Oswal Practice Papers CBSE Class 12 Mathematics Solutions (Practice Paper - 7)

Section-A

1. (d) 4 elements

Explanation :

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

2. (c) zero

Explanation :

The area of triangle formed by three collinear points is zero.

3. (c) cos x

Explanation :

$$\text{In the interval}(0,\frac{\pi}{2}),f(x)=cosx\\⇒ f ^′(x) = – sin x\\\text{which gives }f^′(x) < 0\space in \space(\frac{\pi}{2})\\\text{Hence, f(x) = cos x in decreasing in}\space(\frac{\pi}{2})$$

4. (b) f(x) is discontinuous at x = 0 for any value of l

Explanation :

$$\lim\limits_{x\rarr0^{-}}f(x)=0\space\text{and}\lim\limits_{x\rarr0^{+}}f(x)=1$$

5. (a) 9

Explanation :

f(– 1) = 3(– 1) = – 3
f(2) = (2)2 = 4
f(4) = 2 × 4 = 8
⇒ f(– 1) + f(2) + f(3) = – 3 + 4 + 8 = 9.

6. (b) 2

Explanation :

$$\text{We have,}(1+\frac{dy}{dx})^3=(\frac{d^2y}{dx^2})^2\\\text{So, order = 2}\\\text{and degree = 2}$$

7. (d) all of the above

Explanation :

An optimisation problem may involve finding maximum profit, minimum cost or minimum use of resources etc.

8. $$(\frac{\pi}{4},\frac{5\pi}{4})$$

Explanation :

f(x) = sin x + cos x
f ′(x) = cos x – sin x
Now, f ′(x) = 0 gives
sin x = cos x,
$$\text{which gives that x =}\frac{\pi}{4},\frac{5\pi}{4}=\text{as 0 }≤ x ≤ 2\pi.$$

9. (c) tan x + C

Explanation :

$$\int\frac{dx}{cos2x+sin^x}=\int\frac{dx}{cos^2x-sin^2x+sin^2x}\\=\int\frac{dx}{cos^2x}=\int sec^2 \space x dx\\= tan x + C.$$

10. (a) 20

Explanation :

We have,
$$\begin{vmatrix}1 & 2\\-2 & -b\end{vmatrix}+\begin{vmatrix}a & 4\\3 & 2\end{vmatrix}=\begin{vmatrix}5 & 6\\1 & 0\end{vmatrix}\\⇒\begin{vmatrix}a+1 & 6\\1 & 2-b\end{vmatrix}=\begin{vmatrix}5 & 6\\1 & 0 \end{vmatrix}\\⇒ a + 1 = 5, 2 – b = 0\\⇒ a = 4, b = 2\\⇒ a^2 + b^2 = 42 + 22 = 20$$

11. (b) constraints

Explanation :

The restrictions on the variable of a linear programming problem are called constraints.

12. (a) – 80

Explanation :

Δ = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃
= a₁₁M₁₁ – a₁₂M₁₂ + a₁₃M₁₃
= 1 · (– 40) – 3(– 10) + (– 2) (35)
= – 40 + 30 – 70 = – 80

13. (b) fg – hc

Explanation :

$$A_{21} = (– 1)^{2 + 1} M_{21} = – M_{21}=-\begin{vmatrix}h & g\\f & c\end{vmatrix}\\= – (hc – fg) = fg – hc.$$

14. (a) 14/17

Explanation :

$$P(A ∩ B)=\frac{7}{10}\space and\space P(B)=\frac{17}{20}\\P(A|B)=\frac{P(A ∩ B)}{PB}\\=\frac{7/10}{17/20}=\frac{14}{17}$$

15. (a) 1

Explanation :

$$\frac{d^2y}{dx^2}+(\frac{dy}{dx})^3+6y^5=0 $$

We know that, the degree of a differential equation is exponent highest of order derivative.
Degree = 1.

16. (d)
$$\frac{g(x)}{f(x)}$$

Explanation :

We know that, if f and g are continuous functions, then
(a) f + g is continuous
(b) f – g is continuous
(c) fg is continuous
(d) f\g is continuous at these points, where g(x) ≠ 0.
$$\text{Here,}\space\frac{g(x)}{f(x)}=\frac{\frac{x^2}{2}+1}{2x}=\frac{x^2+2}{4x}\\\text{which is discontinuous at x = 0.}$$

17. (a) (a, 1, c)

Explanation :

$$\text{Given,} x = ay + b, z = cy + d\\y=\frac{x-b}{a},y=\frac{z-d}{c}\\\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}\\\text{Hence direction ratios are (a, 1, c).}$$

18. (c)
$$A(adj A)=(adj A)A=|A|I=\begin{vmatrix}0 & 0\\0 & 0\end{vmatrix}$$

Explanation :

We know, if A is any square matrix of order n, then A(adj A) = (adj A) · A = |A| · I.

19. (d) A is false but R is true.

x² – cos^–1 y = p
⇒ cos^–1 y = (x² – p)
... 0 ≤ cos^–1 y ≤ p
⇒ 0 ≤ x² – p ≤ p
⇒ p ≤ x² ≤ 2p
$$x\varepsilon[-\sqrt{2x},\sqrt{\pi}]\cup[\sqrt{\pi},-\sqrt{2x}]\\\text{Hence domain of f(x) is not real number.}$$

20. (a) Both A and R are true and R is the correct explanation of A.

$$\vec{a}\space and\space\vec{b}\space\text{are reciprocal, then}\\\vec{a}=\lambda\space\vec{b},\lambda\varepsilon\space R{^+}\space and\space|\vec{a}||\vec{b}|=1\\⇒|\vec{a}|=|\lambda||\vec{b}|\\|\lambda|=\frac{|\vec{a}|}{|\vec{b}|}=\frac{1}{|\vec{b}|^{2}}\\\lambda\varepsilon\space R{^+}\\\lambda=\frac{1}{|\vec{b}|^{2}}\\\vec{a}=\frac{\vec{b}}{|\vec{b}|^{2}}\\⇒\vec{a}.\vec{a}=\frac{\vec{b}}{|\vec{b}|^{2}}.\vec{b}=\frac{|\vec{b}|^{2}}{|\vec{b}|^{2}}=1$$

Section-B

21. As principal value of cos^{– 1} q is [0, p]
$$cos^{–1}cos(\frac{13\pi}{6})=cos^{–1}[cos(2\pi+\frac{\pi}{6})]\\cos^{-1}(cos\frac{\pi}{6})\space[cos (2p + q) = cos q]\\=\frac{\pi}{6}∈[0, \pi]\\=cos^{–1}(cos(\frac{13\pi}{6}))=\frac{\pi}{6}$$

$$\text{We know that principal value of}sec^{–1}\theta\space is [0,\pi]-(-\frac{\pi}{2})\\sec^{–1}(-2)=sec^{–1}(-sec\frac{\pi}{3})\\= π–sec^{–1}(sec\frac{\pi}{3})\\=π–\frac{\pi}{3}\\\frac{2\pi}{3}∈ [0,\pi]-(-\frac{\pi}{2})$$

22. (27/8) π(2x+3)²

$$\text{Let V be the volume of the balloon.}\\\text{Radius of spherical balloon}=\frac{3}{4}(2x+3)\\\text{Since}\space V=\frac{4\pi}{3}\text{(radius)}^3\\Then\space V=\frac{4\pi}{3}(\frac{3}{4}(2x+3))^3\\=\frac{9\pi}{16}(2x+3)^3\\\text{On differentiating V with respect to x, we get}\\\frac{dV}{dx}=\frac{9\pi}{16}3(2x+3)^2\frac{d}{dx}(2x+3)\\\text{Hence}\frac{dV}{dx}=\frac{27\pi}{8}(2x+3)^2$$

23. Given equation of line can be written as

$$\frac{x-4}{-2}=\frac{y+3}{2}=\frac{z+2}{1}\\\text{Direction cosines of line parallel to above line are given by}\\\frac{-2}{\sqrt{(-2)^2+(2)^2+(1)^2}},\frac{2}{\sqrt{(-2)^2+(2)^2+(1)^2}},\frac{1}{\sqrt{(-2)^2+(2)^2+(1)^2}}\\=\frac{-2}{\sqrt{4+4+1}},\frac{2}{\sqrt{4+4+1}},\frac{1}{\sqrt{4+4+1}}\\=\frac{-2}{\sqrt{9}},\frac{2}{\sqrt{9}},\frac{1}{\sqrt{9}}\\-\frac{2}{3},\frac{2}{3},\frac{1}{3}\\\text{Hence, required direction cosines of a line parallel to the given line is}(-\frac{2}{3},\frac{2}{3},\frac{1}{3})$$

Given line is 5x – 3 = 15y + 7 = 3 – 10z
Rewritting the equation in standard form :

$$5(x-\frac{3}{5})=15(y+\frac{7}{15})=-10(z-\frac{3}{10})\\i.e.,\space \frac{x-\frac{3}{5}}{\frac{1}{5}}=\frac{y+\frac{7}{15}}{\frac{1}{5}}\\=\frac{z-\frac{3}{10}}{-\frac{1}{10}}\\\text{Thus, the direction ratios of the line are}\frac{1}{5},\frac{1}{15},\frac{-1}{10}i.e., 6, 2, – 3.(\frac{1}{5}×30,\frac{1}{15}×30,\frac{-1}{10}×30)\text{Hence, its direction cosines are}\\ ±\frac{6}{\sqrt{6^2+2^2+(-3)^2}},±\frac{2}{\sqrt{6^2+2^2+(-3)^2}},±\frac{-3}{\sqrt{6^2+2^2+(-3)^2}}i.e.,±\frac{6}{7},±\frac{2}{7},±\frac{3}{7}\\i.e.,\frac{6}{7},\frac{2}{7},\frac{-3}{7}\text{or}\frac{-6}{7},\frac{-2}{7},\frac{3}{7}$$

24. (i) Given differential equation is,

$$\frac{dy}{dx}+\frac{1}{1+x^2}y=sin x\\\text{Integrating factor of above equation is given by}\\e^{\int\frac{1}{1+x^2}dx}=e^{tan^{-1}x}\\\text{(ii) Given differential equation is,}\\x\frac{dy}{dx}+y\space log x = x + y\\⇒\frac{dy}{dx}+\frac{logx}{x}y=\frac{x+y}{x}\\\text{Integrating factor of above equation is given by}\\e^{\int\frac{logx}{x}dx}=e^{\frac{logx}{2}^2}$$

25. We have,

$$\frac{dx}{dt}=-5 cm/min…(i)\\and\space\frac{dy}{dt}=4 cm/min…(ii)\\\text{Now, area of the rectangle (A) = xy}\\⇒\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\\\frac{dA}{dt}= x(4) + y(– 5) \space[using (i) and (ii)]\\\frac{dA}{dt}=4x–5y\space When\space x = 8 cm and y = 6 cm,\\{[\frac{dA}{dt}]}_{atx=8,y=6}\\= 32–30\\=2 cm2/min\\\text{Hence, the rate of change of the area of the rectangle is 2 cm2/min.}$$

Section-C

26. Let

$$I=\int\frac{x+2}{\sqrt{(x-2)(x-3)}}dx\\I=\int\frac{x+2}{\sqrt{x^2-5x+6}}dx\\=\frac{1}{2}\int\frac{2x+4}{\sqrt{x^2-5x+6}}dx=\frac{1}{2}\int\frac{2x+4-5+5}{\sqrt{x^2-5x+6}}dx\\=\frac{1}{2}\int\frac{2x-5}{\sqrt{x^2-5x+6}}dx+\frac{9}{2}\int\frac{dx}{\sqrt{x^2-5x+\frac{25}{4}+6-\frac{25}{4}}}\\=\frac{1}{2}\int(x^2-5x+6)^{-1/2}.(2x-5)dx+\frac{9}{2}\int\frac{dx}{\sqrt{x-(\frac{5}{2})^2-(\frac{1}{2})}}\\=\frac{1}{2}[\frac{(x^2-5x+6)^{-1/2+1}}{-\frac{1}{2}+1}]+\frac{9}{2}\int\frac{dx}{\sqrt{(x-\frac{5}{2})^2}-(\frac{1}{2})^2}\\=(\int[f(x)]^n×f'(x)dx=\frac{f(x)^{n+1}}{n+1}+C)\\I=\sqrt{x^2-5x+6}+\frac{9}{2}log|(x-\frac{5}{2})+\sqrt{x^2-5x+6}|+c$$

27. Let

$$I=\int tan^2\space2x\space tan2x\space sec\space2x\space dx\\=\int(sec^2 2x − 1) tan2x \space sec\space 2x dx\\\text{Put sec 2x} = t\\\text{Differentiate both sides w.r.t. x}\\2 sec 2x\space tan 2x=\frac{dt}{dx}\\sec 2x\space tan\space 2xdx=\frac{dt}{2}\\I=\frac{1}{2}\int(t^2-1)dt\\⇒I=\frac{1}{2}[\frac{t^3}{3}-t]+c\\⇒I =\frac{1}{2}[\frac{sec^3 2x}{3}-sec2x]+c\\⇒I=\frac{sec^32x}{6}-\frac{sec2x}{2}+c$$

28. We have,

$$\int\frac{2x}{(x^2+1)(x^2+3)}dx\\\text{Put} x^2 = t\\\text{Differentiate both side w.r.t.x}\\\text{2x dx = dt}\\I=\int\frac{dt}{(t+1)(t+3)}\\\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}\\\text{Multiply both sides by (t + 1) (t + 3)}\\1 = A(t + 3) + B(t +1) …(i)\\\text{Put t = –1 in equation (i)}\\then 1 = A(–1 + 3)\\⇒ 1 = 2A\\⇒A=\frac{1}{2}\\\text{If t = – 3}\\\text{Put the value in equation (i)}\\\text{then 1 = B(– 3 + 1)}\\1 = – 2B\\B=-\frac{1}{2}\\I=\int\frac{\frac{1}{2}}{t+1}dt+\int\frac{-\frac{1}{2}}{t+3}dt\\I=\frac{1}{2}log|t+1|-\frac{1}{2}log|t+3|+c\\=\frac{1}{2}[log|\frac{t+1}{t+3}|]+c\\⇒I=\frac{1}{2}log|\frac{t+1}{t+3}|+c$$

$$Let\space I=\int\frac{\sqrt{x}}{\sqrt{a^3-x^3}}dx\\=\int\frac{\sqrt{x}}{(\sqrt{a^{3/2}})^2-(x^{3/2})^2}dx\\\text{then put}\space x^{3/2} = t\\\frac{3}{2}\sqrt{x}.dx=dt\\⇒\sqrt{x}.dx=\frac{2}{3}.dt\\\text{Putting the values in (i), we get}\\I=\frac{2}{3}\int\frac{1}{\sqrt{(a^{3/2})^2}-t^2}dt\\⇒I=\frac{2}{3}sin^{-1}(\frac{t}{a^{3/2}})+c\\orI=\frac{2}{3}sin^{-1}(\frac{x}{a})^{3/2}+c$$

29. The given differential equation is
(1 – y²) (1 + log x) dx + 2xy dy = 0

$$\frac{(1+logx)}{x}dx=\frac{-2y}{1-y^{2}}dy\\\text{On integrating both sides, we have}\\\int\frac{1+logx}{x}dx=\int\frac{-2y}{(1-y^2)}dy\\\text{In first integral,}\\put\space 1 + log x = t\\⇒\frac{1}{x}dx=dt\\\text{Also in second integral,}\\\text{put}\space 1 – y^{2} = u\\⇒ – 2y dy = du\\\int t.dt=\int \frac{1}{u}du\\⇒\frac{t^2}{2}-log|u|=c\\or \frac{1}{2}(1+logx)^2-log|1-y^2|=c\\\text{It is given that y = 0 when x = 1}\\So,\frac{1}{2}(1+log1)^2-log|1-0^2|=c\\⇒C=\frac{1}{2}\\\frac{(1+logx)^2}{2}-2log|1-y^2|=1\\or (1 + log x)^2 – 2 log|1 – y^2| = 1\\\text{It is the required particular solution.}$$

$$\frac{dy}{dx}=1 + x^2 + y^2 + x^2y^2\\= (1 + x^2) + y^2 (1 + x^2)\\⇒\frac{dy}{dx}=(1+x^2)(1+y^2)\\⇒\frac{dy}{1+y^2}=(1 + x^2) dx\\\text{On integrating both sides, we have}\\\int\frac{dy}{1+y^2}=\int(1+x^2)dx\\tan^{–1} y =x+\frac{x^3}{3}+c…(i)\\\text{put y = 1 and x = 0 in equation (i),}\\tan^{–1} 1 = 0 + 0 + C\\C=\frac{\pi}{4}\\\text{Equation (i) becomes:}\\tan^{–1} y=x+\frac{x^3}{3}+\frac{|pi}{4}\\⇒ y =tan(x+\frac{x^3}{3}+\frac{\pi}{4})\\\text{is the required particular solution of given equation.}$$

30. Let Bi and Gi denote the apples to be bad and good respectively where i ∈ {R, G}. R stands for red apple
and G for green apple.
Then sample space is S = {G_RG_G, B_RG_G, B_RB_G, G_RB_G}
Let us consider the events
A = both apples are bad
B = red apple is bad
C = green apple is bad
Then, A = {B_RB_G}, B = {B_RG_G , B_RB_G} and C = {B_RB_G, G_RB_G}
and A ∩ B = {B_RB_G}, A ∩ C = {B_RB_G}

$$\text{∴ (i) Required probability = P (A/B)}\\\frac{p(A\cap B)}{P(B)}=\frac{1/4}{2/4}=\frac{1}{2}\\\text{(ii) Required probability = P (A/C)}\\\frac{p(A\cap C)}{P(C)}=\frac{1/4}{2/4}=\frac{1}{2}$$

Let the event in which A fails and B fails be denoted by EA and EB respectively.
Then, P(E_A) = 0·2
P(E_A and E_B) = 0·15
P(B fails alone) = P(E_B) – P(E_A and E_B)
∴ 0·15 = P(E_B) – 0·15
∴ P(E_B) = 0·15 + 0·15 = 0·3.

$$(i) P(E_A/E_B)=\frac{P(E_A\cap E_B)}{P(E_B)}\\=\frac{0.15}{0.3}=0.5.\\\text{(ii) P(A fails alone) = P(EA) – P(EA and EB)}\\= 0·2 – 0·15\\= 0·05.$$

31. Given, maximize Z = x + 2y
Subject to the constraints
x + 2y ≥ 100 2x – y ≤ 0
2x + y ≤ 200 x, y ≥ 0
Converting the inequations into equations we obtain the lines x + 2y = 100, 2x – y = 0, 2x + y = 200.
Then, x + 2y = 100

x	0	100
y	50	0

2x - y = 200

x	10	20
y	20	40

and 2x + y = 200

x	0	100
y	200	0

Plotting these points on the graph, we get the shaded feasible region, i.e., ADECA.

Corner points	Z = x + 2y
A(0, 50)	(0) + 2(50) = 100
D(20, 40)	20 + 2(40) = 100
E(50, 100)	50 + 2(100) = 250
C(0, 200)	0 + 2(200) = 400 maximum

Clearly, the maximum value of Z is 400 at (0, 200).

Section-D

32. L.H.S. = I + A

$$=\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}+\begin{vmatrix}0 & -tan\alpha/2\\tan\alpha/2 & 0\end{vmatrix}\\=\begin{vmatrix}1 & -tan\alpha/2\\tan\alpha/2 & 1\end{vmatrix}\\\text{R.H.S. = (I – A)}\begin{vmatrix}cos\alpha & -sin\alpha\\sin\alpha & cos\alpha\end{vmatrix}\\=(\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}-\begin{vmatrix}0 & -tan\alpha/2\\tan\alpha/2 & 0\end{vmatrix})\begin{vmatrix}cos\alpha & -sin\alpha\\sin\alpha & cos\alpha\end{vmatrix}\\=\begin{vmatrix}1 & tan\alpha/2\\-tan\alpha/2 & 1\end{vmatrix}\begin{vmatrix}cos\alpha & -sin\alpha\\sin\alpha & cos\alpha\end{vmatrix}\\=\begin{vmatrix}cos\alpha+tan\alpha/2sin\alpha & -sin\alpha+tan\alpha/2cos\alpha\\-tan\alpha/2cos\alpha+sin\alpha & tan\alpha/2sin\alpha+cos\alpha\end{vmatrix}\\=\begin{vmatrix}cos\alpha+\frac{sin\alpha/2}{cos\alpha/2}sin\alpha&-sin\alpha+\frac{sin\alpha/2}{cos\alpha/2}cos\alpha\\\frac{-sin\alpha/2}{cos\alpha/2}cos\alpha+sin\alpha&\frac{sin\alpha/2}{cos\alpha/2}sin\alpha+cos\alpha\end{vmatrix}\\=\begin{vmatrix}\frac{cos\alpha cos\alpha/2+sin\alpha/2sin\alpha}{cos\alpha/2} & \frac{-cos\alpha/2sin\alpha+sin\alpha/2cos\alpha}{cos\alpha/2} \\ \frac{-sin\alpha/2cos\alpha+sin\alpha cos\alpha/2}{cos\alpha/2} & \frac{sin\alpha/2sin\alpha+cos\alpha cos\alpha/2}{cos\alpha/2}\end{vmatrix}\\=\frac{1}{cos\alpha/2}\begin{vmatrix}cos(\alpha-\alpha/2) & sin(\alpha/2-\alpha)\\sin(\alpha-\alpha/2) & cos(\alpha-\alpha/2)\end{vmatrix}\\=\frac{1}{cos\alpha/2}\begin{vmatrix}cos\alpha/2 & -sin\alpha/2\\sin\alpha/2 & cos\alpha/2)\end{vmatrix}\\=\begin{vmatrix}1 & -tan\alpha/2\\tan\alpha/2 & 1\end{vmatrix}\\= L.H.S.$$

(iii) Satellite link
(iv) Switch.
(v) MAN (Metropolitan Area Network) as this network can be carried out in a city network.

33. Let AB be the line whose equation is

$$\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\\\text{Let Q be the foot of perpendicular and PQ be the length of perpendicular.Any point Q on the given line is given by}\\\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=λ (say)\\⇒ x = 10λ + 11, y = – 4λ – 2, z = – 11λ – 8\\\text{∴ Coordinates of Q are (10λ + 11, – 4λ – 2, – 11λ – 8)}\\\text{Now, direction ratios of the line PQ are (10λ + 11 – 2, – 4λ – 2 + 1, – 11λ – 8 – 5)}\\\text{∴ Direction ratios of PQ = (10λ + 9, – 4λ – 1, – 11λ – 13)}\\∴ PQ ⊥ AB\\∴ a_1a_2 + b_1b_2 + c_1c_2 = 0\\\text{Where} a_1 = 10λ + 9, b_1 = – 4λ – 1, c_1 = – 11λ – 13\space (Direction ratios of PQ)\\and\space a_2 = 10, b_2 = – 4, c_2 = – 11 (Direction ratios of AB)\\∴ We get\\10(10λ + 9) – 4(– 4λ – 1) – 11(–11λ – 13) = 0\\⇒ 100λ + 90 + 16λ + 4 + 121λ + 143 = 0\\⇒ 237λ + 237 = 0\\⇒ λ = – 1\\\text{Putting λ = – 1 in coordinates of point Q, we get}\\\text{Foot of perpendicular = Q(– 10 + 11, 4 – 2, 11 – 8)}\\= Q(1, 2, 3)\\\text{Also, length of perpendicular PQ is given by}\\PQ=\sqrt{(2-1)^2+(-1-2)^2+(5+3)^2}\\=\sqrt{1+9+4}=\sqrt{14}\space unit$$

$$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1}…(i)\\and\space \frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}…(ii)\\\text{The vector equations of the above two lines are}\\\vec{r}=(\hat{i}-\hat{j}+0\hat{k})+\lambda(2\hat{i}+3\hat{j}+\hat{k})\\\vec{r}=(-\hat{i}+2\hat{j}+2\hat{k})+\mu(5\hat{i}+\hat{j})\\\text{Comparing these equations with}\space\vec{r}=\vec{a_1}+\lambda\vec{b_1}\space and\space\vec{r}=\vec{a_2}+\mu\vec{b_2},\text{we have}\\\vec{a_1}=\hat{i}-\hat{j}\\\vec{a_2}=\hat{i}+2\hat{j}+2\hat{k}\\\vec{b_1}=2\hat{i}+3\hat{j}+\hat{k}\\\space\vec{b_2}=5\hat{i}+\hat{j}\\\text{Now,}\space(\vec{a_2}-\vec{a_1})=2\hat{i}+3\hat{j}+2\hat{k}\\\vec{b_1}×\vec{b_1}=\begin{bmatrix}\hat{i}&\hat{j}& \hat{k}\\2 & 3&1\\5&1 & 0 \end{bmatrix}\\=\hat{i}(0-1)-\hat{j}(0-5)+\hat{k}(2-15)\\=-\hat{i}+5\hat{j}-13\hat{k}\\\text{also}\space|\vec{b_1}×\vec{b_1}|=\sqrt{(-1)^2+(5)^2+(-13)^2}\\\sqrt{1+25+169}=\sqrt{195}\\(\vec{a_2}-\vec{a_1}).(\vec{b_2}×\vec{b_1})=(-2\vec{i}+3\vec{j}+2\vec{k}).(-\hat{i}+5\hat{j}-13\hat{k})\\= 2 + 15 – 26 = – 9\\\text{Therefore, the shortest distance between the lines}\\=|\frac{-9}{\sqrt{195}}|=\frac{9}{\sqrt{195}}≥ 0\\\text{Since},\space d ≠ 0\\\text{Hence, the given lines do not intersect.}$$

34. We have, y = x² and x + y + 2 = 0

⇒ – x – 2 = – x²
⇒ x² – x – 2 = 0
⇒ x² + x – 2x – 2 = 0
⇒ x(x + 1) – 2(x + 1) = 0
⇒ (x – 2) (x + 1) = 0
⇒ x = 2, – 1
Area of shaded region,

$$A=|\int^{2}_{-1}(-x-2+x^2)dx|\\=|\int^{2}_{-1}(x^2-x-2)dx|\\=|[\frac{x^3}{3}-\frac{x^2}{2}-2x]^{2}_{-1}|\\=|[\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2]|\\=|\frac{16-12-24+2+3-12}{6}|\\=|-\frac{27}{6}|=\frac{9}{2}\space\text{sq. units.}$$

35. Let

$$y=tan^{–1}(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}})\\\text{Put} x^2 = cos θ\\θ = cos^{– 1} x^2\\y=tan^{–1}(\frac{\sqrt{1+cos\theta}+\sqrt{1-cos\theta}}{\sqrt{1+cos\theta}-\sqrt{1-cos\theta}})\\=tan^{–1}(\frac{\sqrt{2cos^2\frac{\theta}{2}}+\sqrt{2sin^2\frac{\theta}{2}}}{\sqrt{2sin^2\frac{\theta}{2}}-\sqrt{2sin^2\frac{\theta}{2}}})\\=tan^{–1}(\frac{\sqrt{2}(cos\frac{\theta}{2}+sin\frac{\theta}{2})}{\sqrt{2}(cos\frac{\theta}{2}-sin\frac{\theta}{2})})\\=tan^{–1}(\frac{(cos\frac{\theta}{2}+sin\frac{\theta}{2})}{(cos\frac{\theta}{2}-sin\frac{\theta}{2})})\\y=tan^{–1}(\frac{1+tan\frac{\theta}{2}}{1-tan\frac{\theta}{2}})\\=tan^{–1}[tan(\frac{\pi}{4}+\frac{\theta}{2})]\\=\frac{\pi}{4}+\frac{\theta}{2}=\frac{\pi}{4}+\frac{1}{2}(cos^{-1}x^2)\\\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi}{4})+\frac{1}{2}\frac{d}{dx}\frac{1}{2}(cos^{-1}x^2)\\=0+\frac{1}{2}(-\frac{1}{\sqrt{1-x^4}})×\frac{d}{dx}(x^2)\\=-\frac{1}{2\sqrt{1-x^4}}×2x\\=-\frac{x}{\sqrt{1-x^4}}$$

Given, the function is continuous. For the continuity at x = 2 and x = 10.
Continuity at x = 2 :

$$L.H.L.=\lim\limits_{x\rarr2^{-}}f(x)\\=\lim\limits_{x\rarr2^{-}}5=5\\R.H.L. =\lim\limits_{x\rarr2^{+}}f(x)\\=\lim\limits_{x\rarr2^{+}}(ax+b)\\=\lim\limits_{x\rarr2^{-}}[a(2 + h) + b][x = 2 + h]\\= 2a + b\\\text{Since, f(x) is continuous at x = 2.}\\\text{So, L.H.L. = R.H.L. = f(2)}\\\text{2a + b = 5 ...(i)}\\\text{Continuity at x = 10}\\L.H.L.=\lim\limits_{x\rarr10^{-}}f(x)\\=\lim\limits_{x\rarr10^{-}}f(ax+b)\\=\lim\limits_{h\rarr0}[a(10-h)+b]\space[x=10–h]\\= 10a + b\\R.H.L. =\lim\limits_{x\rarr10^{+}}f(x)\\=\lim\limits_{x\rarr10^{+}}21=21\\\text{Since, f(x) is continuous at x = 10.}\\So, L.H.L. = R.H.L. = f(10)\\10a + b = 21 …(ii)\\\text{Solving (i) and (ii), we get}\\a = 2\space and\space b = 1.$$

Section-E

36. (i) E₁ be event of selecting Sanjay
E₂ be event of selecting Ajay
E₃ be event of selecting Vijay.

$$P(E_1)=\frac{1}{7}\\P(E_2)=\frac{2}{7}\\P(E_3)=\frac{4}{7}\\\text{(ii) A be the event of increasing profits.}\\P(A/E_1) = 0·8\\P(A/E_2) = 0·5\\P(A/E_3) = 0·3\\P(E_2/A) =\frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}\\=\frac{\frac{7}{3}×0.5}{\frac{1}{7}×0.8+\frac{2}{7}×0.5+\frac{4}{7}×0.3}\\=\frac{\frac{1}{7}}{\frac{0.8}{7}+\frac{1}{7}+\frac{1.2}{7}}=\frac{1}{3}\\(iii)\space P(E_1/A) =\frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}\\=\frac{\frac{7}{3}×0.8}{\frac{1}{7}×0.8+\frac{2}{7}×0.5+\frac{4}{7}×0.3}\\=\frac{\frac{0.8}{7}}{\frac{0.8}{7}+\frac{1}{7}+\frac{1.2}{7}}=\frac{0.8}{3}=\frac{8}{30}$$

$$(iii)\space P(E_3/A) =\frac{P(E_3)P(A/E_3)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}\\=\frac{\frac{4}{7}×\frac{7}{10}}{\frac{1}{7}×\frac{2}{10}+\frac{2}{7}×\frac{5}{10}+\frac{4}{7}×\frac{7}{10}}\\=\frac{\frac{28}{70}}{\frac{2}{70}+\frac{10}{70}+\frac{28}{70}}=\frac{28}{40}=\frac{7}{10}.$$

37. (i) Green tea = x
Black tea = 200 – x
f (x) = 120 x + (200 – x) 160
= 120 x + 32000 – 160 x
f (x) = ₹ 32000 – 40 x

(ii) Total revenue 150 × 200 = 30,000
∴ 32,000 – 40x = 30,000
⇒ 2000 = 40 x
⇒ x = 50

(iii) f(400) = 32000 – 40 × 400
= 32000 – 16000
= ₹16000
f(200) = 32000 – 8000
= ₹24000
f(400) – f (200) = 16000 – 24000 = – 8000

g(x) = 2x + 5
Now, g(10) = 2 × 10 + 5 = 25

38. (i) Projection of
$$\vec{a}\space on\space\vec{b}=\frac{\vec{a}.\vec{b}}{|\vec{b}|}\\\vec{a}.\vec{b}=(5\hat{i}+\hat{j}+4\hat{k}).(2\hat{i}+6\hat{j}+3\hat{k})\\= 10 + 6 + 12\\= 28\\\text{Projection =}\frac{28}{7}=4\\\text{(ii) Unit vector in the direction of}\space\vec{a} +\space\vec{b}\space is\\\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{(7\hat{i}+7\hat{j}+7\hat{k})}{\sqrt{7^2+7^2+7^2}}=\frac{7(\hat{i}+\hat{j}+\hat{k})}{\sqrt{147}}\\=\frac{7(\hat{i}+\hat{j}+\hat{k})}{7\sqrt{3}}\\=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}$$