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:

= 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)

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

⇒ 2^y dy = dx

On integrating both sides, we get

∫ 2^y.dy = ∫ dx + C

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

⇒ 2^y = 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,

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

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