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:

= 5 sq. units.

$$Area\space of\space parallelogram=\frac{1}{2}|\overrightarrow{d_1}×\overrightarrow{d_2}|$$

4. (b) (cos x − 2/3)

Explanation: Given,   2x + 3y = sin x

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

(d/dx)(2x+3y)=(d/dx)(sinx)

⇒ 2 + 3(dy/dx)=cosx

⇒ 3(dy /dx)=cosx-2

(dy/dx)=(cos x-2/3).

5. (d) k =(-1/7)

Explanation: Put    5x7 = t
⇒   35x6 dx = dt
⇒  x6 dx =(dt/35)
∴ ∫ x6 sin (5x7 ) dx = ∫ sin t ·(dt/35)

= -(cos t/35)+c

= (-cos 5x7/35)+C

∴ k =(-1/7).

6. (c) tan y – cot x = C

$$Explaination:\space \frac{dy}{dx}=-\left(\frac{1+cos\space 2y}{1-cos\space 2x}\right)$$

$$\Rightarrow\space \frac{dy}{dx}=\frac{-2\space cos^2\space y}{2\space sin^2\space x}$$

⇒ sec2 y dy = – cosec2 x dx

On integrating both sides, we get

tan y = cot x + C

⇒ 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:

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

9. (c) (8 /3)a2 sq. units

$$Area = 2\int^a_0\space y.dx\space = 2\int^a_0\space \sqrt{4ax}\space dx$$

$$=2×2\sqrt{a}×\frac{2}{3}[x^{3/2}]^a_0=\frac{8}{3}a^2sq.\space units.$$

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

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

$$=(a+ib)(a-ib)-(c+id)(c-id)$$

$$=(a^2-i^2b^2)-(c^2-i^2d^2)$$

$$=a^2+b^2-c^2-d^2.$$

13. (d) None of these

Explaination: We have,       $$A=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$$

$$A^{-1}\space exist\space if\space |A|\not= 0.$$

Now,                  $$|A|\space =2(6-5)-\lambda(-5)-3(-2)$$

$$=8+5\lambda$$

But                    $$|A|\not=0$$

$$5\lambda+8\not=0$$

$$\Rightarrow \lambda\not=\frac{-8}{5}$$

$$So, A^{-1} exist\space if\space and\space only\space if \lambda\not=\frac{-8}{5}$$

Explanation: We have,       A

14. (c) (1/3)

Explanation: A = {7, 8, 9}

B = {8}

P(A) =(3/9), P(B) =(1/9), A ∩ B={8}, P(A ∩ B) =(1/9)

P(A) =(3/9), P(B) =(1/9), A ∩ B={8}, P(A ∩ B) =(1/9)

$$P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$$

$$\frac{\frac{1}{9}}{\frac{3}{9}}=\frac{1}{3}$$

15. (a) x2 + 1

Explanation:               $$\frac{dy}{dx}+\frac{2xy}{x^2+1}=\frac{x^2-1}{x^2+1}$$

$$I.F.= e^{\int\frac{2x}{1+x^2}dx}$$

$$e^{log(1+x^2)}$$

$$=1+x^2.$$

16. (c) x = 2

Explanation:    $$\lim\limits_{x \rightarrow 2^-}f(x)=7\space and\space \lim\limits_{x \rightarrow 2^+}\space f(x) =\space 1$$

17.

$$(b)\space \pm \frac{1}{\sqrt{3}}$$ $$Exlanation:\space As \space|p\left(\hat{i} + \hat{j} + \hat{k}\right) | is a unit vector. So,$$ $$\space \space|p\left(\hat{i} + \hat{j} + \hat{k}\right) | \space =\space 1$$ $$\Rightarrow \space \space |\space p\space |\space |\left(\hat{i} + \hat{j} + \hat{k}\right)|\space = \space 1$$ $$\Rightarrow \space \space |\space p\space |\space \sqrt{1^2+1^2+1^2} = \space 1$$ $$\Rightarrow \space \space |\space p\space | \space \sqrt{3} \space = \space 1$$ $$\Rightarrow \space \space |\space p\space | \space = \space \pm \frac{1}{\sqrt{3}}$$

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

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

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

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.

⇒ f(– x) = – f(x)

and g(x) is even ⇒ 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.

$$Explanation:\space \because \space \space \space \space \space \space \space \space \space \space Work\space done, W = \overrightarrow{F}.\overrightarrow{r}$$ $$\therefore \space \space \space \space \space \space \space \space Work\space done\space is\space a\space scalar\space quantity.$$

Section-B

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

21. Given function f : N → N such that

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

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

Case I: When x is odd.

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.

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 =(y − 7/10)

Therefore for any y ∈ R, there exists.

$$x \space =\space \frac{y-7}{10} \space \epsilon \space R\space such \space that$$ $$f(x) \space = \space f\left(\frac{y-7}{10}\right) \space =\space 10 \left(\frac{y-7}{10}\right)+\space 7 \space = \space y-7+7 \space = \space y$$

Therefore, f is onto.

Hence, f is one-one and onto.

22.

$$Given, f(x) \space = | cos\space x \space |$$ $$\enspace \space f'(x)\space = \space \frac{cos\space x}{|\space cos\space x\space |} × (-sin \space x\space ) \space \space \begin{Bmatrix} If & f(x) = | x | \\ Then, & f'(x) = \frac{x}{|x|} \space x’ \end{Bmatrix} \\ \therefore \space \space f’\left(\frac{3\pi}{4}\right) \space =\space \frac{cos\frac{3\pi}{4}}{|cos\frac{3\pi}{4}|}×\left(-sin\frac{3\pi}{4}\right) \\ = \frac{\frac{-1}{\sqrt{2}}×\frac{-1}{\sqrt{2}}}{\begin{vmatrix} \frac{-1}{\sqrt{2}} \end{vmatrix}} \space \space \space \space \space \space \begin{bmatrix} \because cos\frac{3\pi}{4} = \frac{-1}{\sqrt{2}} \\ sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}} \end{bmatrix} \\ = \space \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}}.$$

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

l : m : n = 2 : – 3 : 6

$$Hence, \space \space \space \space l = \frac{a}{\sqrt{a^2+ b^2+ c^2}}, m = \frac{b}{\sqrt{a^2+ b^2+ c^2}}, n = \frac{c}{\sqrt{a^2+ b^2+ c^2}} \\ l = \frac{2}{\sqrt{2^2+ (-3)^2+ 6^2}},\space m = \frac{-3}{\sqrt{2^2+ (-3)^2+ 6^2}},\space n = \frac{6}{\sqrt{2^2+ (-3)^2+ 6^2}} \\ i.e., \space \space l\space = \frac{2}{7}, \space m \space = \space \frac{-3}{7}, \space n \space = \space \frac{6}{7}. \\ OR \\ Given\space line\space is\space \\ \space \space 5x\space – \space 25\space =\space 14\space -\space 7y\space =\space 35z \\ \Rightarrow \space \space 5(x-5)\space =\space -7(y-2)\space =\space 35z \\ \Rightarrow \space \space \frac{x-5}{\frac{1}{5}} = \space \frac{y-5}{\frac{-1}{7}}\space = \space \frac{z-0}{\frac{1}{35}} \\ \Rightarrow \space \space \frac{x-5}{7} =\space \frac{y-2}{-5} =\space \frac{z-0}{1} \\ \therefore \space Vector\space equation\space of\space the\space line\space which\space passes\space through\space the\space point\space A(1,2,-1,) \space and \space whose \space d.r.’s \space are \space proportional \space to \space 7,-5, 1 \space is \\ \space \space \overrightarrow{r} \space = \space \hat{i}+2\hat{j}-\hat{k}+ λ(7\hat{i}-5\hat{j}+\hat{k}).$$

24. Given differential equation is

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

On integrating both sides, we get

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

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

tan– 1 1 = 0 + 0 + C

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

∴ Equation (i) becomes

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

25.

$$Let \space \space \overrightarrow{a} \space = \space \overrightarrow{OA} =\space 2\hat{i}+\hat{j}-\hat{k} \\ \space \space \overrightarrow{b} \space = \space \overrightarrow{OB} =\space 3\hat{i}-2\hat{j}+\hat{k} \\ \space \space \space \overrightarrow{c} \space = \space \overrightarrow{OC} =\space \hat{i}+4\hat{j}-3\hat{k}$$

We know that, if three points with position vectors

are collinear then

$$\overrightarrow{a} × \overrightarrow{b}+ \overrightarrow{b}× \overrightarrow{c}+ \overrightarrow{c}×\overrightarrow{a} = 0 \\ \overrightarrow{a} × \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 3 & -2 & 1 \end{vmatrix}\\ = \space \hat{i}(1-2) – \hat{j}(2+3) + \hat{k}(-4-3) \\ = -\hat{i}-5\hat{j}-7\hat{k} \\$$ $$\overrightarrow{b} × \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 1 & 4 & -3 \end{vmatrix}\\ = \space \hat{i}(6-4) – \hat{j}(-9-1) + \hat{k}(12+2) \\ = 2\hat{i}+10\hat{j}+14\hat{k} \\$$ $$\overrightarrow{c} × \overrightarrow{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -3 \\ 2 & 1 & -1 \end{vmatrix}\\ = \space \hat{i}(-4+3) – \hat{j}(-1+6) + \hat{k}(1-8) \\ = -\hat{i}-5\hat{j}-7\hat{k} \\ Now, \space \space \space \space \space \space \space \space \space \overrightarrow{a} × \overrightarrow{b} + \overrightarrow{b} × \overrightarrow{c} + \overrightarrow{c} × \overrightarrow{a}\space = \space -\hat{i}-5\hat{j}-7\hat{k}+2\hat{i} + 10\hat{j} + 14\hat{k}-\hat{i}-5\hat{j}-7\hat{k}$$

= 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)

26. Put

$$\frac{3x+4}{(x-1)(x+2)(x-3)} =\space \frac{A}{x-1}+\frac{A}{x+2}+\frac{A}{x-3} \space \space \space …(i)$$

⇒  3x + 4 = A(x + 2) (x – 3) + B(x – 1) (x – 3) + C(x + 2) (x – 1) …(ii)

Put x = 1 in (ii)

​⇒ 7 = A(3) (– 2) + B(0) + C(0)

⇒ 7 = – 6A

$$\Rightarrow \space \space A \space = \space-\frac{7}{6}$$

Put x = 3 in (ii)

⇒ 13 = 0 + 0 + C(5) (2)

$$A \space = \space \frac{13}{10}$$

Put x = – 2 in (ii)

⇒ – 2 = B(– 3) (– 5)

$$\Rightarrow \space \space B \space = \space \frac{-2}{15}$$

Put the value in (i)

$$\frac{3x+4}{(x-1)(x+2)(x-3)} \space = \space \frac{-7}{6(x-1)} + \left(\frac{-2}{15}\right)\left(\frac{1}{x+2}\right)+\frac{13}{10}\left(\frac{1}{x-3}\right)$$ $$\therefore \space \space I = \space \int \frac{3x+4}{(x-1)(x+2)(x-3)}dx$$ $$= \int \Bigg[\frac{-7}{6(x-1)}+\left(\frac{-2}{15}\right)\frac{1}{(x+2)}+\frac{13}{10}\left(\frac{1}{x-3}\right)\Bigg]dx$$ $$= \int \Bigg[\frac{-7}{6(x-1)}+\left(\frac{-2}{15}\right)\frac{1}{(x+2)}+\frac{13}{10}\left(\frac{1}{x-3}\right)\Bigg]dx$$ $$=\frac{-7}{6}\int \frac{dx}{x-1}-\left(\frac{2}{15}\right)\int \frac{dx}{x+2}+\frac{13}{10}\int \frac{dx}{x-3}$$ $$= \space \frac{-7}{6}log|x-1|-\frac{2}{15}log|x+2|\space + \space \frac{13}{10}log|x-3|+C$$

27. 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.

$$\Rightarrow \space \space \space P(E_1) = \frac{4000}{24000} = \frac{1}{6}, \space P(E_2) \space = \frac{8000}{24000} = \space \frac{1}{3}, P(E_3) \space = \space \frac{12000}{24000} = \space \frac{1}{2}$$

Let E be the event that accident occurs

$$P\left(\frac{E}{E_1}\right) = \space 0.02,$$ $$P\left(\frac{E}{E_2}\right) = \space 0.06,$$ $$P\left(\frac{E}{E_3}\right) = \space 0.03$$ $$Required\space probability, P \left(\frac{E_2}{E}\right) = \frac{P(E_2).P\left(\frac{E}{E_2}\right)}{P(E_1).P\left(\frac{E}{E_1}\right)+P(E_2).P\left(\frac{E}{E_2}\right)+P(E_3).P\left(\frac{E}{E_3}\right)}$$ $$= \frac{\frac{1}{3}×0.06}{\frac{1}{6}×0.02 +\frac{1}{3}×0.06+\frac{1}{2}×0.03}$$ $$= \frac{0.12}{0.02+0.12+0.09}= \frac{0.12}{0.23} = 0.521$$

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

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

Hence, E and F are independent events.

28.

$$I = \int^{\pi/ 2}_0 log\space sin \space x\space dx \space \space …(i) \\ = \space \int^{\pi/ 2}_0 log\space sin\left(\frac{\pi}{2}-x\right)dx \space \space (by\space property)\\ = \int^{\pi/ 2}_0 log \space cos\space x\space dx \space \space \space …(ii)$$

Adding (i) and (ii), we get

$$2I = \int^{\pi /2}_0 log (sin \space x \space cos\space x)\space dx \\ 2I \space =\space \int^{\pi/2}_0 log \left(\frac{2\space sin \space x \space cos \space x \space}{2}\right)dx \\ 2I =\space \int^{\pi/2}_0 log \left(\frac{sin\space 2x}{2}\right)\space dx \\ 2I =\space \int^{\pi/2}_0 log\space sin\space 2x\space dx-\int^{\pi/2}_0 log \space 2\space dx \\ 2I =\space \int^{\pi/2}_0 log\space sin\space 2x\space dx – \frac{\pi}{2}log\space 2$$

Put

$$2x = t \Rightarrow dx = \frac{dt}{2}$$

When x = 0, t = 0

$$When\space x= \frac{\pi}{2}, \space t = \pi \\ \therefore \space \space \space 2I =\space\frac{1}{2} \int^{\pi/2}_0 log\space sin\space t\space dt – \frac{\pi}{2}log\space 2 \\ 2I =\space\frac{1}{2} \int^{\pi/2}_0 log\space sin\space x\space dx – \frac{\pi}{2}log\space 2$$

$$\\ \Rightarrow \space \space \space 2I = I-\frac{\pi}{2}log\space 2 \space \space \space \space \space [from \space (i)] \\ \Rightarrow \space \space \space \space \space I = \frac{-\pi}{2}log\space 2$$

OR

$$0<x<\frac{1}{2} \\ \Rightarrow \space \space x \space cos \space \pi x>0 \\ \space \space \space \space \space \\ \frac{1}{2}<x<\frac{3}{2} \\ \Rightarrow \space \space \space \space x \space cos \space \pi x<0 \\ \therefore \space \space \space \int^{3/2}_0\space |\space x\space cos\space \pi x|dx \space=\space \int^{1/2}_0\space x\space cos\space \pi x \space dx + \int^{3/2}_{1/2} (-x\space cos\space \pi x)\space dx \\= \left[\frac{x\space sin\space \pi x}{\pi} \right]^{1/2}_0-\int^{1/2}_0 \frac{sin\space \pi x}{\pi}\space dx – \left[\frac{x\space sin\space \pi x}{\pi} \right]^{3/2}_{1/2} + \int^{3/2}_{1/2} \frac{sin\space \pi x}{\pi}dx \\=\space \left[\frac{x}{\pi} sin\space \pi x +\space \frac{1}{\pi^2} cos\space \pi x \right]^{1/2}_0 – \left[\frac{x}{\pi} sin\space \pi x +\frac{1}{\pi^2}cos \space \pi x \right]^{3/2}_{1/2} \\ = \left(\frac{1}{2\pi} – \frac{1}{\pi^2}\right)-\left(\frac{-3}{2\pi}-\frac{1}{2\pi}\right) \\ = \frac{5}{2\pi}-\frac{1}{\pi^2}$$

29.

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

$$\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} = \frac{1+\left(\frac{y}{x}\right)^2}{1+ \left(\frac{y}{x}\right)} \space \space \space …(i)$$

which is a homogeneous differential equation.

Put y = vx

$$\frac{dy}{dx} = \space v + x\frac{dv}{dx} \\ \therefore \space \space \space v + x\frac{dv}{dx}\space =\space \frac{1+v^2}{1+v} \space \space [from \space (i)] \\ \Rightarrow \space \space \space x\frac{dv}{dx} \space = \space \frac{1+v^2}{1+v}-v = \frac{1+v^2-v-v^2}{1+v} \\ \Rightarrow \space \space x\frac{dv}{dx} = \space \frac{1-v}{1+v} \\ \Rightarrow \space \space \space \space \int \frac{1+v}{1-v}dv \space =\space \int \frac{dx}{x}$$

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

OR

Given, differential equation is

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

It is linear differential equation where,

$$P =\space \frac{-1}{x}, Q=\frac{-1}{2} \\ I.F. \space = \space e^{\int \frac{-1}{x}dx} = e^{-log\space x} = e^{log\space x^{-1}} = x^{-1} = \frac{1}{x} \\ \therefore \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space y.I.F. = \int Q.I.F. \space dx$$

30. 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       2x + 5y = 34

 x 0 4 y 12 0
 x 0 17 y 34/5 0

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

 Corner Points Z=10x+6y O(0,0) 0 A(0,34/5) 40.8 B(2,6) 56 → Max. C(4,0) 40

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

31. Given

$$I = \int \frac{x-4}{(x-2)^2}e^x dx \\ = \int \frac{(x-2-2)}{(x-2)^3} e^x dx \\ =\int \left(\frac{x-2}{(x-2)^3}-\frac{2}{(x-2)^3}\right)e^x dx \\ =\int \left(\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3}\right)e^x dx \\ = \frac{1}{(x-2)^2}e^x + C \space \space \left[ \because \frac{d}{dx} \left(\frac{1}{(x-2)^2}\right) = \frac{-2}{(x-2)^3}\right] \\ \left[ \because \int[f(x) + f'(x)e^x \space dx = f(x)e^x \space + C\right]$$

Section-D

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

32.

$$4y=3x^2 \space \space …(i) \\ y= \frac{3}{2}x+6 \space …(ii)$$

On solving (i) and (ii), we get
x = – 2 or 4

$$\therefore \space \space Required \space area = \space \int^{4}_{-2} (y\space of\space line) dx – \space \int^4_{-2}(y\space of\space parabola) \space dx \\ i.e., \space \space A = \space \int^4_{-2}\left[\frac{(3x+12)}{2}-\left(\frac{3x^2}{4}\right)\right]dx \\ = \left[\frac{3}{4}x^2 + 6x-\frac{3}{4}×\frac{x^3}{3}\right]^4_{-2} \\ = \left[\frac{3}{4}x^2 + 6x-\frac{x^3}{3}\right]^4_{-2}\\ = \left[\frac{3}{4}(4)^2 + 24-\frac{64}{4}\right]-\left[\frac{12}{4}-12+\frac{8}{4}\right]$$

= 20 – (– 7) = 27 sq. units.

E° is preferred. But due to the over potential of oxygen, Cl gets oxidised at the anode to produce Cl2 gas.

33.  Let x = a cos θ

$$tan^{-1}\sqrt{\frac{a-x}{a+x}} = tan^{-1}\sqrt{\frac{a-a\space cos\space\theta}{a+a\space cos\space\theta}} \\ =tan^{-1}\sqrt{\frac{1-\space cos\space\theta}{1+\space cos\space\theta}} \\= tan^{-1}\sqrt{\frac{2cos^2\frac{\theta}{2}}{2cos^2\frac{\theta}{2}}} \\ = tan^{-1}\sqrt{tan^2\frac{\theta}{2}} \\ = tan^{-1}\left(tan\frac{\theta}{2}\right) \\= \frac{\theta}{2}$$

·.· x = a cos θ

$$\Rightarrow \space \space cos\space \theta = \frac{x}{a} \\ \Rightarrow \space \space \theta = cos^{-1}\left(\frac{x}{a}\right)\\ \therefore \space \space tan^{-1} \sqrt{\frac{a-x}{a+x}} = \frac{\theta}{2} \\ = \frac{1}{2}cos^{-1}\frac{x}{a}$$

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.

⇒ |a – b| is divisible by 4.

|a – b| = 4k

⇒ a – b = ± 4k   …(i)

Also |b – c| is divisible by 4.

|b – c| = 4m

⇒ b – c = ± 4m   …(ii)

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

a – b + b – c = ± 4(k + m)

⇒ a – c = ± 4(k + m)

∴ (a, c) ∈ R

So, R is transitive.

⇒ R is reflexive, symmetric 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, al + bm + cn = 0   …(i)

and ul2 + vm2 + wn2 = 0 …(ii)

$$Put \space l = \frac{-(bm+cn)}{a}\space from \space (i)\space into \space (ii), \space we\space get \\ \frac{u[-(bm+cn)]^2}{a^2} + vm^2 +wn^2 = 0$$

$$\Rightarrow \space \space \frac{u(b^2m^2 +c^2n^2+2bcmn)}{a^2} + vm^2 + wn^2 = 0 \\ \Rightarrow (ub^2+va^2)m^2 + (uc^2+wa^2)n^2 + 2ubcmn = 0 \\ or \space \space (b^2u+a^2v)\left(\frac{m}{n}\right)^2 + 2ubc\left(\frac{m}{n}\right) + (c^2u+a^2w) = 0 \space \space …(iii)\\ Let \space \frac{m_1}{n_1} \space and \space \frac{m_2}{n_2} \space be \space the\space roots\space of \space (iii). \\ Then \space \space \frac{m_1}{n_1}.\frac{m_2}{n_2} = \frac{c^2u+a^2w}{b^2u+a^2v} \\ \frac{m_1m_2}{c^2u+a^2w} = \frac{n_1n_2}{b^2u+a^2v} = \frac{l_1l_2}{b^2w+c^2v} = k \space \space (by\space symmetry)$$

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
⇒ u2b2c2 = b2c2u2 + b2c2wu + a2vuc2 + a4vw
or

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

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

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

$$\overrightarrow{r}= (4\hat{i}+5\hat{j}+10\hat{k}) + λ[(2\hat{i}+3\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}+10\hat{k})] \\ = (4\hat{i}+5\hat{j}+10\hat{k}) + λ (-2\hat{i}-2\hat{j}-6\hat{k})$$

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

$$\overrightarrow{r}= (2\hat{i}+3\hat{j}+4\hat{k}) + λ[(\hat{i}+2\hat{j}-\hat{k})-(2\hat{i}+3\hat{j}+4\hat{k})] \\ = (2\hat{i}+3\hat{j}+4\hat{k}) + λ (-\hat{i}-\hat{j}-5\hat{k})$$

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)

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

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).

35. Given

$$Given \space \space A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \space and \space B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \\ A+B = \begin{bmatrix} 0+0 & 1-1 \\ 1+1 & 1+0 \end{bmatrix} \\ = \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix} \\and \space \space A-B = \begin{bmatrix} 0-0 & 1+1 \\ 1-1 & 1-0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix} \\ Now, \space \space (A+B)(A-B) = \begin{bmatrix} 0 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ 0 & 1 \end{bmatrix}\\= \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix} \\A^2 = A×A= \begin{bmatrix*}[r] 0 & 1 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 1 \\ 1 & 1 \end{bmatrix*} \\ = \begin{bmatrix*}[r] 0+1 & 0+1 \\ 0+1 & 1+1 \end{bmatrix*} \\=\begin{bmatrix*}[r] 1 & 1 \\ 1 & 2 \end{bmatrix*} \\ B^2 = B×B= \begin{bmatrix*}[r] 0 & -1 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & -1 \\ 1 & 0 \end{bmatrix*} \\ = \begin{bmatrix*}[r] 0-1 & 0+0 \\ 0+0 & -1+0 \end{bmatrix*} \\=\begin{bmatrix*}[r] -1 & 0 \\ 0 & -1 \end{bmatrix*} \\ Now, \space \space A^2-B^2 = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}-\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \\ = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\\ Hence, \space \space (A+B).(A-B) \not= A^2-B^2.$$

Hence Proved.

Section-E

(This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-parts. 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.)

36. (i) Here,

$$I.F. = e^{\int \frac{1}{x}dx} \\ = e^{log \space x} \\ = x$$

(ii) Solution of given differential equation is given by

y.x = ∫ x.3x dx + C         (·.· I.F. = x)

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

(iii) The given differential equation can be rewritten as

$$(1+y^2)\frac{dx}{dy} + (2xy -cot\space y) = 0 \\ \Rightarrow \frac{dx}{dy} + \frac{2y}{1+y^2}.x = \frac{cot\space y}{1+y^2} \\ I.F. = e^{\int P.dy} \\ =\int e^{\int \frac{2y}{1+y^2}.dy} \\ = e^{log|1+y^2|} \\ = 1+y^2 \\ OR \\ Given, \space \space \frac{dy}{dx} = 2^{-y} \\ \frac{dy}{dx} = \frac{1}{2^y}$$

⇒ 2y dy = dx

On integrating both sides, we get

∫ 2y.dy = ∫ dx + C

$$\Rightarrow \space \space \frac{2^y}{log\space 2} = x + C$$

⇒ 2y = x log 2 + 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

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

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

(iii) We have

$$R(x) = 15x-\frac{x^2}{3000} \\ R'(x) = 15-\frac{2x}{3000} \\ Put \space \space R'(x) = 0 \\ 15-\frac{2x}{3000} = 0 \\ \Rightarrow 15 = \frac{2x}{3000}$$

$$\\ \Rightarrow \space \space x = 22,500 \\ Also, \space \space R”(x) = -\frac{1}{1500}<0 \\ So, \space \space Value \space \space of \space x = 22500. \\ OR \\ Let \space \space I = \space \space \int \frac{1}{x^2-16}dx \\ = \space \space \int \frac{1}{x^2-4^2}dx \\ = \int \frac{1}{(x-4)(x+4)}dx \\ = \frac{1}{8} \int \left(\frac{1}{x-4}-\frac{1}{x+4}\right)dx \\ = \frac{1}{8}[log |x-4| – log |x+4|] + C \\ = \frac{1}{8} log \begin{vmatrix}\frac{ x-4}{x+4} \end{vmatrix} + C.$$

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

and F = Event of drawing a second non yellow ball

Here,

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

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

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

Here,

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

CBSE 36 Sample Question Papers Commerce Stream

All Subjects Combined for Class 12 Exam 2023

CBSE 36 Sample Question Papers Science Stream (PCM)

All Subjects Combined for Class 12 Exam 2023

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