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 :    

We have, f(x) = |sin x|
We know that |x| and sin x are continuous for all real x.
So, | sin x | is also continuous for all real x.
|x| is non-differentiable at x = 0
So, |sin x| is non-differentiable when sin x = 0
or x = nπ, n∈Z
Hence, f(x) is continuous everywhere but not differentiable at x = nπ, n∈Z.

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

OR

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

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

OR

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

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

parabolic

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

OR

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

OR

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

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

OR

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 P2 P3 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 
$$⇒ x^3 =\frac{y-7}{4}\\\text{Thus, for any y ∈ R, x =}[\frac{y-7}{4}]^{1/3}\text{is a real number or x ∈ R.}\\\text{Since range of f(x) and codomain of f(x) are both real numbers, f(x) is an onto function.}\\\text{Therefore, f is bijective.}$$

OR

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

OR

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

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). 

Critical points

(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).

OR

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

OR

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

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

CBSE Practice Paper Mathematics Class 12

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