Oswal 36 Sample Question Papers CBSE Class 12 Maths Solutions

Section-A

(Multiple Choice Questions)

Each question carries 1 mark

1. (c) AB and BA both are defined

Explanation:

Let A = [a_ij]_{2 × 3} and B = [b_ij]_{3 × 2}

So, both AB and BA are defined.

2. (b) (n – 1)

Explanation:

Minor of an element is the determinant obtained by leaving the elements of the column and row containing that element. Therefore, minor of an element of a determinant of order n is a determinant of order n – 1.

3. (b) 5 sq. units

Explanation:

$$\text{Area of parallelogram =}\\\frac{1}{2}|\vec{d_{1}}×\vec{d_{2}}|$$

= 5 sq. units.

$$\textbf{4.\space(b)\space}\frac{\textbf{cos x-2}}{\textbf{3}}$$

Explanation:

Given, 2x + 3y = sin x

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

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

$$\textbf{5.\space (d)\space k =}\frac{\textbf{\normalsize-1}}{\textbf{7}}$$

Explanation:

Put 5x⁷ = t

$$\Rarr\space 35x^{6}dx = dt\\\Rarr\space x^{6}dx =\frac{dt}{35}\\\therefore \int x^{6}\text{sin}(5x^{7})dx =\int\space\text{sin t.}\frac{dt}{35}$$

$$=-\frac{cos t}{35} +\text{C}\\=\frac{-\text{cos}\space 5x^{7}}{35} +\text{C}\\\text{k =}\frac{\normalsize-1}{7}.$$

6. (c) tan y – cot x = C

Explanation:

$$\frac{dy}{dx} =-\bigg(\frac{\text{1 + cos}\space\text{2y}}{\text{1 - cos\space 2x}}\bigg)\\\Rarr\space\frac{dy}{dx} =\frac{-2\text{cos}^{2}y}{2\text{sin}^{2}x}\\\Rarr\space\text{sec}^2 \text{y dy = – cosec}^2 \text{x dx}$$

On integrating both sides, we get

tan y = cot x + C

$$\Rarr\space\text{tan y – cot x = C.}$$

7. (b) a function to be optimised

Explanation:

The objective of an LPP is a function to be optimised.

8. (a) tan θ

Explanation:

$$\frac{|\vec{a}×\vec{b}|}{\vec{a}.\vec{b}} =\frac{\text{ab sin}\theta}{\text{ab cos}\theta} =\text{tan}\space\theta.$$

$$\textbf{9.\space (c)\space}\frac{\textbf{8}}{\textbf{3}}\space \textbf{a}^{\textbf{2}}\textbf{sq. units}$$

Explanation:

$$\text{Area = 2}\int^{a}_{0}y.dx = 2\int^{a}_{0}\sqrt{\text{4ax}}\space dx$$

$$= 2 ×2\sqrt{a}×\frac{2}{3}[x^{3/2}]^{a}_{0}\\=\frac{8}{3}\space a^{2}\text{sq. units.}$$

10. (d) 4 elements

Explanation:

Since, 3 × 4 signifies that there are 3 rows and 4 columns. Then, each row has 4 elements.

11. (c) non-negative restrictions

Explanation:

The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

12. (a) a² + b² – c² – d²

Explanation:

$$\Delta =\begin{vmatrix}a + ib &c +id\\ c-id &a -ib\end{vmatrix}$$

= (a + ib) (a – ib) – (c + id) (c – id)

= (a² – i²b²) – (c² – i²d²)

= a² + b² – c² – d².

13. (d) None of these

Explanation:

We have,

$$\text{A =}\begin{bmatrix}2 &\lambda &\normalsize-3\\0 &2 &5\\1 &1 &3\end{bmatrix}$$

A^–1 exists if |A| ≠ 0.

Now, |A| = 2(6 – 5) – λ(– 5) – 3(– 2)

= 8 + 5 λ

But |A| ≠ 0

5 λ + 8 ≠ 0

$$\Rarr\space\lambda \neq\frac{\normalsize-8}{5}$$

So, A^–1 exists if and only

$$\text{if\space}\lambda\neq\frac{\normalsize-8}{5}.$$

$$\textbf{14.\space (c)}\frac{1}{3}$$

Explanation:

A = {7, 8, 9}

B = {8}

$$\text{P(A) =}\frac{3}{9},\space\text{P(B) =}\frac{1}{9},\\\text{A}\cap\text{B} =\lbrace 8\rbrace,\\\text{P(A}\cap\text{B}) =\frac{1}{9}\\\text{P}\bigg(\frac{\text{B}}{\text{A}}\bigg) =\frac{\text{P(A}\cap \text{B})}{\text{P(A)}}\\=\frac{\frac{1}{9}}{\frac{3}{9}}=\frac{1}{3}.$$

15. (a) x² + 1

Explanation:

$$\frac{dy}{dx} + \frac{2xy}{x^{2}+1} \\=\frac{x^{2}-1}{x^{2}+1}\\\text{I.f.} = e^{\int\frac{2x}{1 +x^{2}}}dx$$

= e^{log (1 + x² )}

= 1 + x².

16. (c) x = 2

Explanation:

$$\lim_{x\xrightarrow{}2^{-}}\space f(x) =7\text{and}\lim_{x\xrightarrow{}2^{+}}\space\text{f(x)=1}$$

$$\textbf{17.\space (b)}\pm\frac{1}{\sqrt{3}}$$

Explanation:

$$\text{As\space}|p(\hat{i} +\hat{j} + \hat{k})|\space\\\text{is a unit vector. So,}\\|p(\hat{i} + \hat{j} +\hat{k})|=1\\\Rarr\space |p||\hat{i} + \hat{j} +\hat{k}|=1\\\Rarr\space |p|\sqrt{1^{2}+1^{2}+1^{2}}=1\\\Rarr\space|p|\sqrt{3} =1\\\Rarr\space |p| =\pm\frac{1}{\sqrt{3}}.$$

18. (a) (a, 1, c)

Explanation:

Given, x = ay + b, z = cy + d

$$\therefore y =\frac{x-b}{a}, y =\frac{z-d}{c}\\\therefore \frac{x-b}{a} =\frac{y}{1} =\frac{z-d}{c}$$

Hence, direction ratios are (a, 1, c).

19. (b) Both A and R are true and R is not the correct explanation of A.

Explanation:

∵ f(x) is odd.

$$\Rarr\space\text{f(-x) =-f(x)}\\\text{and g(x) is even}\\\Rarr\space\text{g(– x) = g(x)}$$

Let F(x) = f(x) + g(x)

= f(– x) + g(– x)

= – f(x) + g(x)

≠ ± F(x)

∴ F(x) is neither even nor odd.

Reason is also true, because this a factual statement.

20. (c) A is true and R is false.

Explanation:

$$\because\space\text{Work done, W =}\space\vec{\text{F}}.\vec{\text{r}}$$

∴ Work done is a scalar quantity.

Section-B

[This section comprises of very short answer type-questions (VSA) of 2 marks each]

21. Given function f : N → N such that

$$\text{f(x) =}\begin{cases}\text{x+1,\space \text{if is odd}}\\\text{x-1,\space}\text{if i s even}\end{cases}$$

For one-one: From the given function we observe that

Case I: When x is odd.

Let f(x₁) = f(x₂)

$$\Rarr\space x_{1} +1 = x_{2}+1\\\Rarr\space x_{1} =x_{2}\\\because\space f(x_{1}) = f(x_{2})\\\Rarr\space x_{1}=x_{2}\forall\space x_{1},x_{2}\epsilon\text{N}$$

So, f(x) is one-one.

Case II: When x is even.

Let f(x₁) = f(x₂)

⇒ x₁ – 1 = x₂ – 1

⇒ x₁ = x₂

∵ f(x₁) = f(x₂)

⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N

So f(x) is one-one.

Given, f : R → R is defined as f(x) = 10x + 7.

For one-one: Let f(x) = f(y) where x, y ∈ R

⇒ 10x + 7 = 10y + 7

⇒ x = y

Therefore, f is a one-one function.

For onto: For y ∈ R

Let y = 10x + 7

$$x =\frac{y-7}{10}$$

Therefore for any y ∈ R, there exists.

$$x =\frac{y-7}{10}\epsilon\space\text{such that}\\\text{f(x) = f}\bigg(\frac{y-7}{10}\bigg)\\= 10\bigg(\frac{y-7}{10}\bigg) +7$$

= y – 7 + 7 = y

Therefore, f is onto.

Hence, f is one-one and onto.

22. Given, f(x) = | cos x |

$$\text{f'(x)} =\frac{\text{cos x}}{|\text{cos x}|}×(-\text{sin x})\\\begin{Bmatrix}\text{If\space f(x) =|x|}\\\text{Then,\space f'(x)} =\frac{x}{|x|}x'\end{Bmatrix}\\\text{f'}\bigg(\frac{3\pi}{4}\bigg) =\frac{\text{cos}\frac{3\pi}{4}}{\bigg|\text{cos}\frac{3\pi}{4}\bigg|}×\\\bigg(-\text{sin}\frac{3\pi}{4}\bigg)\\=\frac{\frac{-1}{\sqrt{2}}×\frac{\normalsize-1}{\sqrt{2}}}{\bigg|\frac{\normalsize-1}{\sqrt{2}}\bigg|}\\\begin{bmatrix}\because\space\text{cos}\frac{3\pi}{4} =\frac{\normalsize-1}{\sqrt{2}}\\\text{sin}\frac{3\pi}{4} =\frac{1}{\sqrt{2}}\end{bmatrix} $$

$$=\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} =\frac{1}{\sqrt{2}}.$$

23. If l, m, n are the direction cosines of the line then

l : m : n = 2 : – 3 : 6

$$\text{Hence,\space l}=\frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}},\\m =\frac{b}{\sqrt{a^{2} + b^{2} + c^{2}}},\space n =\frac{c}{\sqrt{a^{2} + b^{2} +c^{2}}}\\l =\frac{2}{\sqrt{2^{2} +(\normalsize-3)^{2} + 6^{2}} }, m =\frac{\normalsize-3}{\sqrt{2^{2} + (-3)^{2} + 6^{2}}},\\n =\frac{6}{\sqrt{2^{2} +(-3)^{2} +6^{2}}}\\\text{i.e.,\space}l =\frac{2}{7}, m =\frac{\normalsize-3}{7},\\n =\frac{6}{7}.$$

Given line is

5x – 25 = 14 – 7y = 35z

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

∴ Vector equation of the line which passes through the point A(1, 2, – 1) and whose d.r.’s are proportional to 7, – 5, 1 is

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

24. Given differential equation is

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

= (1 + x²) (1 + y²)

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

On integrating both sides, we get

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

Put y = 1 and x = 0 in equation (i)

tan^{– 1} 1 = 0 + 0 + C

$$\text{C} =\frac{\pi}{4}$$

∴ Equation (i) becomes

$$\text{tan}^{\normalsize-1}y = x +\frac{x^{3}}{3} +\frac{\pi}{4}. $$

$$\textbf{25.\space}\vec{a} =\vec{\text{OA}} = 2\hat{i} +\hat{j}-\hat{k}\\\vec{b} =\vec{\text{OB}} = 3\hat{i}-2\hat{j}+\hat{k}\\\vec{c}=\vec{\text{OC}} =\hat{i} +4\hat{j}-3\hat{k}$$

We know that, if three points with position vectors

$$\vec{a},\vec{b}\space\text{and}\space\vec{c}\space\text{are collinear then}\\\vec{a}×\vec{b} +\vec{b}+\vec{c} +\vec{c}×\vec{a}= 0\\\vec{a}×\vec{b}=\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\2 &1 &\normalsize-1\\3 &\normalsize-2 &1\end{vmatrix}\\=\hat{i}(1-2)-\hat{j}(2+3) +\vec{k}(-4-3)\\=-\hat{i}-5\hat{j}-7\hat{k}$$

$$\vec{b}×\vec{c}=\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\3 &\normalsize-2 &1\\1 &4 &\normalsize-3\end{vmatrix}\\=\hat{i}(6-4)-\hat{j}(-9-1) +\hat{k}(12+2)\\= 2\hat{i} +10\hat{j}+14\hat{k}\\\vec{c}×\vec{a} =\begin{vmatrix}\hat{i} &\hat{j} &\hat{k}\\1 &4 &\normalsize-3\\2 &1 &\normalsize-1\end{vmatrix}$$

$$=\hat{i}(-4+3)-\hat{j}(-1+6)+\hat{k}(1-8)\\=-\hat{i}-5\hat{j}-7\hat{k}$$

$$\text{Now,\space}\vec{a}×\vec{b}+\vec{b}×\vec{c}+\vec{c}×\vec{a}\\= -\hat{i} -5\hat{j}-7\hat{k}+2\hat{i}+10\hat{j}+14\hat{k}-\\\hat{i}-5\hat{j}-7\hat{k}$$

= 0, which satisfies the condition of collinearity.

Hence, the given points are collinear.

Section-C

[This section comprises of short answer type questions (SA) of 3 marks each]

$$\textbf{26.\space\text{Put}\space}\frac{\textbf{3x+4}}{\textbf{(x-1)(x-2)(x-3)}}=\\\frac{\textbf{A}}{\textbf{x-1}} +\frac{\textbf{B}}{\textbf{x+2}}+\frac{\textbf{C}}{\textbf{x-3}}\\\textbf{...(i)}$$

$$\Rarr\space\text{3x + 4 = A(x + 2) (x – 3) +}\\\text{B(x – 1) (x – 3) + C(x + 2) (x – 1)}\\\text{...(ii)}$$

Put x = 1 in (ii)

$$\Rarr\space\text{7 = A}(3)(\normalsize-2) +\text{B}(0) +\text{C}(0)\\\Rarr\space 7=-6\text{A}\\\Rarr\space \text{A} =-\frac{7}{6}$$

Put x = 3 in (ii)

$$\Rarr\space\text{13 = 0 +0 C(5)(2)}\\\text{C} =\frac{13}{10}$$

Put x = – 2 in (ii)

$$\Rarr\space -2=\text{B(\normalsize-3)(\normalsize-5)}\\\Rarr\space\text{B} =\frac{\normalsize-2}{15}$$

Put the value in (i)

$$\frac{3x+4}{(x-1)(x+2)(x-3)}\\=\frac{\normalsize-7}{6(x-1)}+\bigg(\frac{-2}{15}\bigg)\bigg(\frac{1}{x+2}\bigg)+\\\frac{13}{10}\bigg(\frac{1}{x-3}\bigg)\\\text{I} =\int\frac{3x+4}{(x-1)(x+2)(x-3)}dx\\=\\\int\begin{bmatrix}\frac{-7}{6(x-1)} +\bigg(\frac{-2}{15}\bigg)\frac{1}{(x+2)} + \frac{13}{10}\bigg(\frac{1}{x-3}\bigg)\end{bmatrix}dx$$

$$=\frac{\normalsize-7}{6}\int\frac{dx}{x-1}-\bigg(\frac{2}{15}\bigg)\int\frac{dx}{x+2} +\\\frac{13}{10}\int\frac{dx}{x-3}\\=\frac{\normalsize-7}{6}\text{log|x-1|}-\frac{2}{15}\text{log|x+2|}+\\\frac{13}{10}\text{log}|x-3| +\text{C}.$$

27. We have to, Maximize, Z = 10x + 6y

Subject to constraints 3x + y ≤ 12

2x + 5y ≤ 34

x ≥ 0, y ≥ 0

Now, consider 3x + y = 12

X	0	4
Y	12	0

2x + 5y = 34

X	0	17
Y	$$\frac{34}{5}$$	0

From graph, it is clear that lines intersects at (2, 6).

Corner Points	Z = 10x + 6y
O(0, 0)	0
$$\text{A}\bigg(0,\frac{34}{5}\bigg)$$	40.8
B(2, 6)	56 → Max.
C(4, 0)	40

So, the maximum value of Z is 56 at point (2, 6).

$$\textbf{28.\space}\text{I =}\int^{\frac{\pi}{2}}_{0}\text{log sin xdx}\\\text{...(i)}\\=\int^{\frac{\pi}{2}}_{0}\text{log sin}\bigg(\frac{\pi}{2}-x\bigg)dx\\\text{(by property)}\\=\int^{\frac{\pi}{2}}_{0}\text{log cos x dx}\\\text{...(ii)}$$

Adding (i) and (ii), we get

$$\text{2I =}\int^{\frac{\pi}{2}}_{0}\text{log(sin x cos x)}dx\\\text{2I =}\int^{\frac{\pi}{2}}_{0}\text{log}\\\bigg(\frac{\text{2 sin x cos x}}{2}\bigg)dx\\\text{2I =}\int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\frac{\text{sin 2x}}{2}\bigg)dx\\\text{2I =}\int^{\frac{\pi}{2}}_{0}\text{log sin 2x dx} -\\\int^{\frac{\\\pi}{2}}_{0}\space\text{log 2 dx}\\\text{2I =}\int^{\frac{\pi}{2}}_{0}\text{log sin 2x}-\frac{\pi}{2}\text{log 2}$$

Put 2x = t

$$\Rarr\space dx =\frac{dt}{2}$$

When x = 0, t = 0

$$\text{When x =}\frac{\pi}{2},\space\text{t =}\pi\\\therefore\space \text{2I =}\frac{1}{2}\int^{\pi}_{0}\text{log sin t dt}-\\\frac{\pi}{2}\text{log 2}\\\text{2I =}\int^{\frac{\pi}{2}}_{0}\text{log sin x dx}-\frac{\pi}{2}\text{log 2}\\\Rarr\space\text{2I = 1 -}\frac{\pi}{2}\text{log 2}\\\text{from (i)}\\\Rarr\space\text{I =}\frac{-\pi}{2}\text{log 2.}$$

$$0\lt x\lt\frac{1}{2}\\\Rarr\space \text{x cos}\pi x\gt 0\\\frac{1}{2}\lt x\lt\frac{3}{2}\\\Rarr\space\text{x cos}\pi x\lt 0\\\therefore\space \int^{\frac{3}{2}}_{0}|\text{x cos}\pi x|dx =\int^{\frac{1}{2}}_{0}\text{x cos}\pi dx +\\\int^{\frac{3}{2}}_{\frac{1}{2}}\space(-\text{x cos}\pi x)dx\\=\begin{bmatrix}\frac{\text{x sin}\pi x}{\pi}\end{bmatrix}^{\frac{1}{2}}_{0}-\int^{\frac{1}{2}}_{0}\frac{\text{sin}\space\pi x}{\pi}dx-\\\begin{bmatrix}\frac{x\space sin\pi x}{\pi}\end{bmatrix}^{\frac{3}{2}}_{\frac{1}{2}} +\int^{\frac{3}{2}}_{\frac{1}{2}}\frac{\text{sin}\pi x}{\pi}dx$$

$$=\begin{bmatrix}\frac{x}{\pi}\text{sin\space}\pi x + \frac{1}{\pi^{2}}\text{cos}\pi x\end{bmatrix}^{\frac{1}{2}}_{0}-\\\begin{bmatrix}\frac{x}{\pi}\text{sin}\space\pi x + \frac{1}{\pi^{2}}\text{cos}\pi x\end{bmatrix}^{\frac{3}{2}}_{\frac{1}{2}}\\=\bigg(\frac{1}{2\pi} -\frac{1}{\pi^{2}}\bigg) -\bigg(\frac{-3}{2\pi} -\frac{1}{2\pi}\bigg)\\=\frac{5}{2\pi}-\frac{1}{\pi^{2}}.$$

29. Given, (x² + xy) dy = (x² + y²) dx

$$\frac{dy}{dx} =\frac{x^{2} + y^{2}}{x^{2} +xy}\\=\frac{1 +\bigg(\frac{y}{x}\bigg)^{2}}{1 +\bigg(\frac{y}{x}\bigg)}\\\text{...(i)}$$

which is a homogeneous differential equation.

Put y = vx

$$\frac{dy}{dx} = v+x\frac{dv}{dx}\\\therefore\space\text{v+x}\frac{dv}{dx} =\frac{1 + v^{2}}{1 + v}\\\lbrack\text{from (i)}\rbrack\\\Rarr\space x\frac{dv}{dx} =\frac{1+v^{2}}{1+v}-v\\=\frac{1 + v^{2}-v-v^{2}}{1+v}\\\Rarr\space x\frac{dv}{dx} =\frac{1-v}{1+v}\\\Rarr\space\int\frac{\text{1+v}}{\text{1-v}}dv =\int\frac{dx}{x}$$

$$\Rarr\space\frac{2}{\text{1 - v}}dv -\int\frac{\text{1 - v}}{\text{1 - v}}dv \\=\space\text{log x}\\\Rarr\space\text{-2 log|1 - v|-v}\\=\text{log x + C}\\\Rarr\space \text{-2 log}\begin{vmatrix}1 -\frac{y}{x}\end{vmatrix}-\frac{y}{x} =\text{log x + C}\\\Rarr\space\text{-2 log}\begin{vmatrix}\frac{x-y}{y}\end{vmatrix}-\frac{y}{x}\\=\text{log x + C}.$$

Given, differential equation is

$$2x^{2}\frac{dy}{dx}-2yx +x^{2} = 0\\\Rarr\space \frac{dy}{dx}-\frac{2xy}{2x^{2}} =\frac{-x^{2}}{2x^{2}}\\\Rarr\space\frac{dy}{dx}-\frac{1}{x}y=-\frac{1}{2}\\\text{which is of the form}\space\\\frac{dy}{dx} +\text{Py} =\text{Q}$$

It is linear differential equation where,

$$\text{P =}\frac{\normalsize-1}{x},\text{Q} =\frac{\normalsize-1}{2}$$

$$\text{I.F. =} e^{\int\frac{\normalsize-1}{x}dx} = e^{-\text{log x}}\\e^{\text{log x}^{\normalsize-1}} = x^{\normalsize-1}=\frac{1}{x}\\\therefore\space\text{y.I.F.} =\int\text{Q.I.F.\space dx}\\\Rarr\space\frac{y}{x} =\int\frac{\normalsize-1}{2}×\frac{\normalsize-1}{x}dx\\\Rarr\space\frac{y}{x} =\frac{\normalsize-1}{2}\text{log x + log C}\\\Rarr\space \frac{y}{x} =\text{log}\frac{\text{C}}{\sqrt{x}}\\\Rarr\space\frac{\text{C}}{\sqrt{x}} = e^{\frac{y}{x}}\\\Rarr\space \text{C} =\sqrt{x}.e^{y/x}\\\text{or\space}\sqrt{x}.e^{y/x} =\text{C}.$$

30. Let E₁ be the probability that he is a cyclist, E₂ be the probability that he is a scooter driver and E₃ be the probability that he is a car driver.

$$\Rarr\space\text{P(E}_{1}) =\frac{4000}{24000} =\frac{1}{6},\\\text{P(E}_{2}) =\frac{8000}{24000}=\frac{1}{3},\\\text{P(E}_{3})=\frac{12000}{24000}=\frac{1}{2}$$

Let E be the event that accident occurs

$$\text{P}\bigg(\frac{\text{E}}{\text{E}_{1}}\bigg) =0.02,\\\text{P}\bigg(\frac{\text{E}}{\text{E}_{2}}\bigg) = 0.06,\\\text{P}\bigg(\frac{\text{E}}{\text{E}_{3}}\bigg) =0.03$$

$$\text{Required probability,}\space\text{P}\bigg(\frac{\text{E}_{2}}{\text{E}}\bigg)\\= \\\frac{\text{P(E}_{2}).\text{P}\bigg(\frac{\text{E}}{\text{E}_{2}}\bigg)}{\text{P}(\text{E}_{1}).\text{P}\bigg(\frac{\text{E}}{\text{E}_{1}}\bigg) + \text{P(E}_{2}).\text{P}\bigg(\frac{\text{E}}{\text{E}_{2}}\bigg) + \text{P}(\text{E}_{3}).\text{P}\bigg(\frac{\text{E}}{\text{E}_{3}}\bigg)}$$

$$=\\\frac{\frac{1}{3}×0.06}{\frac{1}{6}×0.02 + \frac{1}{3}×0.06 +\frac{1}{2}×0.03}\\=\frac{0.12}{0.02 +0.12+ 0.09}\\=\frac{0.12}{0.23}= 0.521$$

When a coin is tossed three times, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}

Now E = event that the first throw results in a head

∴ E = {HHH, HHT, HTH, HTT}

and F = event that the last throw results in a tail

∴ F = {HHT, THT, HTT, TTT}

So, E ∩ F = {HHT, HTT}

Clearly, n(E) = 4, n(F) = 4, n(E ∩ F) = 2

and n(S) = 8

$$\therefore\space\text{P(E) =}\frac{\text{n(E)}}{\text{n(S)}} =\frac{4}{8}=\frac{1}{2}\\\text{P(F) = }\frac{\text{n(F)}}{\text{n(S)}} =\frac{4}{8} =\frac{1}{2}\\\text{and\space}\text{P(E}\cap\text{F}) =\frac{\text{n(E}\cap\text{F})}{\text{n(S)}}\\=\frac{2}{8}=\frac{1}{4}$$

Now, P(E ∩ F) = P(E) × P(F)

$$=\space\frac{1}{2}×\frac{1}{2}=\frac{1}{4}$$

Hence, E and F are independent events.

31. Given

$$\text{I =}\int\frac{x-4}{(x-2)^{3}}e^{x}dx\\=\int\frac{(x-2-2)}{(x-2)^{2}}e^{x}dx\\=\int\bigg(\frac{x-2}{(x-2)^{3}} -\frac{2}{(x-2)^{3}}\bigg)e^{x}dx\\ =\int\bigg(\frac{1}{(x-2)^{2}} -\frac{2}{(x-2)^{3}}\bigg)e^{x}dx\\=\frac{1}{(x-2)^{2}}e^{x} +\text{C}\\\begin{bmatrix}\because\space\frac{d}{dx}\bigg(\frac{1}{(x-2)^{2}}\bigg) =\frac{\normalsize-2}{(x-2)^{3}}\end{bmatrix}\\\begin{bmatrix}\because\space\int\lbrack f(x) + f'(x)\rbrack \space \\e^{x} dx =f(x)e^{x} +\text{C}\end{bmatrix}$$

Section-D

[This section comprises of long answer type questions (LA) of 5 marks each]

32.

4y = 3x² …(i)

$$y =\frac{3}{2}x + 6\space\text{...(ii)}$$

On solving (i) and (ii), we get

x = – 2 or 4

∴ Required area =

$$\int^{4}_{\normalsize-2}(\text{y of line})dx -\\\int^{4}_{\normalsize-2}\space\text{(y of parabola)}\space dx\\\text{i.e.\space}\text{A =}\int^{4}_{\normalsize-2}\begin{bmatrix}\frac{(3x+12)}{2} -\bigg(\frac{3x^{2}}{4}\bigg)\end{bmatrix}dx\\=\begin{bmatrix}\frac{3}{4}x^{2} +6x -\frac{3}{4}×\frac{x^3}{3}\end{bmatrix}^{4}_{\normalsize-2}\\=\begin{bmatrix}\frac{3}{4}x^{2} +6x -\frac{x^{3}}{4}\end{bmatrix}^{4}_{\normalsize-2}\\=\begin{bmatrix}\frac{3}{4}(4)^{2} +24-\frac{64}{4}\end{bmatrix}-\\\begin{bmatrix}\frac{12}{4}-12 +\frac{8}{4}\end{bmatrix}$$

= 20 – (– 7) = 27 sq. units.

33. Let x = a cos θ

$$\therefore\space\text{tan}^{-1}\sqrt{\frac{a-x}{a+x}} =\\\text{tan}^{\normalsize-1}\sqrt{\frac{\text{a - a cos}\space\theta}{\text{a +a}\space\text{cos}\theta}}$$

$$=\space\text{tan}^{\normalsize-1}\sqrt{\frac{\text{1 - cos}\space\theta}{\text{1 + cos}\space\theta}}\\=\text{tan}^{\normalsize-1}\sqrt{\frac{\text{2 sin}^{2}\frac{\theta}{2}}{\text{2 cos}^{2}\frac{\theta}{2}}}\\=\text{tan}^{\normalsize-1}\sqrt{\text{tan}^{2}\frac{\theta}{2}}\\=\text{tan}^{\normalsize-1}\bigg(\text{tan}\frac{\theta}{2}\bigg)\\=\frac{\theta}{2}\\\because\space\text{x = a cos}\space\theta\\\Rarr\space\text{cos}\space\theta =\frac{x}{a}\\\Rarr\space\theta =\text{cos}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)$$

$$\therefore\space\text{tan}^{\normalsize-1}\sqrt{\frac{a-x}{a +x}} =\frac{\theta}{2}\\=\frac{1}{2}\text{cos}^{\normalsize-1}\frac{x}{a}.$$

Given, R = {(a, b) : a, b ∈ A, |a – b| is divisible by 4}

Reflexivity: For any a ∈ A

| a – a| = 0, which is divisible by 4

⇒ (a, a) ∈ R

So, R is reflexive.

Symmetric: Let (a, b) ∈ R

⇒ |a – b| is divisible by 4.

⇒ |b – a| is divisible by 4.

⇒ (b, a) ∈ R

So, R is symmetric.

Transitivity: Let (a, b) ∈ R and (b, c) ∈ R.

$$\Rarr\space\text{|a – b| is divisible by 4.}\\\text{|a-b| = 4k}\\\Rarr\space\text{a-b =}\pm4k\space\text{...(i)}$$

Also |b – c| is divisible by 4.

|b – c| = 4m

$$\Rarr\space\text{b - c} =\pm\text{4m}\\\text{...(ii)}$$

Adding equations (i) and (ii), we get

a – b + b – c = ± 4(k + m)

$$\Rarr\space\text{a – c =}\pm 4(k +m)\\\therefore\space\text{(a,c)}\epsilon\text{R}$$

So, R is transitive.

$$\Rarr\space\text{R is reflexive, symmetric}\\\text{and transitive.}$$

∴ R is an equivalence relation.

Let x be the element of R such that (x, 1) ∈ R then |x – 1| is divisible by 4.

x – 1 = 0, 4, 8, 12, … [∵ x ≤ 12]

∴ Set of elements of A which are related to 1 are {1, 5, 9}.

Equivalence of class 2 i.e.,

[2] = {(a, 2) : ∈ A, |a – 2| is divisible by 4}

⇒ |a – 2| = 4 [k is whole number, k ≤ 3]

a = 2, 6, 10

Therefore, equivalence class [2] is [2, 6, 10].

34. Given

$$\text{A =}\begin{bmatrix}0 &1\\1 &1\end{bmatrix}\space\text{and B =}\begin{bmatrix}0 &\normalsize-1\\1 &0\end{bmatrix}\\\text{A + B} =\begin{bmatrix} 0+0 &1-1\\ 1+1 &1+0\end{bmatrix}\\=\begin{bmatrix}0 &0\\2 &1\end{bmatrix}\\\text{and\space A - B =}\begin{bmatrix}0-0 &1+1\\ 1-1 &1-0\end{bmatrix} =\\\begin{bmatrix}0 &2\\0 &1\end{bmatrix}\\\text{Now,\space (A+B) (A-B)} =\\\begin{bmatrix}0 &0\\2 &1\end{bmatrix}\begin{bmatrix}0 &2\\0 &1\end{bmatrix}$$

$$=\begin{bmatrix}0 &0\\0 &5\end{bmatrix}\\\text{A}^{2} = \text{A×A} =\\\begin{bmatrix}0 &1\\1 &1\end{bmatrix}\begin{bmatrix}0 &1\\1 &1\end{bmatrix}\\ =\begin{bmatrix}0+1 &0+1\\0+1 &1+1\end{bmatrix}\\=\begin{bmatrix}1 &1\\1 &2\end{bmatrix}\\\text{B}^{2} =\text{B×B}=\\\begin{bmatrix}0 &\normalsize-1\\1 &0\end{bmatrix}\begin{bmatrix}0 &\normalsize-1\\1 &0\end{bmatrix}\\=\begin{bmatrix}0-1 &0+0\\0+0 &-1+0\end{bmatrix}$$

$$=\begin{bmatrix}-1 &0\\0 &\normalsize-1\end{bmatrix}\\\text{Now, A}^2 – \text{B}^2 =\\\begin{bmatrix}1 &1\\1 &2\end{bmatrix}\begin{bmatrix}\normalsize-1 &0\\0 &\normalsize-1\end{bmatrix}\\=\begin{bmatrix}2 &1\\1 &3\end{bmatrix}$$

Hence, (A + B)·(A – B) ≠ A² – B².

Hence Proved.

35. Given, al + bm + cn = 0 …(i)

and ul² + vm² + wn² = 0 …(ii)

$$\text{Put l =}\frac{-(\text{bm +cn})}{a}$$

from (i) into (ii), we get

$$\frac{u\lbrack-\text{(bm + cn)}\rbrack^{2}}{a^{2}} +\\vm^{2} +wn^{2} = 0\\\Rarr\space \frac{u(b^{2}m^{2} +c^{2}n^{2} + 2bcmm)}{a^{2}}+\\\text{vm}^{2} +wn^{2} = 0\\\Rarr\space(ub^{2} +va^{2})m^{2} +\\(uc^{2} +wa^{2})n^{2} + 2ubcmn =0\\\text{or}\space (b^{2}u +a^{2}v)\bigg(\frac{m}{n}\bigg)^{2} \\+ 2ubc\bigg(\frac{m}{n}\bigg) +(c^{2}u + a^{2}w) = 0\\\text{...(iii)}\\\text{Let\space}\frac{m_{1}}{n_{1}}\text{and}\frac{m_{2}}{n_{2}}\\\text{be the roots of (iii).}$$

$$\text{Then\space}\frac{m_{1}}{n_{1}}.\frac{m_{2}}{n_{2}} =\frac{c^{2}u + a^{2}w}{b^{2}u +a^{2}v}\\\therefore\space \frac{m_{1}m_{2}}{c^{2}u +a^{2}w} =\frac{n_{1}n_{2}}{b^{2}u + a^{2}v}\\=\frac{\text{I}_{1}\text{I}_{2}}{b^{2}w + c^{2}w} = k\space\\\text{(by symmetry)}$$

So, l₁l₂ + m₁m₂ + n₁n₂

= k(b²w + c²v + c²u + a²w + b²u + a²v)

For perpendicularity, we have

l₁l₂ + m₁m₂ + n₁n₂ = 0

i.e., a²(v + w) + b²(w + u) + c²(u + v) = 0

For the given lines to be parallel, the d.c.’s must be equal and the roots of (iii) must be equal.

Then, we must have

4u²b²c² – 4(b²u + a²v) × (c²u + a²w) = 0

$$\Rarr\space u^{2}b^{2}c^{2} =b^{2}c^{2}u^{2} + b^{2}c^{2}wu +\\a^{2}vuc^{2} +a^{4}vw\\\text{or\qquad}\frac{a^{2}}{u} + \frac{b^{2}}{v} +\frac{c^{2}}{w} = 0.$$

First, we find vector equation of AB, where A(4, 5, 10) and B(2, 3, 4).

We know that two points vector formula of line is given by

$$\vec{r} = \vec{a} +\lambda(\vec{b} - \vec{a})\\\text{...(i)}\\\text{where}\space\vec{a}\space\text{and}\space\vec{b}\space\text{are the position}\\\text{vectors of points through.}\\\text{Here\qquad}\vec{a} =\vec{\text{OA}} \\=4\hat{i} + 5\hat{j} +10 \hat{k}\\\vec{b} =\vec{\text{OB}} = 2\hat{i} +3\hat{j} +4\hat{k}$$

∴ Using equation (i) required equation of line AB is

$$\vec{r} =(4\hat{i} +5\hat{j} +10\hat{k}) +\lambda(2\hat{i} +3\hat{j} +4\hat{k})-\\(4\hat{i} +5\hat{j} +10\hat{k})\\=(4\vec{i} +5\hat{j} +10 \hat{k})+\lambda(-2\hat{i} -2 \hat{j}-6\hat{k})$$

Similarly, vector equation line BC where B(2, 3, 4) and C(1, 2, – 1) is

$$\vec{r} =(2\hat{i} +3\hat{j} + 4\hat{k}) +\lambda\lbrack(\hat{i}+2\hat{j}-\hat{k})-\\(2\hat{i} +3\hat{j} +4\hat{k})\rbrack\\=(2\hat{i}+ 3\hat{j} +4\hat{k}) +\lambda(-\hat{i} -\hat{j}-5\hat{k})$$

Now, let the coordinates of D be (x, y, z).

Mid-point of diagonal BD = Mid-point of diagonal AC

(∵ Diagonals of parallelogram bisect each other)

$$\therefore\space \bigg(\frac{x+2}{2},\frac{y+3}{2},\frac{z+4}{2}\bigg)\\=\bigg(\frac{4+1}{2},\frac{5+2}{2},\frac{10-1}{2}\bigg)$$

Comparing corresponding coordinates of

$$\frac{x+2}{2} = \frac{5}{2},\frac{y+3}{2}=\frac{7}{2},\\\frac{z+4}{2}=\frac{9}{2}$$

x = 3, y = 4, z = 5

∴ Coordinates of point D(x, y, z) = (3, 4, 5).

Section-E

[This section comprises of 3 case-study/passage-based questions of 4 marks each with sub-parts. The First two case study questions have three sub-parts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two sub-parts of 2 marks each.]

$$\textbf{36.\space}(\text{i}\space)\space\text{Here,\space} \text{I.F. = e}^{\int\frac{1}{x}}dx\\ = e^{\text{log x}}$$

= x

(ii) Solution of given differential equation is given by

y.x = ∫ x.3x dx + C (∵ I.F. = x)

$$\Rarr\space y.x =\frac{3x^{3}}{3} +\text{C}\\\Rarr\space y = x^{2} +\frac{\text{C}}{x}$$

(iii) The given differential equation can be rewritten as

$$(1 +y^{2})\frac{dx}{dy} + (\text{2xy - cot y}) = 0\\\Rarr\space \frac{dx}{dy}+\frac{2y}{1 +y^{2}}.x =\\\frac{\text{cot y}}{\text{1 +y}^{2}}\\\text{I.F = e}^{\int\text{P.dy}}\\=\int e^{\int\frac{2y}{1 +y^{2}}}.dy$$

= e^{log|1 + y²|}

= 1 + y².

$$\text{Given,\space}\frac{dy}{dx} = 2^{\normalsize-y}\\\frac{dy}{dx} =\frac{1}{2^{y}}\\\Rarr\space 2^{y}\text{dy = dx}$$

On integrating both sides, we get

$$\int2^{y}.dy =\int dx +\text{C}\\\Rarr\space \frac{2^{y}}{\text{log 2}} = \text{x + C}\\\Rarr\space \text{2}^{y} = x log 2 +\text{C'}$$

where C′ = log 2.

37. (i) Let p be the price per ticket and x be the number of tickets sold.

So, revenue function, R(x) = px

$$= \bigg(15 -\frac{x}{3000}\bigg)x\\ = 15x -\frac{x^{2}}{3000}$$

(ii) Since, more that 36,000 tickets cannot be sold. So, range is [0, 36000].

(iii) We have

$$\text{R(x) = 15x -}\frac{x^{2}}{3000}\\\text{R'(x) = 15 -}\frac{2x}{3000}$$

Put R′(x) = 0

$$15 -\frac{2x}{3000} = 0\\\Rarr\space 15 =\frac{2x}{3000}\\\Rarr\space x = 22,500\\\text{Also\space R''(x)} =-\frac{1}{1500}\lt 0$$

So, Value of x = 22500.

$$\text{Let\space I =}\int\frac{1}{x^{2}-16}dx\\=\int\frac{1}{x^{2}-4^{2}}dx\\=\int\frac{1}{(x-4)(x+4)}dx\\=\frac{1}{8}\int\bigg(\frac{1}{x-4} -\frac{1}{x+4}\bigg)dx\\=\frac{1}{8}\lbrack\text{log|x-4|} -\text{log}|x+4|\rbrack\\+\text{C}\\ =\frac{1}{8}\text{log}\begin{vmatrix}\frac{\text{x-4}}{\text{x+4}}\end{vmatrix} +\text{C}.$$

38. (i) Let E = Event of drawing a first green ball

and F = Event of drawing a second non yellow ball

$$\text{Here,\space P(E) =}\frac{5}{35}\space\\\text{and P}\bigg(\frac{\text{F}}{\text{E}}\bigg) =\frac{25}{34}\\\therefore\space \text{P(F}\cap \text{E}) =\text{P(E).P}\bigg(\frac{\text{F}}{\text{E}}\bigg)\\\frac{5}{35}×\frac{25}{34} =\frac{1}{7}×\frac{25}{34}\\=\frac{25}{238}.$$

(ii) Let E = Event of drawing a first non-blue ball

F = Event of drawing a second non-blue-ball

Here,

$$\text{P(E)} =\frac{23}{38}\space\text{and P(F) =}\frac{22}{34}\\\therefore\space \text{P(F}\cap \text{E}) =\text{P(E).P}\bigg(\frac{\text{F}}{\text{E}}\bigg)\\=\frac{23}{35}×\frac{22}{34}=\frac{253}{595}$$

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