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 = [aij]2 × 3 and B = [bij]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:

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

ds_math_ans9

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

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:

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

= elog (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.

OR

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

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

OR

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
ds_cbse math_ans27

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

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

OR

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

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

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

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

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

OR

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

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

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

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

OR

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.

ds_cbse math_ans32

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

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

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

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

OR

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

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

OR

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

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

= elog|1 + y2|

= 1 + y2.

OR

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

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.

OR

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

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,

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

CBSE 36 Sample Question Papers Commerce Stream

All Subjects Combined for Class 12 Exam 2024

CBSE 36 Sample Question Papers Science Stream (PCM)

All Subjects Combined for Class 12 Exam 2024

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