ B = {4}

$$\text{and}\space \text{P(A ∩ B)}=\frac{1}{6}\\\text{We know,}\space\text{P(A ∪ B) = }\\\text{P(A) + P(B) – P(A ∩ B)}\\=\frac{3}{6} +\frac{4}{6}-\frac{1}{6} =\frac{6}{6} = 1$$

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

Explanation :

y = [x(x – 2)]2

= (x2 – 2x)2

$$\therefore\space\frac{dy}{dx} = 2(x^{2}- 2x)(2x-2)\\\text{or}\space\frac{dy}{dx}= 4x(x-1)(x-2)\\\text{given}\space\frac{dy}{dx} = 0$$

4x(x – 1) (x – 2) = 0

x = 0, x = 1, x = 2

Intervals are

(– ∞, 0), (0, 1) (1, 2) and (2, ∞).

∴ f(x) is increasing in (0, 1) ∪ (2, ∞).

Hence, both statements are true.

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

Explanation :

$$f(0) = \text{RHL} =\lim_{x\to 0^{+}}f(x) =\lim_{h\to0} f(h)$$

$$=\lim_{h\to0}\space \frac{(27 -2h)^{\frac{1}{3}}-3}{9-3(243 + 5h)^{\frac{1}{5}}}\\=\frac{1}{3}\space\lim_{h\to 0}\frac{\frac{(27-2h)^{\frac{1}{3}}- (27)^{\frac{1}{3}}}{(27-2h) - 27}× (-2h)}{\frac{(243)^{\frac{1}{5}} -(243 + 5h)^{\frac{1}{5}}}{243-(243 + 5h)}×(-5h)}$$

$$=\frac{1}{3}.\frac{2}{5}.\frac{\frac{1}{3}(27)^{\frac{-2}{3}}}{\frac{1}{5}(243)^{\frac{-4}{5}}}$$

$$\bigg[\because\frac{x^{n}- a^{n}}{x-a} = nx^{n-1}\bigg]\\=\frac{2}{15}.\frac{5}{3}.\frac{3^{\normalsize-2}}{3^{\normalsize-4}} =\frac{2}{9}.9 = 2$$

(xi) 1

$$\because\space \begin{bmatrix}2 &1\\2 &2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}\\\Rarr\space \begin{bmatrix}2x+y\\ 2x + 2y\end{bmatrix} =\begin{bmatrix}2 \\3\end{bmatrix}$$

On comparing, we get

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

and 2x + 2y = 3 ...(ii)'

On solving equations (i) and (ii), we get
y = 1

(xii) Let A be the event of getting a multiple of 2 on first die and B be the event of getting a prime number on second die. Then

A = {2, 4, 6} and B = {2, 3, 5}

∴ Required probability = P(A | B)

$$=\frac{\text{(P∩B)}}{\text{P(B)}}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}.$$

(xiii) (c) y = cosec–1 x

(xiv) 0

y = esin x

$$\therefore\space \frac{dy}{dx} = \text{cos x e}^{sinx}\\\Rarr\space\frac{dy}{dx}|_{x =\frac{\pi}{2}} =\text{cos}\frac{\pi}{2}e^{\frac{sin\pi}{2}}$$

= 0 × e1= 0

(xv) Zero

$$\int^{3}_{3}\text{sin x dx} =0\\\bigg[\because\space\int^{a}_{a}\text{f(x)dx} =0\bigg]$$

Solution 2.

(i) We observe the graph that 1 and – 1 ∈ R such that f(– 1) = f(1) i.e., there are two distinct elements in R which have the same image. So f is not one-one since f(x) assumes only non-negative values.

So, no negative real number in R (co-domain) has its pre-image in domain of f i.e., R, consequently f is not onto.

OR

(ii) R : {(x, y) : | x2 – y2 | < 1}

A = {1, 2, 3, 4, 5}

∴ R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}.

Ans.

Solution 3.

Given, tan–1 (x – 1) + tan–1 x + tan–1 (x + 1) = tan–1 3x

$$\Rarr\space \text{tan}^{-1}(x-1) + \text{tan}^{-1}(x+1)\\=\text{tan}^{\normalsize-1} 3x -\text{tan}^{-1} x$$

$$\Rarr\space \begin{Bmatrix}\frac{(x+1) + (x-1)}{1- (x+1)(x-1)}\end{Bmatrix}\\=\text{tan}^{\normalsize-1}\frac{3x-x}{1 + 3x.x}\\\Rarr\space \frac{2x}{1 - (x^{2}-1)}\\=\frac{2x}{1 + 3x^{2}}\\\Rarr\space \frac{x}{2-x^{2}} =\frac{x}{1 + 3x^{2}}$$

⇒ x(1 + 3x2 ) = x(2 – x2 )

⇒ x + 3x3 = 2x – x3

⇒ 4x3 – x = 0

⇒ x(4x2 – 1) = 0

⇒ x = 0 or 4x2 – 1 = 0

$$\Rarr\space x = 0\space \text{or}\space 4x^{2}-1=0\\\Rarr\space x =0\space\text{or}\space x^{2}=\frac{1}{4}\\\Rarr\space x =0\space\text{or}\space x^{2} =\frac{1}{4}\\\Rarr\space x =0\space\text{or}\space x =\pm\frac{1}{2}\\\therefore\space \text{x = 0,}\pm\frac{1}{2}\space \textbf{Ans.}$$

Solution 4.

$$\text{Given,\space}\\\text{tan}^{-1}2x + \text{tan}^{\normalsize-1}3x =\frac{\pi}{4}\\\Rarr\space \text{tan}^{\normalsize-1}\bigg[\frac{2x+3x}{1-2x× 3x}\bigg]=\frac{\pi}{4}\\\Rarr\space \text{tan}^{\normalsize-1}\bigg[\frac{5x}{1-6x^{2}}\bigg]=\frac{\pi}{4}\\\Rarr\space \frac{5x}{1-6x^{2}}=\text{tan}\frac{\pi}{4}=1$$

$$\Rarr\space 5x = 1-6x^{2}\\\Rarr\space 6x^{2}+5x-1 =0\\\Rarr\space(6x-1)(x+1) =0\\\Rarr\space x =\frac{1}{6},\normalsize-1\\\textbf{Ans.}$$

Solution 5.

$$\text{(i)\space}\int\frac{\sqrt{x}}{a^{3}-x^{3}}dx\\\text{Let x}^{\frac{3}{2}} =t\\\frac{3}{2}x^{\frac{1}{2}} dx=dt\\\sqrt{x}dx=\frac{2}{3}\text{dt}\\\therefore\space \int\frac{\sqrt{x}}{a^{3}-x^{3}}dx =\\\int\frac{\frac{2}{3}}{a^{3} -t^{2}}dt =\frac{2}{3}\int\frac{dt}{(a^{\frac{3}{2}})^{2}-t^{2}}$$

$$=\frac{2}{3}.\frac{1}{2a^{\frac{3}{2}}}\text{log}\bigg|\frac{a^{\frac{3}{2}}+t}{a^{\frac{3}{2}}-t}\bigg| + c\\=\frac{1}{3a^{\frac{3}{2}}}\text{log}\bigg|\frac{a^{\frac{3}{2}} + x^{\frac{3}{2}}}{a^{\frac{3}{2}}-x^{\frac{3}{2}}}\bigg| + c.$$

Ans.

OR

(ii) Let

$$\text{I} =\int\frac{dx}{\sqrt{\text{sin}^{3}x \text{sin}(x+\alpha)}}\\\text{I}=\int\frac{dx}{\sqrt{\text{sin}^{3}x(\text{sin x cos}\alpha + \text{cos} x\text{sin}\alpha)}}\\\text{I} =\int\frac{dx}{\sqrt{\text{sin}^{4}x \text{sin}\alpha\bigg(\frac{\text{cos}\alpha}{\text{sin}\alpha} + \frac{\text{cos x}}{\text{sin x}}\bigg)}}\\\text{I} =\int\frac{dx}{\text{sin}^{2}x\sqrt{\text{sin}\alpha(\text{cot}\alpha + \text{cot x})}}\\\text{I} =\int\frac{\text{cosec}^{2}xdx}{\sqrt{sin}\alpha\sqrt{\text{cot}\alpha + \text{cot}x}}$$

Let cot α + cot x = t2

$$\Rarr\space -\text{cosec}^{2}x =2t\frac{dt}{dx}\\\Rarr\space\text{cosec}^{2}xdx =- 2dt\\\therefore\space \text{I}=\int\frac{-2tdt}{\sqrt{\text{sin}\space\alpha}\sqrt{t^{2}}}\\\text{I}=\frac{\normalsize-2}{\sqrt{\text{sin}\alpha}}\int dt\\\text{I}=\frac{-2t}{\sqrt{\text{sin}\space\alpha}} +\text{C}\\\text{I} =\frac{-2\sqrt{\text{cot}\alpha + \text{cot}x}}{\sqrt{\text{sin}\alpha}} +\text{C}\\\text{I} = -2\sqrt{\frac{\text{cot} \alpha + \text{cot} x}{\text{sin}\space\alpha}} + \text{C}\\\text{I} =-2\sqrt{\frac{\text{cos}\alpha}{\text{sin}\alpha} + \frac{\text{cos}x}{\text{sin}x}} +\text{C}\\\text{I}=-2\sqrt{\frac{\text{cos}\alpha \text{sin} x +\text{cos x sin}\alpha}{\text{sin}^{2}\alpha sin x}}+ \text{C}$$

$$\text{I}=\frac{\normalsize-2}{\text{sin}\space\alpha}\sqrt{\frac{\text{sin(x +}\alpha)}{\text{sin x}}}+ \text{C}\\\text{I}=-2\text{cosec}\space\alpha\sqrt{\frac{\text{sin(x +} \alpha)}{\text{sin x}}} +\text{C}\\\textbf{Ans.}$$

Solution 6.

The given differential equation is,

$$2x^{2}\frac{dy}{dx}-2xy +x^{2} =0\\\Rarr\space\frac{dy}{dx}-\frac{2xy}{2x^{2}} +\frac{x^{2}}{2x^{2}} =0\\\Rarr\frac{dy}{dx}-\frac{y}{x} +\frac{1}{2} = 0\\\Rarr\space \frac{dy}{dx}-\frac{y}{x}=-\frac{1}{2}$$

This is the linear differential equation of the form

$$\frac{dy}{dx}+Py = Q.\\\text{where\space}\text{P}=-\frac{1}{x},\text{Q}=-\frac{1}{2}\\\therefore\space \text{Integrating factor (I.F.) =} e^{\int\text{Pdx}}\\=e^{\int-\frac{1}{x}dx}$$

= e– log x = elog x–1 = x–1

$$=\frac{1}{x}$$

The solution of the differential equation is given as :

$$\text{y.I.F} =\int\text{Q.I.F.dx + c}\\\Rarr\space y×\frac{1}{x}=\int\bigg(-\frac{1}{2}×\frac{1}{x}\bigg)dx + \text{C}\\\Rarr\space\frac{y}{x}=\int\bigg(-\frac{1}{2}×\frac{1}{x}\bigg)dx + \text{C}\\\frac{y}{x} =-\frac{1}{2}\text{log x + C}$$

where C is the constant of integration.

Ans.

Solution 7.

$$\text{Let\space} \text{I} = \int^{3}_{\normalsize-6}|x +3|dx\\\text{Here}\space |x + 3|=\\\begin{cases}-(x+3), x\lt\normalsize-3\\(x+3), x\gt-3\end{cases}\\\text{I}=\int^{\normalsize-3}_{\normalsize-6}-(x+3)dx +\int^{3}_{\normalsize-3}(x+3)dx$$

$$=-\bigg[\frac{x^{2}}{2} +3x\bigg]^{3}_{-6} +\bigg[\frac{x^{2}}{2} + 3x\bigg]^{3}_{-3}$$

$$=-\bigg[\bigg(\frac{9}{2}-9\bigg)-\bigg(\frac{36}{2}-18\bigg)\bigg]+\\\bigg[\bigg(\frac{9}{2} +9\bigg) -\bigg(\frac{9}{2}-9\bigg)\bigg]\\=\frac{9}{2} +18=\frac{45}{2}.\space\textbf{Ans.}$$

Solution 8.

The given curve is

x2 = 4y ...(i)

Let (x1, y1) be the required point on curve

∴ x1 2 = 4y1 ...(ii)

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

$$2x=4\frac{dy}{dx}\\\Rarr\space\frac{dy}{dx} =\frac{x}{2}\\\therefore\space \text{Slope of tangent,}\\{\text{m} = }\frac{dy}{dx}|(x_1,y_1)=\frac{x_1}{2}\\\text{and\space}\text{Slope of normal, m′ =}-\frac{1}{m}\\=-\frac{2}{x_1}$$

∴ Equation of normal at (x1, y1) is, y – y1 = m′ (x – x1)

It passes through (1, 2),

$$2-y_1=-\frac{2}{x_1}(1 -x_1)$$

⇒ 2x1 – x1y1 = –2 + 2x1

⇒ x1y1 = 2

Put value of y1 from equation (ii)

$$x_{1}\frac{x_{1}^{2}}{4}=2\\\Rarr\space x_{1}^{3} =8\Rarr\space x_1=2\\\therefore\space\text{From (ii),}\space y_{1}\frac{x_1^{2}}{4}=\frac{2^{2}}{4}=1\\\therefore\space\text{Point on curve is (2, 1) and}\\ m =\frac{2}{2}=1, m' =-1$$

Equation of normal at (2, 1) and m′ = –1 is

y – 1 = –(x – 2)

⇒ x + y = 3

Equation of tangent at (2, 1) and m = 1

y – 1 = (x – 2)

⇒ x – y = 1.  Ans.

Solution 9.

(i) Consider,

$$\int^{1}_{0}\frac{xe^{x}}{(1 +x^{2})}dx =\int^{1}_{0}\frac{(x+1)-1}{(1 +x^{2})}e^{x}dx$$

$$=\int^{1}_{0}\frac{(x+1)e^{x}}{(x+1)^{2}}dx -\int^{1}_{0}\frac{e^{x}}{(x+1)^{2}}dx\\=\int^{1}_{0}\frac{e^{x}}{x+1}dx-\int^{1}_{0}\frac{e^{x}}{(x+1)^{2}}dx$$

Integrating by parts, we get

$$=\bigg[e^{x}\frac{1}{x+1}\bigg]^{1}_{0}-\int^{1}_{0}e^{x}\bigg[-\frac{1}{(x+1)^{2}}\bigg]dx-\\\int^{1}_{0}\frac{e^{x}}{(x+1)^{2}}dx$$

$$=\bigg[\frac{1}{2} e^{1}-1.e^{0}\bigg] +\int^{1}_{0}\frac{1}{(x+1)^{2}}.e^{x}dx\\-\int^{1}_{0}\frac{1}{(x+1)^{2}e^{x}}dx\\=\frac{1}{2}e-1=\frac{e}{2}-1\space\textbf{Ans.}$$

OR

$$\text{(ii)\space}\text{I}=\int\frac{\sqrt{1 +x^{2}}}{1 - x^{2}}dx\\\Rarr\space \text{I}=\int\frac{\sqrt{1 +x^{2}}}{1-x^{2}}×\frac{\sqrt{1 +x^{2}}}{\sqrt{1 + x^{2}}}dx\\\Rarr\space \text{I}=\int\frac{1+x^{2}}{(1-x^{2})\sqrt{1 + x^{2}}}dx\\\Rarr\space \text{I}=\int\frac{2-(1-x^{2})}{(1-x^{2})\sqrt{1 +x^{2}}}dx\\\Rarr\space \text{I}=\int\frac{2}{(1-x^{2})\sqrt{1 + x^{2}}}dx-\\\int\frac{1-x^{2}}{(1-x^{2})\sqrt{1 + x^{2}}}dx$$

$$\Rarr\space\text{I =2}\int\frac{dx}{\sqrt{1 +x^{2}}(1-x^{2})}-\int\frac{dx}{\sqrt{1 + x^{2}}}\\\Rarr\space\text{I = 2}\int\frac{dx}{\sqrt{1 + x^{2}}(1-x^{2})}-\\\text{log}\bigg|x + \sqrt{1 +x^{2}}\bigg| + \text{C}_{1}\\\text{...(i)}\\\Rarr\space\text{I} = 2\text{I}_{1}-\text{log}\bigg|x + \sqrt{1 + x^{2}}\bigg| + \text{C}_{1}\\\Rarr\space \text{I}_{1}=\int\frac{dx}{\sqrt{1 + x^{2}}(1-x^{2})}\\\text{Put\space} x =\frac{1}{t}\\\Rarr\space dx =-\frac{1}{t^{2}}dt$$

$$\text{I}_{1}=\int\frac{-\frac{1}{t^{2}}dt}{\sqrt{1 + \frac{1}{t^{2}}}\bigg(1-\frac{1}{t^{2}}\bigg)}\\\text{I}_{1}=\int\frac{-t}{\sqrt{t^{2}+1}(\text{t}^{2}-1)}$$

Put t2+ 1 = u2

⇒ 2t dt = 2u du

⇒ tdt = udu

$$\text{I}_{1}=\int\frac{-u}{u(u^{2}-2)}du =-\int\frac{du}{u^{2}-(\sqrt{2})^{2}}\\=-\frac{1}{2\sqrt{2}}\text{log}\bigg|\frac{u-\sqrt{2}}{u + \sqrt{2}}\bigg|+\text{C}_{2}\\=-\frac{1}{2\sqrt{2}}\text{log}\bigg|\frac{\sqrt{t^{2}+1}-\sqrt{2}}{\sqrt{t^{2}+1} + \sqrt{2}}\bigg|+ \text{C}_{2}\\=-\frac{1}{2\sqrt{2}}\text{log}\bigg|\frac{\sqrt{1 + x^{2}}-\sqrt{2}x}{\sqrt{1 + x^{2}} + \sqrt{2}x}\bigg|-\\\text{log}\bigg|x + \sqrt{x^{2}+1}\bigg| + \text{C}_{1}+\text{C}_{2}\\\therefore\space \text{I}=\frac{\normalsize-1}{\sqrt{2}}\text{log}\bigg|\frac{\sqrt{1 +x^{2}}-\sqrt{2}x}{\sqrt{1 + x^{2}} + \sqrt{2}x}\bigg|-\\\text{log}\bigg|x + \sqrt{x^{2}+1}\bigg|+\text{C}$$

Where C = C1 + C2. Ans.

Solution 10.

(i) Let A and B denote the events that the student has failed in mathematics and chemistry, respectively.

Then, P (A) = 30/100 = 3/10, P (B) = 20/100 = 2/10 and, P (A ∩ B) = 10/100 = 1/10

(a) The probability that the student has failed either in chemistry or in mathematics is

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

$$=\frac{3}{10} +\frac{2}{10}-\frac{1}{10}\\=\frac{4}{10}=\frac{2}{5}\space\textbf{Ans.}$$

(b) The probability that the student has failed in mathematics, knowing that he has failed in chemistry is

$$\text{P(A|B)}=\frac{\text{P(A∩B)}}{\text{P(B)}}\\=\frac{\frac{1}{10}}{\frac{2}{10}}=\frac{1}{2}\space\textbf{Ans.}$$

OR

$$\text{(ii) Given:P\text{(A)}}=\frac{5}{6},\text{P(B)}=\frac{4}{5}\\\text{and P(C)}=\frac{3}{4}\\\text{So,\space}\text{P}(\bar{\text{A}}) = 1-\text{P(A)}\\= 1-\frac{5}{6}=\frac{1}{6}\\\text{P}(\bar{\text{B}}) = 1-\text{P(B)}\\=1-\frac{4}{5}=\frac{1}{5}\\\text{P}(\bar{\text{C}}) =1-\text{P(C)}\\= 1-\frac{3}{4}=\frac{1}{4}$$

(a) Probability (exactly two persons hit the target)

$$=\text{P(A)}.\text{P(B)}.\text{P}(\bar{\text{C}}) + \text{P(A).\text{P}}\bar{(\text{B})}.\text{P(C)}+\\\text{P}(\bar{\text{A}}).\text{P(B)}.\text{P(C)}\\=\frac{5}{6}×\frac{4}{5}×\frac{1}{4} +\frac{5}{6}×\frac{1}{5}×\frac{3}{4}+\\\frac{1}{6}×\frac{4}{5}×\frac{3}{4}\\=\frac{1}{6}+ \frac{1}{8}+\frac{1}{10}\\=\frac{20 + 15 + 12}{120}=\frac{47}{120}\space\textbf{Ans.}$$

(b) Probability (at least one person hits the target)

= 1 – No one will hit the target

$$= 1 -(\text{P}(\bar{\text{A}})\text{P}(\bar{\text{B}})\text{P}(\bar{\text{C}}))\\=1 -\bigg(\frac{1}{6}×\frac{1}{5}×\frac{1}{4}\bigg)\\=1-\frac{1}{120}\\=\frac{120-1}{120}\\=\frac{119}{120}\space\textbf{Ans.}$$

Solution 11.

$$\text{L.H.S}=\begin{vmatrix}x &x^{2}& 1 + px^{3}\\y &y^{2} &1 + \text{py}^{3}\\z &z^{2} &1 + pz^{3}\end{vmatrix}\\=\begin{vmatrix}x &x^{2} &1\\y &y^{2} &1\\z &z^{2} &1\end{vmatrix} + \begin{vmatrix}x &x^{2} &px^{3}\\y &y^{2} &py^{3}\\z &z^{2} &pz^{3}\end{vmatrix}\\=\begin{vmatrix}x &x^{2} &1\\y &y^{2} &1\\z &z^{2} &1\end{vmatrix}+ p\begin{vmatrix}x &x^{2} &x^{3}\\y &y^{2} &y^{3}\\ z&z^{2} &z^{3}\end{vmatrix}\\=\begin{vmatrix}1 &x &x^{2}\\1 &y &y^{2}\\ 1 &z &z^{2}\end{vmatrix}+ pxyz\begin{vmatrix}1 &x &x^{2}\\ 1&y &y^{2}\\1 &z &z^{2}\end{vmatrix}\\=(1 + pxyz)\begin{vmatrix}1 & x & x^{2}\\ 1& y & y^{2}\\ 1 &z &z^{2}\end{vmatrix}$$

Applying R1 → R1 – R2, R2 → R2 – R3,

$$=(1 + pxyz)\begin{vmatrix}0 &x-y &x^{2}-y^{2}\\0 &y-z &y^{2}-z^{2}\\1 &z &z^{2}\end{vmatrix}\\=(1 + pxyz)(x-y)(y-z)\begin{vmatrix}0 &1 &x+y\\0 &1 &y+z\\1 &z &z^{2}\end{vmatrix}$$

= (1 + pxyz) (x – y) (y – z).1[(y + z) – (x + y)]

= (1 + pxyz) (x – y) (y – z) (z – x)

= R.H.S. Hence Proved.

Solution 12.

(i) Put x = tan  θ

⇒ dx = sec2  θ d θ

and  θ = tan–1 x

When x = 1,

$$\theta =\text{tan}^{\normalsize-1}1=\frac{\pi}{4}\\\text{When x = 0,}\space\\\theta=\text{tan}^{\normalsize-1}0=0\\\therefore\space \text{I} =\int^{\frac{\pi}{4}}_{0}\text{sin}^{\normalsize-1}\bigg(\frac{2\text{tan}\theta}{1 +\text{tan}\theta}\bigg).\text{sec}^{2}\theta d\theta\\\bigg[\because\space \text{sin 2A} =\frac{2\text{tan A}}{\text{1 + tan}^{2}\text{A}}\bigg]\\=\int^{\frac{\pi}{4}}_{0}\text{sin}^{\normalsize-1}(\text{sin}2 \theta)\text{sec}^{2}\theta d\theta\\=\int^{\frac{\pi}{4}}_{0} 2\theta\text{sec}^{2}d\theta = 2\int^{\frac{\pi}{4}}_{0}\space\underset{\text{I}}{\theta}\space\underset{\text{II}}{\text{sec}^{2}\theta d\theta}$$

$$= 2\begin{bmatrix}\theta\int^{\frac{\pi}{4}}_{0}\text{sec}^{2}d\theta -\int^{\frac{\pi}{4}}_{0}1.\text{tan}\theta d\theta\end{bmatrix}\\= 2[\theta\text{tan}\theta +\text{log}|\text{cos}\space\theta|]^{\frac{\pi}{4}}_{0}\\= 2\begin{Bmatrix}\frac{\pi}{4}\text{tan}\frac{\pi}{4} + \text{log}\begin{vmatrix}\text{cos}\frac{\pi}{4}\end{vmatrix}\end{Bmatrix}-\\\lbrace0 +\text{log 1}\rbrace\\= 2\begin{bmatrix}\frac{\pi}{4} + \text{log}\frac{1}{\sqrt{2}}\end{bmatrix}=\frac{\pi}{2} + 2\text{log 2}^{-\frac{1}{2}}\\=\frac{\pi}{2}- 2×\frac{1}{2}\text{log 2}\\=\frac{\pi}{2}-\text{log 2}.$$

OR

$$\text{(ii)\space Let\space}\text{I =}\int\text{tan}^{\normalsize-1}\space\sqrt{x}dx\\\text{Let}\space \sqrt{x} = t\\\Rarr\space \frac{1}{2\sqrt{x}}dx= dt\\\Rarr\space dx = 2tdt\\\therefore\space \text{I} =\int\underset{\text{II}}{2t}\underset{\text{I}}{\text{tan}^{\normalsize-1}}t dt$$

On integrating by parts,

$$= 2\begin{bmatrix}\text{tan}^{\normalsize-1}t.\frac{t^{2}}{2}-\int\frac{1}{\text{1+ t}^{2}}.\frac{t^{2}}{2}dt\end{bmatrix}\\= t^{2}\text{tan}^{\normalsize-1}t -\int\frac{1 + t^{2}-1}{1 + t^{2}}dt\\= t^{2}\text{tan}^{\normalsize-1}t-\int\frac{1 + t^{2}}{1 +t^{2}}dt + \int\frac{1}{1 + t^{2}}dt\\= t^{2}\text{tan}^{\normalsize-1}t-\int\frac{1 + t^{2}}{1 + t^{2}}dt +\int\frac{1}{1 + t^{2}}dt$$

= t2 tan–1 t – t + tan–1 t + c

$$= x\text{tan}^{\normalsize-1}\sqrt{x}-\sqrt{x}\\+ \text{tan}^{\normalsize-1}\sqrt{x} + c.\space\textbf{Ans.}$$

Solution 13.

(i) The given curve is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1$$

Differentiating w.r.t. x, we get

$$\frac{2x}{a^{2}}-\frac{2y}{b^{2}}.\frac{dy}{dx}=0\\\Rarr\space \frac{dy}{dx}=\frac{\frac{x}{a^{2}}}{\frac{y}{b^{2}}}=\frac{b^{2}}{a^{2}}.\frac{x}{y}\\\text{At}\space\text{P}(\sqrt{2}a, b)$$

$$\text{Slope of the tangent,}\\\text{m}_1 = \frac{b^{2}}{a^{2}}.\frac{\sqrt{2a}}{b} =\sqrt{2}.\frac{b}{a}$$

$$\text{and slope of the normal,}\\ \text{m}_{2}=-\frac{a}{2\sqrt{2}}$$

∴ Equation of the tangent at P is

$$y-b =\sqrt{2}\frac{b}{a}(x - \sqrt{2}a)\\\Rarr\space y =\frac{\sqrt{2}bx}{a}-2b+b\\\Rarr\space y =\frac{\sqrt{2}bx}{a}-b\\\Rarr\space \sqrt{2}bx-ay-ab = 0$$

Also, equation of the normal at P is

$$y-b =\frac{-a}{\sqrt{2}b}(x-\sqrt{2}a)\\\Rarr\space y =\frac{-ax}{\sqrt{2}.b} + b + \frac{a^{2}}{b}\\\Rarr\space \sqrt{2}\space\text{by}-\sqrt{2}b^{2} =-ax +\sqrt{2a^{2}}\\\Rarr\space ax +\sqrt{2}\space by-\sqrt{2}(a^{2} + b^{2})=0$$

Ans.

OR

(ii) Given function is

f (x) = (x + 1)3 (x – 3)3

On differentiating f (x) with respect to x, we get

$$\Rarr\space f'(x)=\begin{Bmatrix}3(x+1)^{2}\frac{d}{dx}(x+1)\end{Bmatrix}(x-3)^{3} +\\(x+1)^{3}.\begin{Bmatrix}3(x-3)^{2}\frac{d}{dx}(x-3)\end{Bmatrix}$$

⇒ f ′(x) = 3(x + 1)2 (x – 3)3 + 3(x + 1)3
(x – 3)2

⇒ f ′(x) = 3(x + 1)2 (x – 3)2 [x + 1 + x – 3]

⇒ f ′(x) = 6(x + 1)2(x – 3)2(x – 1)

For increasing value of f(x) we must have

f ′(x) > 0

Therefore, 6(x + 1)2(x – 3)2 (x – 1) > 0

⇒ x – 1 > 0 and x ≠ – 1, 3

⇒ x > 1 and x ≠ – 1, 3

⇒ x ∈ (1, 3) ∪ (3, ∞)

So, f(x) is increasing in (1, 3) ∪ (3, ∞).

For f(x) to be decreasing, we must have

f ′(x) < 0

⇒ 6(x + 1)2 (x – 3)2 (x – 1) < 0

⇒ x – 1 < 0 and x ≠ – 1, 3

⇒ x < 1 and x ≠ – 1, 3

⇒ x ∈ (– ∞, – 1) ∪ (– 1, 1)

So, f(x) is decreasing in (– ∞, – 1) ∪ (– 1, 1). Ans.

Solution 14.

Let x% students pass in both the subjects

Students who pass in Physics = 70%

$$\Rarr\space \text{P(P)}=\frac{70}{100}$$

Students who pass in Maths = 75%

$$\Rarr\space \text{P(M)}=\frac{75}{100}$$

Students who fail in both = 10%

Students who pass in both at least in one = 100 – 10%

= 90%

$$\Rarr\space \text{P(M ∪ P)}=\frac{90}{100}=0.90\%$$

∴ P (M) + P (P) – P (M  ∩ P) = P (M ∪ P)

0.75 + 0.70 – P (M  ∩ P) = 0.90

$$\text{P(M ∩ P)} = 0.55\text{or}\frac{55}{100}\\\Rarr\space \text{P(M ∩ P)}=\frac{55}{100}$$

(i) P (passes Physics and Mathematics) =

$$\frac{55}{30}=\frac{11}{20}$$

= 0.55 Ans.

$$\textbf{(ii)\space}\text{P(M|P)}=\frac{\text{P(M ∩ P)}}{\text{P(P)}}\\=\frac{55/100}{70/100}\\=\frac{55}{70}=\frac{11}{14}\space\textbf{Ans.}\\\textbf{(iii)\space}\text{P(P|M)}=\frac{\text{P(M ∩ P)}}{\text{P(M)}}\\=\frac{55/100}{75/100}\\=\frac{55}{75}=\frac{11}{15}\space\textbf{Ans.}$$

## Section-B

Solution 15.

(i) (a) (2, – 3)

$$\vec{a} =\hat{i} + x\hat{j} + 3\hat{k}\\\vec{b}=3\hat{i} + 4\hat{J} + 7\hat{k}\\\vec{c} = y\hat{i}-2\hat{j}-5\hat{k}\\\therefore\space \vec{\text{AB}} = \vec{b}-\vec{a} \\= 3\hat{i} + 4\hat{j} +7\hat{k}-\hat{i}-x\hat{j}-3\hat{k}\\\Rarr\space \vec{\text{AB}} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}\\\vec{\text{BC}} = \vec{c}- \vec{b} = \\y\hat{i}-2\hat{j}-5\hat{k}-3\hat{i}-4\hat{j}-7\hat{k}\\\Rarr\space \vec{\text{BC}} = (y-3)\hat{i}-6\hat{j}-12\hat{k}$$

If points A, B, C are collinear, then

$$\vec{\text{AB}} =\lambda\vec{\text{BC}}\\\frac{2}{y-3}=\frac{4-x}{-6}=\frac{4}{\normalsize-12}\\\Rarr\space \frac{2}{y-3} =\frac{\normalsize-1}{3}\\\Rarr\space y+3=6\\\Rarr\space y =-3\\\text{and}\space \frac{4-x}{\normalsize-6}=\frac{\normalsize-1}{3}\\\Rarr\space 12-3x =+6\\\Rarr\space x = 2\\\therefore\space x= 2 , y=-3$$

$$\text{(ii)(c)}\frac{1}{6}$$

2x + 2y – z + 2 = 0

and 4x + 4y – 2z + 5 = 0

$$\Rarr\space 2x + 2y-z +\frac{5}{2} = 0$$

The distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is

$$d=\begin{vmatrix}\frac{|d_1 - d_2|}{\sqrt{a^{2} + b^{2} +c^{2}}}\end{vmatrix}\\ d =\begin{vmatrix}\frac{2-\frac{5}{2}}{\sqrt{2^{2} + 2^{2} + (\normalsize-1)^{2}}}\end{vmatrix}\\=\begin{vmatrix}\frac{\normalsize-1}{6}\end{vmatrix} =\frac{1}{6}.$$

(iii) Plane ABCD passes through the point P(4, – 2, – 5) and

$$\text{plane is perpendicular to}\space\vec{\text{OP}.}$$

Position vector of point P is

$$\vec{a} = 4\hat{i}-2\hat{j}-5\hat{k}\\\text{and normal to plane,}\\\space\vec{n} =\vec{\text{OP}}=4\hat{i}-2\hat{j}-5\hat{k}.$$

The vector equation of plane 'ABCD' is

$$\vec{r}.\vec{n} =\vec{a}.\vec{n}\\\Rarr\space \vec{r}.(4\hat{i}- 2\hat{j}-5\hat{k})\\= (4\hat{i}-2\hat{j}-5\hat{k}).(4\hat{i}-2\hat{j}-5\hat{k})\\\Rarr\space \vec{r}.(4\hat{i}- 2\hat{j}-5\hat{k})\\ 16 + 4+25\\\Rarr\space \vec{r}.(4\hat{i}-2\hat{j}-5\hat{k})=45$$

Reduction to Cartesian form:

$$\text{Putting}\space \vec{r} =x\hat{i} + y\hat{j}+z\hat{k}\space\\\text{in (1),\space \text{we obtain}}$$

$$(x\hat{i} + y\hat{j} + z\hat{k}).(4\hat{i}-2\hat{j}-5\hat{k})= 45$$

$$\Rarr\space 4x-2y-5z = 45,\\\text{which is the required.}$$

Cartesian equation of the plane ABCD.

$$\text{(iv)\space Given}|\vec{a}|=8,|\vec{b}|=3,\\|\vec{a}×\vec{b}| =12\\\vec{a}×\vec{b} =|a||b|\text{sin}\space\theta\hat{n}\\\Rarr\space |\vec{a}×\vec{b}| =|a||b|\text{sin}\theta\\\Rarr\space 12 = 8×3\space\text{sin}\space\theta\\\text{sin}\theta =\frac{12}{24} =\frac{1}{2}\\\therefore\space\theta = 30\degree\\\vec{a}.\vec{b} =|\vec{a}||\vec{b}|\text{cos}\theta$$

= 8 × 3 × cos 30°

$$\vec{a}.\vec{b}=\frac{24×\sqrt{3}}{2} = 12\sqrt{3}\\\therefore\space\vec{a}.\vec{b} = 12\sqrt{3}\space\textbf{Ans.}\\\text{(v)\space}\text{Sum of the vectors}\space \vec{a},\vec{b},\vec{c}\space\text{is}\\\vec{a} +\vec{b} +\vec{c} =(\hat{i}-2\hat{j} + \hat{k}) + (2\hat{i}-4\hat{j} + 5\hat{k})+\\ (\hat{i}- 6\hat{j}-7\hat{k})\\= 4\hat{i}-12\hat{j}-\hat{k}$$

Solution 16.

$$\text{(i)\space\text{Let}\space}\vec{\text{OC}} =\vec{\text{C}}\\\text{Position vector of point A =}\space\vec{a}\\\text{Position vector of point B = }\space\vec{b}\\\text{and}\\\text{Position vector of point C = }\vec{c}$$

Since B is mid-point of side AC,

$$\text{position vector of point B is\space}\frac{\vec{a} + \vec{c}}{2}.\\\text{But position vector of point\space}\vec{b,}\\\therefore\space \frac{\vec{a} + \vec{c}}{2} =\vec{b}\\\Rarr\space \vec{c} = 2\vec{b}-\vec{a}\\\text{Hence}\\\vec{\text{OC}} = 2\vec{b}-\vec{a}.\space\textbf{Ans.}$$

OR

$$\text{(ii)\space Given :}\space\vec{a} =\hat{i} + 2\hat{j} +3\hat{k}\\\text{and}\space \vec{b} = 2\hat{i} + 4\hat{j}- 5\hat{k}\\\text{Let\space}\vec{d}_{1}\space\text{and}\space \vec{d}_{2}\space\text{be the diagonal}\\\text{vectors of the parallelogram.}$$

$$\therefore\space\vec{d_{1}}=\vec{a} + \vec{b}\\\text{and\space}\vec{d_{2}}=\vec{a} -\vec{b}\\\text{So,\space}\vec{d}_{1} = (\hat{i} + 2\hat{j} + 3\hat{k}) +\\(2\hat{i} + 4\hat{j}-5\hat{k})\\= 3\hat{i} + 6\hat{j}-2\hat{k}\\\text{and\space} \vec{d}_{2} = \vec{a} - \vec{b}\\\vec{d}_{2} = (\hat{i} + 2\hat{j} + 3\hat{k})-\\(2\hat{i} + 4\hat{j}-5\hat{k})\\=-\hat{i}-2\hat{j} + 8\hat{k}$$

$$\therefore\space \text{Unit vector parallel to,}\\\vec{d}_{1}=\frac{\vec{d}_{1}}{|\vec{d}_{1}|}\\=\frac{3\hat{i} + 6\hat{j}- 2\hat{k}}{\sqrt{3^{2} + 6^{2} + (-2)^{2}}}\\=\frac{3}{7}\hat{i} + \frac{6}{7}\hat{j}-\frac{2}{7}\hat{k}$$

and, unit vector parallel to,

$$\vec{d}_{2} =\frac{\vec{d}_{2}}{|\vec{d}_{2}|}\\=\frac{-\hat{i}-2\hat{j} + 8\hat{k}}{\sqrt{(-1)^{2} + (\normalsize-2)^{2} + 8}}\\=-\frac{1}{\sqrt{69}}\hat{i}-\frac{2}{\sqrt{69}}\hat{j} +\frac{8}{\sqrt{69}}\hat{k}\\\text{Hence, the unit vectors are}\frac{3}{7}\hat{i} +\frac{6}{7}\hat{j}-\frac{2}{7}\hat{k}\\\text{and}\space\\-\frac{1}{\sqrt{69}}\hat{i}-\frac{2}{\sqrt{69}}\hat{j} + \frac{8}{\sqrt{69}}\hat{k.}\\\textbf{Ans.}$$

Solution 17.

(i) Given,

$$\vec{r} = (1 - t)\hat{i} + (t-2)\hat{j}+ (3-2t)\hat{k}\\\text{and}\\\vec{r} =(s +1)\hat{i} + (2s-1)\hat{j}-(2s+1)\hat{k}\\\text{Vectors equations can be rearranged as,}\\\vec{r} (\hat{i}-2\hat{j} + 3 \hat{k}) +t(-\hat{i} + \hat{j}-2\hat{k})\\\text{and}\space \vec{r} = (\hat{i}-\hat{j}-\hat{k}) + s(\hat{i}+ 2\hat{j}-2 \hat{k})\\\therefore\space \vec{a}_{1} =\hat{i}-2\hat{j} +3\hat{k}\\\text{and}\space \vec{a}_{2}=\hat{i}-\hat{j}-\hat{k}\\\vec{b_{1}} =-\hat{i} + \hat{j}-2\hat{k}\space\text{and}\\ \vec{b}_{2} =\hat{i} + 2\hat{j}-2\hat{k}\\\vec{a}_{2}-\vec{a}_{1} =\hat{i}-\hat{j} +\hat{k}-\hat{i} +2\hat{j}-3\hat{k}\\=0\hat{i} + \hat{j}-4\hat{k}\\$$

$$\vec{b_{1}}×\vec{b_{2}}=\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\-1 &1 &\normalsize-2\\1 &2 &\normalsize-2\end{vmatrix}\\=\hat{i}(-2 +4)-\hat{j}(2 + 2) + \hat{k}(-2-1)$$

Shortest distance between two lines =

$$\begin{vmatrix}\frac{(\vec{a_{2}-a_{1}}).(\vec{b}_{1}×\vec{b}_{2})}{|\vec{b}_{1}×\vec{b}_{2}|}\end{vmatrix}\\=\begin{vmatrix}\frac{(0\hat{i}+ \hat{j}-4\hat{k}).(2\hat{i}-4\hat{j}-3\hat{k})}{\sqrt{29}}\end{vmatrix}\\=\begin{vmatrix}\frac{0- 4 + 12}{\sqrt{29}}\end{vmatrix}\\=-\frac{8}{\sqrt{29}}\text{units}\space\textbf{Ans.}$$

OR

(ii) Given lines are :

$$\vec{r} =\vec{a} +\lambda\vec{b}\space\text{...(i)}\\\vec{r} =\vec{b} +\mu\vec{a}\space\text{...(ii)}\\\text{We know that the lines}\\\vec{r} =\vec{a}_{1} +\lambda\vec{b}_{1},\vec{r} =\vec{a}_{2} + \mu\vec{b}_{2}\\\text{are coplanar only if}\\(\vec{a}_{2}-\vec{a}_{1}).(\vec{b}_{1}×\vec{b}_{2}) = 0\\\text{Here\space}\vec{a}_{1} =\vec{a}, \vec{b}_{1} =b,\vec{a}_{2} = b,\vec{b}_{2} =\vec{a}.\\\therefore\space (\vec{a}_{2} -\vec{a}_{1}).(\vec{b}_{1}×\vec{b}_{2}) \\= (\vec{b} - \vec{a}).(\vec{b}× \vec{a})\\=\vec{b}.(\vec{b}×\vec{a})-\vec{a}.(\vec{b}×\vec{a})\\=\bigg[\vec{b}\vec{b}\vec{a}\bigg]-\bigg[\vec{a}\vec{b}\vec{a}\bigg]$$

= 0 – 0 = 0

[When two of three vectors are identical, then their scalar triple product is 0]

Hence given lines are coplanar.

Also equation of plane containing the given lines is :

$$(\vec{r}-\vec{a}).(\vec{b}_{1}×\vec{b}_{2}) =0\\\Rarr\space (\vec{r}-\vec{a}).(\vec{b}×\vec{a}) =0\\\Rarr\space \vec{r}.(\vec{b}×\vec{a})-\vec{a}(\vec{b}×\vec{a})\\\Rarr\space \vec{r}.(\vec{b}×\vec{a})=0\\\Rarr\space -\vec{r}.(\vec{a}×\vec{b}) =0\\\Rarr\space \vec{r}.(\vec{a}×\vec{b}) =0\space\textbf{Hence Proved.}$$

Solution 18.

Given curve is,

x = 4y – y2

i.e., y2 – 4y = – x

y2– 4y + 4 = – x + 4

(y – 2)2 = – (x – 4)

Which represents a parabola with vertex at A (4, 2).

The parabola meets Y-axis i.e., x = 0

4y – y2 = 0

i.e., at y = 0, 4.

The area enclosed between the curve and the Y-axis is

$$\int^{4}_{0} x\space dy =\int^{4}_{0}(4y-y^{2})dy\\=\bigg[4.\frac{y^{2}}{2}-\frac{y^{3}}{3}\bigg]^{4}_{0}\\=\bigg[32 -\frac{64}{3}\bigg]-[0-0]\\=\bigg[\frac{32}{3}\bigg]\space\text{sq. units}\\=\frac{32}{3}\text{sq. units}\space\textbf{Ans.}$$

## Section-C

(i) (c) 50

x = 100 – 4p

⇒ 4p = 100 – x

$$\Rarr\space \text{p}=\frac{100-x}{4}\\= 25-\frac{x}{4}\\\text{R(x) = px =}\bigg(25 -\frac{x}{4}\bigg)x \\= 25x-\frac{x^{2}}{4}\\\text{R(x) = px}=\bigg(25 -\frac{x}{4}\bigg)x\\= 25x-\frac{x^{2}}{4}\\\therefore\space \text{MR} =\frac{d}{dx}\bigg(25x -\frac{x^{2}}{4}\bigg)$$

$$= 25 -\frac{2x}{4} = 25-\frac{x}{2}\\\because\space \text{MR = 0}\\\Rarr\space 25 - \frac{x}{2} = 0\\\Rarr\space x = 50.$$

(ii) (b) 0

$$\text{Regression lines are}\\ y -\bar{y} = r\frac{\sigma_y}{\sigma_x}(x - \bar{x}),\\ x -\bar{x} =r\frac{\sigma x}{\sigma y}(y -\bar{y})\\\text{Now, if r = 0}\\\Rarr\space y -\bar{y} = 0, x-\bar{x} = 0\\\Rarr\space y =\bar{y}, x = \bar{x}$$

Which are parallel to coordinate axes.

$$\text{(iii)\space C(x)}=\frac{x^{3}}{3} + 5x^{2} -16x +2\\\Rarr\space \text{MC}=\frac{d}{dx}[\text{C(x)}]\\\Rarr\space \text{MC}=\frac{3x^{2}}{3} + 10 x - 16$$

= x2 + 10x - 16.  Ans.

(iv) MR = 11 – 3x + 4x2

$$\therefore\space \text{R(x)} =\int (11 - 3x + 4x^{2})dx\\= 11x-\frac{3x^{2}}{2} +\frac{4x^{3}}{4} +c.$$

Ans.

(v) Assuming 4x + 10y = 9 to be regression line of y on x.

10y = – 4x + 9

$$\Rarr\space y =-\frac{4}{10}x + \frac{9}{10}\\\Rarr\space b_{yx}=\frac{\normalsize-2}{5}$$

Then, 6x + 3y = 4 will be the regression line of x on y.

∴ 6x = – 3y + 4

$$\Rarr\space x =-\frac{3}{6}y +\frac{4}{6}\\\Rarr\space b_{xy}=-\frac{1}{2}\\\text{Now,\space} b_{yx}×b_{xy}\\=-\frac{2}{5}×-\frac{1}{2}=\frac{1}{5}\lt 1$$

which is true. Hence, our assumption is correct, i.e., the regression line of y on x is 4x + 10y = 9. Ans.

Solution 20.

(i) Let the no. of units produced be x.

∴ R(x) = px = 8x

$$\therefore\space \text{C(x)} = 24000 + \frac{25}{100}×8x\\ = 24000 + 2x$$

At break-even point, R(x) = C(x)

⇒ 8x = 24000 + 2x

⇒ 6x = 24000

⇒ x = 4000 units. Ans.

OR

(ii) Given,

$$\text{C(x)} =\frac{3}{4}x^{2} - 7x + 27\\\therefore\space \text{AC} =\frac{\text{C(x)}}{x}=\frac{3}{4}x - 7 + \frac{27}{x}\\\therefore\space \text{MC} =\frac{d}{dx}[\text{C(x)}] =\frac{3}{4}×2x-7\\=\frac{3x}{2}-7\\\because\space \text{MC = AC}$$

$$\Rarr\space \frac{3x}{2}-7 = \frac{3}{4}x-7 + \frac{27}{x}\\\Rarr\space 4x×\frac{3x}{2}-4x×7\\= 4x×\frac{3}{4}x-4x×7 +\\4x×\frac{27}{x}\\\Rarr\space 6x^{2}-28x\\=3x^{2}-28x + 108\\\Rarr\space 3x^{2} = 108\\\Rarr\space x^{2} =\frac{108}{3}=36\\\Rarr\space x = \sqrt{36} = 6\space\text{units}.\space\textbf{Ans.}$$

Solution 21.

Here, number of candidates n = 10.

We construct the following table by according to given graph:

 Name of Candidate Marks English x Marks Maths y $$(x -\bar{x})$$ $$(y -\bar{y})$$ $$(x -\bar{x})^{2}$$ $$(x -\bar{x})(y -\bar{y})$$ A 20 17 4 -1 16 -4 B 13 12 -3 -6 9 18 C 18 23 2 5 4 10 D 21 25 5 7 25 35 E 11 14 -5 -4 25 20 F 12 8 -4 -10 16 40 G 17 19 1 1 1 1 H 14 21 -2 3 4 -6 I 19 22 3 4 9 12 J 15 19 -1 1 1 -1 Total 160 180 110 125

$$\bar{x} =\frac{\Sigma x}{n} =\frac{160}{10} = 16,\\\bar{y} =\frac{\Sigma y}{n} =\frac{180}{10} = 18$$

As we want to find the probable score for mathematics when the marks in English are 24, so we need the equation of the line of regression of y on x.

$$b_{yx}=\frac{\Sigma(x-\bar{x})(y - \bar{y})}{\Sigma(x - \bar{x})^{2}}\\=\frac{125}{110}=\frac{25}{22}$$

The line of regression of y on x is given by

$$y -\bar{y} = b_{yx}(x-\bar{x})\\\Rarr\space y - 18 =\frac{25}{22}(x-16)\\\Rarr\space 22y - 396 = 25 x - 400\\\Rarr\space 25 x - 22y - 4 =0\space\text{...(i)}$$

When x = 24, from (i), we get

25 × 24 – 22y – 4 = 0

$$\Rarr\space 596 - 22y =0\\\Rarr\space y =\frac{596}{22} = 27.09$$

Thus, when the marks in English are 24, then the probable score in mathematics = 27.1.

Ans.

Solution 22.

(i) Let x and y be the units of products A and B respectively.

 Products Quantity Teakwood Plywood Rosewood Profit A B x y 2 1 1 2 1 1 48 40 90 80 50 Z

$$\therefore \space\text{Maximise Z =} 48x + 40 y$$

Subject to constraints :

$$\text{2x + y}\leq 90\\ x + 2y\leq 80\\\text{x + y}\leq 50 , x,y\geq 0$$

∴ The corresponding equations are

2x + y = 90 …(i)

x + 2y = 80 …(ii)

x + y = 50 …(iii)

2x + y = 90

 x 0 45 y 90 0

x + 2y = 80

 x 0 80 y 40 0

x + y = 50

 x 0 50 y 50 0

∴ The feasible region is represented by the shaded region OABCD.

∴ The corner points are O(0, 0), A(45, 0), B(40, 10), C(20, 30) and D(0, 40).

 Corner Points Z = 48x + 40y O (0, 0) Z = 48 × 0 + 40 × 0 = 0 A (45, 0) Z = 48 × 45 + 40 × 0 = ₹ 2160 B (40, 10) Z = 48 × 40 + 40 × 10 = ₹ 2320 C (20, 30) Z = 48 × 20 + 40 × 30 = ₹ 2160 D (0, 40) Z = 48 × 0 + 40 × 40 = ₹ 1600

∴ The maximum profit is ₹ 2320 at B(40, 10).

∴ 40 units of product A and 10 units of product B must be produced and sold. Ans.

OR

(ii) The given inequations are :

3y – 2x < 4 …(i)

x + 3y > 3 …(ii)

and x + y ≤ 5 …(iii)

To draw the graph of 3y – 2x < 4 :

Draw the straight line 3y – 2x = 4

$$\text{which passes through the points}\\(-2,0)\space\text{and}\space\bigg(0,\frac{4}{3}\bigg).$$

The line divides the plane into two parts. Further as O (0, 0) satisfies the inequation 3y – 2x < 4,

(˙.˙ 3.0 – 2.0 = 0 < 4). Therefore, the graph consists of that part of the plane divided by  the line 3y – 2x = 4, which contains the origin.

Similarly, draw the graphs of other two in equations x + 3y > 3 and x + y ≤ 5. Taking the equations x + 3y = 3 and x + y = 5 which passes through (0, 1), (3, 0) and (0, 5), (5, 0) respectively.

$$\text{The corner points are}\space\text{A}\bigg(-\frac{1}{3},\frac{10}{9}\bigg),\\\text{B(6,-1)}\space \text{and C}\bigg(\frac{11}{5},\frac{14}{5}\bigg)\\\text{by solving inequations as equation.}$$

Shade the common part of the graphs of all the three given inequations (i), (ii) and (iii).

The solution set consists of all the points in the shaded part of the coordinate plane shown in the figure.

The points on the line segment BC are included in the solution. Ans.