Oswal Practice Papers CBSE Class 12 Mathematics Solutions (Practice Paper - 9)

Section-A

1. (a)
$$\frac{\pi}{2}$$

Explanation :

$$cos^{-1}(cos\frac{3\pi}{2})≠\frac{3\pi}{2}\space as\space\frac{3\pi}{2}\epsilon[0,\pi]\\cos^{-1}cos(2\pi-\frac{\pi}{2})=\frac{\pi}{2}$$

2. (c)
$$-8\hat{i},3\hat{j}$$

Explanation :

Let, the initial point of the vector is A (6, 4) and the terminal point of the vector is B(– 2, 7).
∴ The required vector is,
$$\vec{AB}=(-2-6)\hat{i}+(7-4)\hat{j}\\-8\hat{i},-3\hat{j}\\\text{Hence, the required vector components of the vector are:}-8\hat{i},\space and\space3\hat{j}$$

3. (a) √157

Explanation :

$$\text{Magnitude of}\space3\hat{i}+2\hat{j}+12\hat{k}\\=\sqrt{3^2+2^2+12^2}=\sqrt{157}$$

4. (b) f is everywhere continuous but not differentiable at x = nπ, n ∈ Z

Explanation :

5. (a)
$$a(\frac{u^3}{3})+b(\frac{u^2}{2})+cu+C$$

Explanation :

$$\text{The given integral is}\int(au^2 + bu + c) du .\\\int(au^2 + bu + c) du=\int au^2du+\int budu+\int cdu\\=a\int u^2du+b\int udu+c\int du\\=a(\frac{u^3}{3})+b(\frac{u^2}{2})+cu+C$$

6. (b)
$$y=e^x+ sin\space x+\frac{x^2}{2}+ log\space sec\space x+C$$

Explanation :

$$\frac{dy}{dx}=e^x + cos x + x + tan x\\\text{On integrating both sides, we get}\\y=e^x+ sin\space x+\frac{x^2}{2}+ log\space sec\space x+C$$

7. (b) half plane that neither contains the origin nor the points on the line 2x + 3y = 6

Explanation :

Since, origin (0, 0) does not satisfy the given inequality (i.e., 0 + 0 > 6 is not true). Also,
2x + 3y > 6 (i.e., the points on the line 2x + 3y = 6 are not included).
The graph is half plane that neither contains the origin nor the points on the line 2x + 3y = 6.

8. (a) $$\frac{1}{3}e^{x^3}+C$$

Explanation :

$$\text{Let}\space I=\int x^2e^{x^3}dx\\\text{Put}\space x^3=t\\⇒3x^2dx=dt\\⇒x^2dx=\frac{1}{3}dt\\I=\frac{1}{3}\int e^tdt\\I=\frac{1}{3}e^t+C\\I=\frac{1}{3}e^{x^3}+C$$

9. (b)
$$\frac{(\vec{a}.\vec{b})\vec{b}}{|\vec{b}|^2}$$

Explanation :

$$\text{Projection of}\space \vec{a}\space on\space\vec{b}\\=\frac{(\vec{a}.\vec{b})}{\vec{b}}=\frac{(\vec{a}.\vec{b})\vec{b}}{\vec{b}.\vec{b}}\\=\frac{(\vec{a}.\vec{b})\vec{b}}{|\vec{b}|^2}$$

10. (a) Corner points of the feasible region

Explanation :

By putting the corner points in objective function, we get the optimal value.

11. (a)
$$\Delta=\frac{1}{2}\begin{vmatrix}x_1& y_1&1\\x_2& y_2&1\\x_3& y_3&1 \end{vmatrix}$$

Explanation :

$$\text{Area of a triangle with vertices}(x_1, y_1),(x_2, y_2),(x_3, y_3)is=\frac{1}{2}\begin{vmatrix}x_1& y_1&1\\x_2& y_2&1\\x_3& y_3&1\end{vmatrix}$$

12. (d) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Explanation :

Δ = Sum of product of elements of any row (or column) with their correspoding cofactors.
a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

13.(d) Diagonal matrix

Explanation :

Matrix with all elements zero except diagonal is called diagonal matrix.

14. (c) 7/8

Explanation :

$$P(A)=\frac{4}{5},P(A\cap B)=\frac{7}{10}\\P(B|A) =\frac{P(A\cap B)}{P(A)}=\frac{7/10}{4/5}=\frac{7}{8}$$

15. (c) 4

Explanation :

Number of arbitrary constants in the general solution of differential equation is equal to the order of differential equation.

16. (c)
$$\frac{2}{1+x^2}$$

Explanation :

$$\text{Put x}=tan\theta\\y=sin^{-1}(\frac{2 tan\theta}{a+tan^2\theta})\\= sin^{–1} (sin 2\theta)\\y=2\theta=2tan{–1}x\\\frac{dy}{dx}=\frac{2}{1+x^2}$$

17. (c) 60°

Explanation :

$$\text{We have,}\vec{a}.\vec{b}=\frac{1}{2}|\vec{a}||\vec{b}|\\⇒|\vec{a}||\vec{b}|cos\theta=\frac{1}{2}|\vec{a}||\vec{b}|,\\\text{where q is the required angle.}\\⇒cos\theta=\frac{1}{2}\\⇒\theta=60°.$$

18. (c)
$$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$$

Explanation :

$$\text{Since, On x-axis coordinates of y and z are zero.}\\\text{Equation of x-axis becomes}\\\frac{x-0}{1-0}=\frac{y-0}{0-0}=\frac{z-0}{0-0}\\⇒\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$$

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

Explanation :

$$A=\begin{bmatrix}a&0&0&0\\0&b&0&0\\0&0&c&0\end{bmatrix}\text{is not a diagonal matrix.}\\\text{Since, A is not a square matrix.}$$

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

Explanation :

$$AA^{–1} = A^{–1}.A = I\\\text{... Reason is true.}\\\text{On applying this property in assertion.}\\\text{If |A| = 5 then |A–1|} =\frac{1}{|A|}=\frac{1}{5}\\\text{Thus, assertion is true and the reason is the correct explanation of the assertion.}$$

Section-B

21. Given equation is
$$cos(2sin^{-1}x)=\frac{1}{9},x > 0 ...(i)\\\text{We put}\space sin^{–1}x = y\\⇒ x = sin y\\\text{Equation (i) becomes}\\cos 2y =\frac{1}{9}\\⇒ 1 – 2sin^2 y =\frac{1}{9}\\⇒ 2sin^2 y = 1-\frac{1}{9}\\=\frac{8}{9}\\⇒ sin^2 y=\frac{4}{9}\\⇒ x^2=\frac{4}{9}(... x = sin y)\\x = ±\frac{2}{3}\\\text{But given that x > 0}\\ x =\frac{2}{3}$$

$$\text{Principal value of}\space tan^{–1}\theta\space is(-\frac{\pi}{2},\frac{\pi}{2})\\\text{The principal value of}\space tan^{–1}(tan\frac{3\pi}{4})\\=tan^{–1}[tan(\pi-\frac{\pi}{4})]\\=tan^{–1}tan(\pi-\frac{\pi}{4})[... tan (π – θ)=–tan θ]\\= tan^{–1} tan(-\frac{\pi}{4})[... – tanθ=tan(– θ)]\\=-\frac{\pi}{4}\varepsilon(-\frac{\pi}{2},\frac{\pi}{2})$$

22. Let r be the radius and A be the area of the circular wave at any time t.
$$\text{Then, area of circular wave = A =}\pi r^2 and\frac{dr}{dt}= 3·5 cm/sec\\Now, A = \pi r^2\\\text{On differentiating with respect to t, we get}\\\frac{dA}{dt}=\pi(2r\frac{dr}{dt})\\⇒\frac{dA}{dt}=2r\pi\frac{dr}{dt}\\⇒\frac{dA}{dt}= 2\pi r(3·5)=7pr\text{(given)}\\⇒(\frac{dA}{dt})_{r=7·5}= 7\pi(7·5)\\= 52·5\pi cm^2/sec.$$

23. Direction ratios of
$$\vec{OP}\text\space{are – 1, 2, – 2.}\\\text{Therefore, direction cosines of}\vec{OP}\space are\\\frac{-1}{\sqrt{(-1)^2+2^2+(-2)^2}},\frac{2}{\sqrt{(-1)^2+2^2+(-2)^2}},\frac{-2}{\sqrt{(-1)^2+2^2+(-2)^2}}\\i.e.,\frac{-1}{3},\frac{2}{3},\frac{-2}{3}\\\text{Hence}\space\vec{OP}=|\vec{OP}|(l\hat{i}+m\hat{j}+n\hat{k})\\=3(\frac{-1}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k})\\=-1\hat{i}+2\hat{j}-2\hat{k}\\\text{Therefore, coordinates of P are (– 1, 2, – 2).}$$

$$\text{Given equation of line in Cartesian form is}\\\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\\\text{The point on line is (5, – 4, 6) and direction ratios are (3, 7, 2).}\\\text{We know that vector equation of a line passing through}\space\vec{a}\space\text{and parallel to}\space\vec{b}\space is\\\vec{r}=\vec{a}+\lambda\vec{b}\\\text{Here,}\vec{a}= (5, – 4, 6) and\space\vec{b}= (3, 7, 2)\\\text{Its equation in vector form is}\\\vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda(3\hat{i}+7\hat{j}+2\hat{k})$$

24. Given differential equation is,
$$\frac{dy}{dt}+y=1\\⇒\frac{dy}{dt}=1-y\\⇒\frac{dy}{1-y}=dt\\\text{Integrating both sides}\\– log (1 – y) = x + C\\⇒ log (1 – y) = – x – C\\⇒ 1 – y = e^{–x– C} = e^{–x}. e^{–C}\\⇒ 1 – y = Ae^{–x}, where A = e^{–C}\\⇒ y = 1 – Ae^{–x}\\\text{which is the required general solution.}$$

25. Here,
$$\vec{a}-\vec{b}=(\hat{i}+2\hat{j}-\hat{k})-(3\hat{i}+\hat{j}-5\hat{k})\\=-2\hat{i}+\hat{j}+4\hat{k}\\\text{Let,}\space\vec{a}-\vec{b}=\vec{c}\\\text{Now, unit vector in the direction of}\space\vec{c}\\\frac{\vec{c}}{|\vec{c}|}=\frac{=-2\hat{i}+\hat{j}+4\hat{k}}{\sqrt{(-2)^2+(1)^2+(4)^2}}\\=\frac{-2\hat{i}+\hat{j}+4\hat{k}}{\sqrt{21}}\\=\frac{-2}{\sqrt{21}}\hat{i}+\frac{1}{\sqrt{21}}\hat{j}+\frac{4}{\sqrt{21}}\hat{k}$$

Section-C

26. Let

$$\frac{2}{(1-x)(1+x^2)}=\frac{A}{(1-x)}+\frac{Bx+C}{1+x^2}\\2=A(1 + x^2)+(Bx + C)(1 – x)\\Put x = 1\\⇒ 2 = 2A\\⇒ A = 1\\Put\space x = 0\\⇒ 2 = A + C\\⇒ 2 = 1 + C\\⇒ C = 1\\Put x = 2\\⇒ 2 = 5A + (2B + C) (1 – 2)\\⇒ 2 = 5A – 2B – C\\⇒ 2 = 5 – 2B – 1\\⇒ 2 = 4 – 2B\\⇒ B =\frac{4-2}{2}\\⇒ B = 1\\I=\int\frac{2dx}{(1-x)(1+x^2)}\\=\int\frac{dx}{1-x}+\int\frac{x}{1+x^2}dx+\int\frac{dx}{1+x^2}\\=\int\frac{dx}{1-x}+\frac{1}{2}\int\frac{2x}{1+x^2}+\int\frac{dx}{1+x^2}\\I = – log |1 – x|+\frac{1}{2}log|1 + x^2| + tan^{–1} x + C.$$

27. We have, x² = y and y = x + 2

⇒ x² = x + 2
⇒ x² – x – 2 = 0
⇒ x² – 2x + x – 2 = 0
⇒ x(x – 2) + 1(x – 2) = 0
⇒ (x + 1) (x – 2) = 0
x = – 1, 2
Required area of shaded region
$$=\int^{2}_{-1}(x+2-x^2)dx=[\frac{x^2}{2}+2x-\frac{x^3}{3}]^{2}_{-1}\\=[\frac{4}{2}+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3}]\\=6+\frac{3}{2}-\frac{9}{3}\\=\frac{36+9-18}{3}=\frac{27}{6}\\=\frac{9}{2}\text{sq. units.}$$

28. Let A, B, C be the given students and let E1, E2, E3 be the events that the problem is solved by A, B, C respectively.
$$Then,\={E_1} , \={E_2} , \={E_3}\space \text{are the events that the given problem is not solved by A, B, C respectively.Then,}\\P(E_1)=\frac{1}{2},P(E_2)=\frac{1}{3}\space and P(E_3)=\frac{1}{3}\\P(\={E_1})=(1-\frac{1}{2})=\frac{1}{2};P(\={E_1})=(1-\frac{1}{3})=\frac{2}{3}\\andP(\={E_3})=(1-\frac{1}{4})=\frac{3}{4}\\\text{P(exactly one of them solves the problem)}=P[({E_1}\cap\={E_2}\cap\={E_3})\space or\space({\=E_1}\cap {E_2}\cap\={E_3})\\or\space({\=E_1}\cap\={E_2}\cap {E_3})]=P({E_1}\cap\={E_2}\cap+\={E_3})\space +P\space({\=E_1}\cap {E_2}\cap\={E_3})+P\space({\=E_1}\cap\={E_2}\cap {E_3})]\\\text{Now since}(E_1,\={E_2},\={E_3});(\={E_1},{E_2},\={E_3});\={E_1}\={E_2}{E_3}\space\text{are independent events respectively}\\\text{∴ P(only one of A, B, C solve the problem correctly)}=[P(E_1)×P(\={E_2})×P(\={E_3})]\\+[P\={E_1)}×P({E_2})×P(\={E_3})]+[P\={E_1)}×P(\={E_2})×P({E_3})]\\=(\frac{1}{2}×\frac{2}{3}×\frac{3}{4})+(\frac{1}{2}×\frac{1}{3}×\frac{3}{4})+(\frac{1}{2}×\frac{2}{3}×\frac{1}{4})\\=\frac{6+3+2}{24}=\frac{11}{24}\\\text{Hence, the required probability is}\frac{11}{24}.$$

Let A and B be the events that problem is solved by first and second student respectively.
Probability of solving the problem by A
$$P(A)=\frac{1}{4}\\\text{Probability of solving the problem by B}\\P(B)=\frac{1}{5}\\\text{Since the problem is being solved independently}\\P(A ∩ B)=P(A)·P(B)=\frac{1}{4}×\frac{1}{5}=\frac{1}{20}\\P(\={A}) =1–P(A)=1-\frac{1}{4}=\frac{3}{4}\\P(\={B})=1–P(B)=1-\frac{1}{5}=\frac{4}{5}\\\text{(i) Probability that the problem is solved=P(A ∪ B)}\\=P(A)+P(B)–P(A ∩ B)\\=\frac{1}{4}+\frac{1}{5}-\frac{1}{20}=\frac{5+4}{20}-\frac{1}{2}\\=\frac{8}{20}=\frac{5}{20}\\\text{(ii) Probability that exactly one of them solves the problem is given by}\\P(A).P{\=(B)} + P(B).P\={(A)} =\frac{1}{4}×\frac{4}{5}+\frac{1}{5}×\frac{3}{4}=\frac{4+3}{20}\\=\frac{7}{20}$$

29.
$$I=\int\frac{(3sin\theta-2)cos\theta}{5-cos^2\theta-4sin\theta}d\theta\\⇒I=\int\frac{(3sin\theta-2)cos\theta}{4+1-cos^2\theta-4sin\theta}d\theta\\⇒I=\int\frac{(3sin\theta-2)cos\theta}{4+sin^2\theta-4sin\theta}d\theta[... 1–cos^2\theta=sin^2\theta]\\⇒I=\int\frac{(3sin\theta-2)cos\theta}{(sin\theta-2)^2}d\theta\\\text{Put}\space sin\theta=t\\⇒ cos\theta.d\theta= t\\I=\int\frac{(3t-2)}{(t-2)^2}.dt\\Consider,\\\frac{(3t-2)}{(t-2)^2}=\frac{A}{(t-2)}+\frac{B}{(t-2)^2}\\⇒ A(t – 2) + B = 3t – 2\\\text{On comparing, we get}\\A = 3 and B = 4\\I=\int(\frac{3}{t-2}+\frac{4}{(t-2)^2})dt\\⇒ I = 3 log |t – 2| –\frac{4}{t-2}+C\\⇒ I = 3 log |sin\theta – 2| –\frac{4}{sin\theta-2}+C\\or I = 3 log |2 – sin \theta|+\frac{4}{2-sin\theta}+C$$

$$Let\space I=\int\frac{2x}{(x^2+1)(x^2+2)^2}dx\\Put\space x^2 = t\\⇒ 2x dx = dt\\I=\int\frac{2x}{(t+1)(t+2)^2}\\\int\frac{dt}{(t+1)(t+2)^2}\\Let=\int\frac{1}{(t+1)(t+2)^2}=\frac{A}{(t+1)}+\frac{B}{(t+2)}+\frac{C}{(t+2)^2}...(A)\\\frac{1}{(t+1)(t+2)^2}=\frac{A(t+2)^2+B(t+1)(t+2)+C(t+1)}{(t+1)(t+2)^2}\\1 = A(t + 2)^2 + B(t + 1) (t + 2) + C(t + 1)\\\text{Equating coefficients,}t^2,\text{t and constant terms on both sides, we get}\\A + B = 0 ...(i)\\4A + 3B + C = 0 ...(ii)\\4A + 2B + C = 1 ...(iii)\\\text{Subtracting equation (iii) from (ii), we get}\\B = – 1\\\text{Substitute in equation (i)}\\⇒ A = 1\\\text{Substitute the value of A and B in (ii), we get}\\4 – 3 + C = 0\\C = –1\\\text{From equation (A)}\\\frac{1}{(t+1)(t+2)^2}=\frac{1}{(t+1)}+(\frac{-1}{t+1})+(\frac{1}{(t-2)^2})\\\int\frac{1}{(t+1)(t+2)^2}dt=\int\frac{1}{(t+1)}dt-\int(\frac{1}{t+1})dt-\int(\frac{1}{(t-2)^2})dt\\= log|t+1|–log|t+2|+\frac{1}{t+2}+C\\\text{Where C is constant of integration.}\\\int\frac{2x}{(x^2+1)(x^2+2)^2}dx= log|\frac{x^2+1}{x^2+2}|+\frac{1}{x^2+2}+C$$

30. Given differential equation is
$$xy\frac{dy}{dx}= (x + 2) (y + 2)\\⇒\frac{y}{y+2}dy=\frac{(x+2)}{x}dx\\\text{On integrating both sides, we have}\\\int\frac{y}{y+2}dy=\int\frac{(x+2)}{x}dx\\\text{Put y + 2 = t ⇒ y = t – 2}\\\text{Differentiate w.r.t. y}\\\text{1=}=\frac{dt}{dy}\\... dy = dt\\\text{Equation (i) becomes:}\\\int\frac{t-2}{t}dt=\int1dx 2\int\frac{dx}{x}\\⇒\int(1-\frac{2}{t})dt = x + 2 log x + C\\⇒ t – 2 log t = x + 2 log x + C\\⇒ y + 2 – 2 log (y + 2) = x + 2 log x + C …(ii)\\\text{When x = 1 and y = – 1}\\– 1 + 2 – 2 log 1 = 1 + 2 log 1 + C\\⇒ 1 = 1 + C\\... C = 0\\\text{Equation (ii) becomes:}\\y + 2 – 2 log (y + 2) = x + 2 log x\\\text{which is the particular solution of the given differential equation.}$$

Given differential equation is
$$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}\\{dy}=(\frac{1-cosx}{1+cosx}){dx}\\{dy}=(\frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}){dx}\\dy = tan^2\frac{x}{2}dx\\dy =(sec^2\frac{x}{2}-1)dx\\\text{Integrating both sides:}\\y =\int(sec^2\frac{x}{2}-1)dx\\y =(\frac{tan\frac{x}{2}}{\frac{1}{2}}-x)dx\\y = 2 tan\frac{x}{2}– x + C\\\text{which is the required general solution.}$$

31. The given LPP can be re-written as :
Maximize or Minimize
Z = 3x + 5y
Subject to constraints :
3x – 4y ≥ – 12
2x – y ≥ – 2
2x + 3y ≥ 12
x ≤ 4
y ≥ 2
x ≥ 0
Converting the inequations into equations, we obtain the following equations
3x – 4y = – 12, 2x – y = – 2, 2x + 3y = 12, x = 4, y = 2 and x = 0.
These lines are drawn on suitable scale. The shaded region P₁ P₂P₃ P₄ P₅ represents the feasible region of the given LPP.

The values of the objective function at these points are given in the following table :

Point (x, y)	Value of the objective function Z = 3x + 5y
P₁ (3, 2)	Z = 3 × 3 + 5 × 2 = 19
P₂ (4, 2)	Z = 3 × 4 + 2 × 5 = 22
P₃ (4, 6)	Z = 3 × 4 + 5 × 6 = 42
P₄(4/5,18/5)	Z = 3 ×(4/5)+5×(18/5)=102/5
P₅(3/4,7/2)	Z = 3 ×(3/4)+5×(7/2)=79/4

Clearly Z assumes its minimum value 19 at x = 3 and y = 2.and the maximum value of Z is 42 at x = 4 and y = 6.

Section-D

32. The vertices of the ∆ABC are A(– 2, 1), B(0, 4) and C(2, 3)

Equation of the side AB is
$$y–y_1=\frac{y_2-y_1}{x_2-x_1}(x_1 – x_1)\\⇒ y–1=\frac{4-1}{0-(-2)}(x–(– 2))\\=\frac{3}{2}(x + 2)\\⇒y=\frac{3}{2}x+4\\\text{Equation of the side BC is}\\y–4=\frac{3-4}{2-0}(x– 0)=\frac{-1}{2}x\\⇒ y =\frac{-1}{2}x+4\\\text{Equation of the side AC is}\\y-1=\frac{3-1}{2-(-2)}(x–(– 2))\\=\frac{1}{2}(x+2)\\y =\frac{x}{2}+ 2\\\text{Required area = Shaded area}\\=\int^{0}_{-2}(\frac{3}{2}x+4)dx +\int^{2}_{0}(\frac{-1}{2}x+4)dx-\int^{2}_{-2}(\frac{1}{2}x+2)dx\\=[\frac{3}{2}\frac{x{^2}}{2}+4x]^{0}_{-2}=[\frac{-1}{2}\frac{x{^2}}{2}+4x]^{2}_{0}-[\frac{1}{2}\frac{x{^2}}{2}+2x]^{2}_{-2}\\=(0+0)-(\frac{3}{2}×\frac{4}{2}-8)+(\frac{-1}{2}×\frac{4}{2}+8)-(0+0)-(\frac{1}{2}×\frac{4}{2}+4)-(\frac{1}{2}×\frac{4}{2}-4)\\= –3 + 8 –1 + 8 – 0 – 8 = 4 sq. units.$$

33. f : R → R is given by
f(x) = 4x³ + 7
For injective :
Let f(x) = f(y)
⇒ 4x³ + 7 = 4y³ + 7
⇒ x³ = y³
⇒ x³ – y³ = 0
⇒ (x – y)(x² + xy + y²) = 0
⇒ x – y = 0 (as the other term cannot be zero)
⇒ x = y
Thus, f(x) = f(y)
⇒ x = y ∀ x, y ∈ R
For surjective :
Let y ∈ R
Then, f(x) = y
⇒ 4x³ + 7 = y
$$⇒ x^3 =\frac{y-7}{4}\\\text{Thus, for any y ∈ R, x =}[\frac{y-7}{4}]^{1/3}\text{is a real number or x ∈ R.}\\\text{Since range of f(x) and codomain of f(x) are both real numbers, f(x) is an onto function.}\\\text{Therefore, f is bijective.}$$

One-one : Let
f (x₁) = f (x₂), ∀ x₁, x₂ ∈ W
Case I : When x₁, x₂ both are even, then
Let f (x₁) = f (x₂), ∀ x₁, x₂ ∈ W
⇒ x₁ + 1 = x₂ + 1
⇒ x₁ = x₂
f (x₁) = f (x₂)
⇒ x₁ = x₂, ∀ x₁, x₂ ∈ W
Case II : When both x₁, x₂ are odd, then
Let f (x₁) = f (x₂), ∀ x₁, x₂ ∈ W
⇒ x₁ – 1 = x₂ – 1
⇒ x₁ = x₂
f (x₁) = f (x₂)
⇒ x₁ = x₂, ∀ x₁, x₂ ∈ W
Hence, from case I and case II, we observe that
f (x₁) = f (x₂)
⇒ x₁ = x₂, ∀ x₁, x₂ ∈ W
f(x) is a one-one function.
Onto :
Case I : Let n = 2m
⇒ f(2m) = 2m + 1 = n + 1
Case II : Let n = 2m + 1
f (2m + 1) = (2m + 1) – 1 = 2m
f (2m + 1) = 2m = n (even terms)
Also, f (1) = 0
f : W → W is onto function.
Since, f(x) is both one-one and onto function, so it is a bijective function.

34. Given y = | x + 3 |
$$\text{Interest from first bond =}\frac{10×x×1}{100}=\frac{10}{100}\\\text{Interest from second bond =}\frac{12×y×1}{100}=\frac{12y}{100}\\\text{Interest received by trust = ₹2,800}\\\text{According to the question,}\\\frac{10x}{100}+\frac{12y}{100}= 2,800\\⇒ 10x + 12y = 2,80,000 …(i)\\\frac{12x}{100}+\frac{10y}{100}= 2,800\\⇒ 12x + 10y = 2,70,000 …(ii)\\\text{This system of equations can be written in matrix form as follows :}\\\begin{bmatrix}10&12\\12&10\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2,80,000\\2,70,000\end{bmatrix}\\or AX = B\\A=\begin{bmatrix}10&12\\12&10\end{bmatrix},X=\begin{bmatrix}x\\y\end{bmatrix}\\B=\begin{bmatrix}2,80,000\\2,70,000\end{bmatrix}\\Now, |A| =\begin{vmatrix}10&12\\12&10\end{vmatrix}\\= 100–144=–44≠0\\So,A^{–1}\text{exists and the solution of the given system of equations is given by}\\X = A^{– 1}B\\Let c_{ij}\text{be the cofactor of} a_{ij} in A=[a_{ij}].Then,\\c_{11} = 10, c_{12} = – 12, c_{21} = – 12, c_{22} = 10\\\ ad_{j}A=\begin{bmatrix}10&-12\\-12&10\end{bmatrix}^T=\begin{bmatrix}10&-12\\-12&10\end{bmatrix}\\So,A^{–1}\frac{1}{|A|}=(adj A)=-\frac{1} {44}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}\\\text{Hence, the solution is given by}\\X=A^{–1}B\\=-\frac{1} {44}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}\begin{bmatrix}2,80,000\\2,70,000\end{bmatrix}\\\begin{bmatrix}x\\y\end{bmatrix}=-\frac{1}{44}\begin{bmatrix}28,00,000-32,40,000\\-33,60,000+27,00,000\end{bmatrix}\\⇒\begin{bmatrix}x\\y\end{bmatrix}=-\frac{1}{44}\begin{bmatrix}-4,40,000\\-6,60,000\end{bmatrix}=\begin{bmatrix}10 000\\15 000\end{bmatrix}\\⇒ x=10,000 and y=15,000\\⇒A=x+y=10,000+15,000\\=₹25,000\\\text{Hence, the amount invested by the trust is}=₹25,000$$

35. Given equation of lines are
$$\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{6}\\and\space \frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}\\\text{Now, the vector equation of lines are}\\\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda(2\hat{i}+3\hat{j}+6\hat{k})…(i)\\\text{[Vector form of equation of line is}\vec{r}=\vec{a}+\lambda\vec{b}]\\\vec{r}=(3\hat{i}+3\hat{j}-5\hat{k})+\mu(4\hat{i}+6\hat{j}+12\hat{k})…(i)\\\text{Here,}\space\vec{a_1}=\hat{i}+2\hat{j}-4\hat{k}+\vec{b_1}=2\hat{i}+3\hat{j}+6\hat{k}\\\vec{a_2}=3\hat{i}+3\hat{j}-5\hat{k},\vec{b_2}=4\hat{i}+6\hat{j}+12\hat{k}\\\text{Now,}\vec{b_2}×\vec{b_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&6\\4&6&12\end{vmatrix}\\\hat{i}((36-36)+\hat{j}(24-24)-\hat{k}(12-12)\\=0\hat{i}+0\hat{j}-0\hat{k}\\\vec{b_1}×\vec{b_2}= 0\\⇒ Vector\space b_1 \text{is parallel to}\space b_2\space[\vec{a}×\vec{b}=0,then\space \vec{a}||\vec{b}]\\\text{Therefore, the two lines are parallel}\\\vec{b}=(2\hat{i}+3\hat{j}+6\hat{k})...(iii)\\\text{[Since, direction ratios of given lines are proportional]}\\\text{Since, the two lines are parallel, we use the formula for shortest distance between two parallel lines.}\\\text{We know that}\\d =|\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}|…(iv)\\\text{...From equations (iii) and (iv), we get}\\d =|\frac{(2\hat{i}+3\hat{j}+6\hat{k})×(2\hat{i}+\hat{j}-\hat{k})}{\sqrt{(2)^2+(3)^2+(6)^2}}|…(v)\\\text{Now}(2\hat{i}+3\hat{j}+6\hat{k})×(2\hat{i}+\hat{j}-\hat{k})\\=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&6\\2&1&-1\end{vmatrix}\\\hat{i}(-3-6)+\hat{j}(-2-12)+\hat{k}(2-6)\\-9\hat{i}+14\hat{j}-4\hat{k}\\\text{From equation (v), we get}\\d=|\frac{-9\hat{i}+14\hat{j}-4\hat{k}}{\sqrt{49}}|\\=\frac{\sqrt{(-9)^2+(14)^2+(-4)^2}}{7}\\d =\frac{\sqrt{81+196+16}}{7}=\frac{\sqrt{293}}{7}\space\text{units}$$

$$Let\space\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-1}{4}=\lambda\\x=2\lambda+1,y=–\lambda–1,z=4\lambda+3\\\text{Therefore, a general point on the line (i) is}\\P(2\lambda + 1, – l – 1, 4\lambda + 3)\\Let\space\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+1}{2}=\mu…(ii)\\⇒ x=3\mu+1,y=4\mu–6,z=2\mu–1\\\text{Therefore, a general point on the line (ii) is}\\Q(3\mu + 1, 4\mu – 6, 2\mu – 1)\\\text{Let PQ be the shortest distance between the two given lines. Then, the line of shortest distance passes through P and Q.}\\\text{Direction ratios of PQ are}\\(3\mu – 2l, 4\mu + l – 5, 2\mu – 4\lambda – 4)\\\text{Since, PQ is perpendicular to the line (i),}\\2(3\mu–2\lambda)–1(4\mu+\lambda–5)+4(2\mu–4\lambda–4)=0\\⇒ 10\mu – 21\lambda – 11 = 0 …(iii)\\\text{Also, PQ is perpendicular to the line (ii).}\\\ 3(3\mu – 2\lambda) + 4(4\mu + \lambda – 5) + 2(2\mu – 4\lambda – 4) = 0\\⇒ 29\mu – 10\lambda – 28 = 0 …(iv)\\\text{Solving equations (iii) and (iv), we get}\\\frac{\lambda}{-280 - 319}=\frac{\mu}{110-588}=\frac{1}{-690+100}\\⇒\frac{\lambda}{39}=\frac{\mu}{-478}=\frac{1}{-509}\\⇒\lambda=\frac{-39}{509}\\and\space\mu=\frac{478}{509}\\\text{So, the required points, where the line of shortest distance meets the given lines are}\\P(\frac{431}{509},\frac{-470}{509},\frac{1371}{509})\text{and Q}(\frac{1943}{509},\frac{-1142}{509},\frac{447}{509})\\\text{Therefore, shortest distance = PQ}\\=\sqrt{(\frac{1943}{509}-\frac{431}{509})^2+(\frac{-1142}{509}+\frac{470}{509})^2+(\frac{447}{509}}+\frac{1371}{509})^2\\=\sqrt{(\frac{1512}{509})^2+(\frac{672}{509}+)^2+(\frac{-924}{509}})^2\\=\frac{1}{\sqrt{509}}\sqrt{2286144+451584+853776}\\=\frac{1}{\sqrt{509}}\sqrt{3591504}\\=\frac{4}{\sqrt{509}}\sqrt{224469}\text{units}\\\text{Also, the line of shortest distance is}\\\frac{x-\frac{431}{509}}{\frac{1512}{509}}=\frac{y+\frac{470}{509}}{\frac{-672}{509}}=\frac{z-\frac{1371}{509}}{\frac{-924}{509}}\\i.e.,\frac{509x-431}{512}=\frac{509y-470}{-612}=\frac{509y-1371}{-924}$$

Section-E

36. (i) f(x) = 3x⁴ – 4x³ – 12x² + 5
f ′(x) = 12x³ – 12x² – 24x + 0
= 12x(x² – x – 2)
= 12x(x² – 2x + x – 2)
= 12x[x(x – 2) + 1(x – 2)]
= 12x(x + 1) (x – 2)
For critical points put f ′(x) = 0 ⇒ x = 0, x = – 1, x = 2
∴ Critical points are (0, – 1, + 2).

(ii)

Value of x	x	(x + 1)	(x – 2)	Sing of f'(x)	inc/dec
x < – 1	–ve	–ve	–ve	–ve	↓
–1 < x < 0	–ve	+ve	–ve	+ve	↑
0 < x < 2	+ve	+ve	–ve	–ve	↓
x > 2	+ve	+ve	+ve	+ve	↑

f(x) is strictly increasing in (– 1, 0) ∪ (2, ∞).

(iii) Based on the table given in (ii), f(x) is strictly decreasing in the interval (– ∞, – 1) ∪ (0, 2).

For local maxima find f ′′(x) = 36x2 – 24x – 24
f ′′(0) = – 24 < 0 local maxima at x = 0
and Maxima value f(0) = 5
f ′′(– 1) = 36 + 24 – 24 = 36 > 0
Local Minima f ′′(2) = 36 × 22 – 24 × 2 – 24
144 – 48 – 24 = 72 > 0 local minima
∴ Local maxima at x = 0.

37. (i) Ram’s height in function h = 3 + 14t – 5t²
Differentiating h, w.r.t. to t,
$$\text{We get}\frac{dh}{dt}= 14 – 10t\\\text{Put,}\frac{dh}{dt}=0\\14 – 10t = 0\\10t = 14\\t = 1·4\\\text{(ii) Maximum height covered by Ram’s ball}\\h = 3 + 14(1·4) – 5(1·4)^2\\h = 12·8 m\\\text{(iii) Rahul will take time T, when function H is maximum}\\H = 15T^2 – 4T + 3\\\frac{dH}{dT}= 30T – 4\\Put\space\frac{dH}{dT}= 0\\30T – 4 = 0\\30T = 4\\T=\frac{4}{30}=\frac{2}{15}$$

$$\text{Height H = 15}(\frac{2}{15})^2-4×\frac{2}{15}+3=\frac{97}{15}$$

38. (i) Total probability is P(E₁) P(A/E₁) + P(E₂) P(A/E₂)
$$P(E_1)=\frac{60}{100}\\P(E_2)=\frac{40}{100}\\=\frac{60}{100}×\frac{2}{100}+\frac{40}{100}×\frac{1}{100}\\=\frac{120+40}{10000}=\frac{160}{10000}=\frac{4}{250}=\frac{2}{125}\\(ii)\space P(E_1/A) =\frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}\\=\frac{0.6×0.02}{0.6×0.02+0.4×0.01}\\=\frac{0.012}{0.012+ 0.004}\\=\frac{0.012}{0.016}\\=\frac{3}{4}$$