# Oswal Practice Papers CBSE Class 12 Mathematics Solutions (Practice Paper - 8)

## Section-A

1. (c) 1

Explanation :

f(x) = x2 – 8x + 17
f (x) = 2x – 8
So, f (x) = 0, gives x = 4
Now, f ′′(x) = 2 > 0, ∀ x
So, x = 4 is the point of local minima.
Minimum value of f(x) = at x = 4.
f(4) = 4 × 4 – 8 × 4 + 17 = 1.

2. (c) f(0) = 0

Explanation :

$$f(x)=\begin{vmatrix}0 & x-a&x-b\\x+a & 0&x-c\\x+b & x-c&0\end{vmatrix}\\\text{put x = 0,}\\⇒ f(0) =\begin{vmatrix}0 & -a&-b\\a & 0&-c\\b & c&0\end{vmatrix}\\\text{which is skew-symmetric determinant of order 3}\\\text{Hence, f(0) = 0}$$

3. (d) B + D

Explanation :

Only B + D is defined because matrices of the same order can only be added.

4. (d) Not defined

Explanation :

The given differential equation is not a polynomial equation in terms of its derivatives, so its degree is not defined.

$$5.\space(a)\frac{x^{52}}{52}(tan^{-1}x+cot^{-1}x)+C$$

Explanation :

$$\int{x^{51}}(tan^{-1}x+cot^{-1}x)dx\\=\int{x^{51}}.\frac{\pi}{2}dx\\{ie...(tan^{-1}x+cot^{-1}x)=\frac{\pi}{2}}\\\frac{\pi x^{52}}{104}+C=\frac{x{52}}{52}(tan^{-1}x+cot^{-1}x)+C$$

6. (d) None of these

Explanation :

x – [x] = 0 when x is an integer, so that f(x) is discontinuous for all x ∈ I i.e., f(x) is
discontinuous at infinite number of points.

7. (b) 6 +  5

Explanation :

y = 5 sin2 x + 7 cos2 x – 4 sin x cos x
= (2 sin x – cos x)2 + sin2 x + 6 cos2 x
= (2 sin x – cos x)2 + (sin2 x + cos2 x) + 5 cos2 x
ymax =( 22+(1)2)+1+5
=6 +

8. (c) All of the given constraints

Explanation :

Feasible region is the set of points which satisfies all the given constraints.

9. (c) unit matrix

Explanation :

If square matrix in which all diagonals elements are 4 and rest are 0, is called unit matrix.

10. (a)
$$\frac{2}{3}u^{3/2}+2u^{1/2}+C$$

Explanation :

$$\int(\sqrt{u}+\frac{1}{\sqrt{u}})du=\int(\sqrt{u})du+\int(\frac{1}{\sqrt{u}})du\\=\int(u^{1/2})du+\int(\frac{1}{u^{1/2}})du\\=\frac{u^{3/2}}{\frac{3}{2}}+\frac{u^{1/2}}{\frac{1}{2}}+C\\=\frac{2}{3}u^{3/2}+2u^{1/2}+C$$

11. (c) Concave region

Explanation :

Optimal solution, feasible solution and objective function are the terms used in linear
programming problem.

12. (a) [0, 1]

Explanation :

We know that cos– 1 x is defined for x ∈ [– 1, 1]
f(x) = cos– 1 (2x – 1) is defined if
– 1 ≤ 2x – 1 ≤ 1
⇒ 0 ≤ 2x ≤ 2
⇒ 0 ≤ x ≤ 1.

13. (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.

14. (b) 6, 9, 6

Explanation :

The number of elements in m × n matrix is equal to mn.
∴ elements in A = 3 × 2 = 6
∴ elements in B = 3 × 3 = 9
∴ elements in C = 2 × 3 = 6

15. (c) 2, 1

Explanation :

Given that,
$$[1+(\frac{dy}{dx})^2]=\frac{d^2y}{dx^2}\\\text{\ Order = 2 and degree = 1.}$$

16. (c)
$$(x=(2n+1)\frac{\pi}{2};nεZ)$$

Explanation :

$$\text{when tan}(2n+1)\frac{\pi}{2}=\text{tan}(n\pi+\frac{\pi}{2})=-\text{cot} n\pi\\\text{It is not defined at the integral points (n∈Z)}\\\text{Hence, f(x) is discontinuous at}(2n+1)\frac{\pi}{2}$$

17. (b) It has local minimum at x = 0

Explanation :

Given f(x) = x3 sin x
f (x) = x3 cos x – 3x2 sin x
Now, f ′′(x) = – x3 sin x + 3x2 cos x – 3x2 cos x – 6x sin x
For critical points, f (x) = 0
⇒ x3 cos x – 3x2 sin x = 0
⇒ x2(x cos x – sin x) = 0
x = 0
At x = 0
⇒ f ′′(x) = 0
⇒ Function has local minimum at x = 0.

18. (b) sin x = C sec y

Explanation :

$$\frac{dy}{dx}= \text{cot x cot y}\\⇒ \text{cot x dx – tan y dy = 0}\\\text{Integrating on both sides, we get}\\⇒\int \text{cot x dx} −\int \text{tan y dy = 0}\\\text{log sin x – log sec y = log C}\\⇒log(\frac{\text{sinx}}{\text{sexy}})= log C\\⇒ \text{sin x = C sec y.}$$

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

Explanation :

$$\text{A is false but R is true.}\\\text{Area of triangle with vertices (1, 1), (0, 2), (k, 0) is 3 sq. units.}\\\frac{1}{2}\begin{vmatrix}1&1&1\\0&2&1\\k&0&1\end{vmatrix}=3\\⇒ 1(2 – 0) – 1(0 – k) + 1(0 – 2k) = 6\\⇒ 2 + k – 2k = 6\\⇒ – k = 6 – 2\\⇒ k = 4\\\text{Value of k is : 4.}$$

20. (c) A is true but R is false.

Explanation :

$$\text{Work done, W} =\vec{F}.\vec{r}\\\text{∴ Work done is a Scalar quantity.}$$

## Section-B

21. We know that the principal value of cos–1 q is [0, π]
$$cos^{-1}(cos\frac{7\pi}{6})≠\frac{7\pi}{6}as\frac{7\pi}{6}ε[0,\pi]\\cos^{-1}(cos\frac{7\pi}{6})=cos^{-1}[cos(2\pi-\frac{5\pi}{6})][... cos (2\pi – \theta) = cos \theta]\\=cos^{-1}[cos\frac{5\pi}{6}]\\=\frac{5\pi}{6}ε[0,\pi]$$

OR

$$\text{Principal value of tan}^{-1}\theta\space is(-\frac{\pi}{2},\frac{\pi}{2})\text{and that of sin}^{-1}{\theta}\space is[-\frac{\pi}{2},\frac{\pi}{2})].\\\tan^{-1}(1)+sin^{-1}(-\frac{1}{2})\\=\tan^{-1}(tan\frac{\pi}{4})+sin^{-1}(-sin\frac{\pi}{6})\\=\tan^{-1}\tan\frac{\pi}{4}+\sin^{-1}\sin-\frac{\pi}{6}\\=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$$

22.Let r be the radius of the sphere and Δr be the error in measuring the radius. Then r = 9 cm and Δr = 0.03 cm. Now, the volume V of the sphere is given by
$$V=\frac{4}{3}\pi r^3\\or\space\frac{dV}{dr}=4\pi r^2\\dV=(\frac{dV}{dr})Δr=(4\pi r^2)Δr\\= 4π(9)^2 (0·03)\\=9·72\text{π cm}^3\\\text{Thus, the approximate error in calculating the volume is 9.72π cm}^3.$$

23. Let length of a side of the square cut out = x cm
Volume, V = (24 – 2x)2.x
dv/dx= 2(24 – 2x) (– 2)x + (24 – 2x)2
= (24 – 2x) (– 4x + 24 – 2x)
= (24 – 2x) (24 – 6x)
= 12(12 – x) (4 – x)
Put dv/dx= 0 i.e., (12 – x) (4 – x) = 0
Since, x ≠ 12,
∴ x = 4
Hence, the volume will be maximum when length of a side of the square cut out = 4 cm.

OR

$$\text{Given function is}\\f(x) =\frac{x^4}{4}-x^3-5x^2+24x+12\\⇒ f ′(x) =\frac{4x^3}{4}-3x^2-10x+24\\\text{For critical points, put f ′(x) = 0}\\∴ x^3 – 3x^2 – 10x + 24 = 0\\(x – 2) (x^2 – x – 12) = 0\\(x – 2) (x – 4) (x + 3) = 0\\⇒ x = 2, 4, – 3\\\text{Therefore, we have the intervals (– ∞, – 3), (– 3, 2), (2, 4) and (4, ∞).}$$

$$\text{Since f'(x) > 0 in (– 3, 2) ∪ (4, ∞).}\\∴ \text{f(x) is increasing in interval (– 3, 2) ∪ (4, ∞).}\\\text{And f′(x) < 0 in (– ∞, – 3) ∪ (2, 4)}\\∴ \text{f(x) is decreasing in (– ∞, – 3) ∪ (2, 4).}$$

24. Given differential equation is,
$$e^x\text{tan y dx + }(1 – e^x)\text{sec}^2\space y\space dy = 0\\⇒ (1 – ex) sec^2\space y\space dy = – e^x\text{tan y dx}\\⇒\frac{sec^2\space y}{tan\space y}dy=-\frac{e^x}{1-e^x}dx\\\text{Integrating both sides}\\log|tan y| = log|1 – e^x|+ C\\⇒ log|tan\space y|– log|1 – ex| = C\\⇒ log|\frac{tan\space y}{1-e^x}|=C\\⇒\frac{tan\space y}{1-e^x}= e^C\\⇒ tan\space y = e^C (1 – e^x).\\\text{which is the required solution.}$$

25. Let at any instant,
Length of string = y m
and horizontal distance of the kite = x m
$$then\frac{dx}{dt}= 5·2 m/sec\\Then, y^2 = x^2 + (120)^2\\\text{Differentiate above equation w.r.t. t}\\⇒ 2y·\frac{dy}{dt}=2x.\frac{dx}{dt}\\⇒\frac{dy}{dt}=\frac{x}{y}.\frac{dx}{dt}=\frac{x}{y}(5.2)...(i)\\\text{When y = 130 m, then}\\x =\sqrt{(130)^2-(120)^2}=\sqrt{2500}\\\text{Hence the rate at which the string is being pulled out at the instant when string’s length is 130 m.}\\=\frac{50}{130}×(5.2)=\frac{5×52}{130}= 2 m/sec$$

## Section-C

26. Let

$$I=\int\frac{x^3\space sin(tan^{-1}x^4)}{1+x^8}dx\\\text{Put}\space\text{tan}^{–1} x^4 = t\\\text{Differentiate w.r.t. x}\\⇒\frac{1}{1+x^8}.4x^3=\frac{dt}{dx}\\\frac{1}{1+x^8}.dx=\frac{dt}{dx}\\I=\frac{1}{2}\int sin t\space dt=\frac{1}{4}(-cos t)+C\\⇒I=-\frac{1}{4}(-cos)(tan^{-1}x^4)+C$$

27. We first convert the inequalities into equations to obtain lines
2x + 4y = 8 …(i)
3x + y = 6 …(ii)
x + y = 4 …(iii)
x = 0
and y = 0
We need to maximise the objective function
Z = 2x + 5y
These lines are drawn and the feasible region of the L.P.P. is the shaded region.
The point of intersection of equations (i) and (ii) is B(1·6, 1·2).
The coordinates of the corner points of the feasible region are O(0, 0), A(0, 2), B(1·6, 1·2) and C(2, 0).

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

 Corner Points Objective function Z = 2x + 5y O(0, 0) 2 × 0 + 5 × 0 = 0 A(0, 2) 2 × 0 + 5 × 2 = 10 ← Maximum B(1·6, 1·2) 2 × 1·6 + 5 × 1·2 = 9·2 C(2, 0) 2 × 2 + 5 × 0 = 4

Out of these values of Z, the maximum value of Z is 10 which is attained at the point (0, 2). Thus the maximum value of Z is 10.

28. Let

$$I=\int\frac{dx}{x(x^5+3)}\\\text{Multiplying num. and deno. by} x^4=\int\frac{x^4}{x^5(x^5+3)}dx\\\text{Put} x^5 = t\\\text{Differentiate both sides w.r.t. x}\\5x^4=\frac{dt}{dx}⇒ dx =\frac{dt}{5x^4}\\I=\frac{1}{5}\int\frac{dt}{x^5(t+5)}=\frac{1}{5}\int\frac{dt}{t(t+3)}\\⇒\frac{1}{t(t+3)}=\frac{A}{t}+\frac{B}{t+3}\\⇒1 = A(t + 3) + Bt\\If t = 0, 1 = 3A\\⇒ A =\frac{1}{3}\\If t = – 3\\1 = – 3B\\⇒ B =-\frac{1}{3}\\I =\frac{1}{5}[\int\frac{\frac{1}{3}}{t}dt+\int\frac{\frac{-1}{3}}{t+3}dt]\\=\frac{1}{15}log|t|-\frac{1}{15}log|t+3|+C\\=\frac{1}{15}log|\frac{t}{t+3}|+C\\I=\frac{1}{15}log|\frac{x^5}{x^5+3}|+C$$

OR

$$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\\3x–1 = A(x – 2)(x – 3)+B(x – 1)(x – 3)+C(x –1)(x –2)\\If\text{x = 1, then}\\2 = A(– 1) (– 2)\\⇒ 2A = 2\\⇒ A = 1\\\text{If x = 2, then}\\5 = B(1) (– 1)\\⇒ B = – 5\\\text{If x = 3, then}\\8 = C(2) (1)\\⇒ C = 4\\\int\frac{3x-1}{(x-1)(x-2)(x-3)}dx=\int\frac{dx}{x-1}+\int\frac{-5}{x-2}dx+\int\frac{4}{x-3}dx\\⇒\int\frac{3x-1}{(x-1)(x-2)(x-3)}dx= \text{log |x – 1| – 5 log |x – 2| + 4 log |x – 3| + C.}$$

29. Given differential equation is,
$$(1+x^2)\frac{dy}{dx}+y=tan^{-1}x\\\text{Divide by}(1 + x^2)\text{in above equation, we get}\\\frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{tan^{-1}x}{1+x^2}...(i)\\\text{The above equation is of the form}\frac{dy}{dx}+py=Q\\\text{Where, P =}\frac{1}{1+x^2}\text{and Q =}\frac{tan^{-1}x}{1+x^2}\\I.F. = e^{∫} P dx\\=e^{∫\frac{1}{1+x}dx}\\= e^{tan–1}x\\\text{Solution of the equation is given by}\\y.I.F = ∫ Q × I.F. dx + C\\\\ye^{tan–1} x=\int\frac{e^{tan-1}x\space tan^{-1}x}{1+x^2}dx+C...(ii)\\\text{Put}\space tan^{–1} x = t\\\text{Differentiate both sides w.r.t.x}\\\frac{1}{1+x^2}dx=dt\\\text{Equation (ii) becomes}\\y^{etan–1} x = ∫ e^t. t dt + C\\y^{etan–1} x = t e^t–∫e^t. dt + C\\= t e^t – e^t + C\\= tan^{–1} x e^{tan–1} x – e^{tan–1 x} + C\\⇒ ye^{tan–1} x = e^{tan–1x} (tan^{–1} x – 1) + C\\⇒ y = tan^{–1} x – 1 + Ce^{–tan–1} x$$

OR

$$\text{Given differential equation is,}\\x log x\frac{dy}{dx}+y=2logx\\⇒\frac{dy}{dx}+\frac{1}{xlogx}y=\frac{2}{x}...(i)\\Here P =\frac{1}{xlogx},Q=\frac{2}{x}\\I.F. = e^{∫P dx}\\= e^{∫\frac{1}{xlogx}dx}\\= e^{∫\frac{1/x}{xlogx}dx}\\= e^{log (log x)} = log x\\\text{Solution of given equation will be given by,}\\y.I.F. = ∫Q I.F. dx + C\\⇒ y log x = 2 ∫log x .\frac{1}{x}dx + C\\⇒ y\space log x =\frac{2(logx)^2}{2}+C\space [\int[f(x)^n.f^{'}(x)dx]=\frac{[f(x)]^{n+1}}{n+1}+C]\\⇒ y = log x +\frac{C}{logx}.$$

30. Let A be event that the sum of observations is 8.
∴ A = {(2, 6), (3, 5), (5, 3), (4, 4), (6, 2)}
⇒ n(A) = 5
⇒ P(A)=5/36
Let B be event that the observation on red die is less than 4.
∴ B = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1, 3), (2, 3), (3, 3), (4, 3),
(5, 3), (6, 3)}
⇒ n(B) = 18
$$⇒ P(B) =\frac{18}{36}=\frac{1}{2}\\Clearly A ∩ B = {(5, 3), (6, 2)}\\∴ n(A ∩ B) = 2\\∴ P(A ∩ B) =\frac{2}{36}=\frac{1}{18}\\P(A/B) =\frac{(P ∩ B)}{P(B)}=\frac{\frac{1}{18}}{\frac{1}{2}}\\\frac{2}{18}=\frac{1}{9}$$

OR

There are three possible cases. They are :
A1 : The bag has 2 white balls and 2 balls of other colour
A2 : The bag has 3 white balls and 1 ball of other colour
A3 : The bag has 4 white balls
Now, P(A1) = P(A2) = P(A3) =1/3
Let E be the event of selecting 2 white balls.
$$P(\frac{E}{A_1})=\frac{^2C_2}{^4C_2}=\frac{1}{6}\\P(\frac{E}{A_2})=\frac{^3C_2}{^4C_2}=\frac{1}{2}\\P(\frac{E}{A_3})=\frac{^4C_2}{^4C_2}=1\\\text{Applying Bayes’ theorem,}\\P(\frac{A_3}{E})=\frac{P(A_3)P(\frac{E}{A_3})}{[P(A_1)P(\frac{E}{A_1})+P(A_2)P(\frac{E}{A_2})+P(A_3)P(\frac{E}{A_3})]}\\=\frac{\frac{1}{3}×1}{\frac{1}{3}(\frac{1}{6}+\frac{1}{2}+1)}=\frac{3}{5}$$

31. Let
$$I=\int\frac{e^x}{(e^x-1)^2(e^x+2)}dx\\\text{Putting}\space e^x = t\space and\space e^x \text{dx = dt, we get}\\I=\int\frac{dt}{(t-1)^2(t+2)}\\Let,\frac{1}{(t-1)^2(t+2)}=\frac{A}{t-1}+\frac{B}{(t-1)^2}+\frac{C}{t+2}\\⇒ 1 = A(t – 1) (t + 2) + B(t + 2) + C(t – 1)^2\\\text{Putting, t =1}\\⇒ 1 = 3B ⇒ B = 1/3\\Putting, t = – 2\\⇒ 1 = 9 C ⇒ C = 1/9\\Putting, t = 0\\⇒ 1 = –2A + 2B + C\\⇒ 1 = –2A +\frac{2}{3}+\frac{1}{9}\\⇒A=-\frac{1}{9}\\I=\frac{-1}{9}\int\frac{1}{(t-1)}dt+\frac{1}{3}\int\frac{1}{(t-1)^2}dt+\frac{1}{9}\int\frac{1}{(t+2)}dt\\I =\frac{-1}{9}log|t-1|-\frac{1}{3(t-1)}+\frac{1}{9}log|t+2|+C\\I=\frac{1}{9}log|\frac{e^x+2}{e^x-1}|-\frac{1}{3(e^x-1)}+C$$

## Section-D

32. Let H denote the event that students who read Hindi newspaper and E denote the students who read English newspaper It is given that,

$$P (H) = 60\%=\frac{60}{100}=\frac{3}{5}\\P (E) = 40\%=\frac{40}{100}=\frac{2}{5}\\P (H ∩ E) = 20\%=\frac{20}{100}=\frac{1}{5}\\\text{(i) Probability that a student reads neither Hindi nor English newspaper is}\\P (H ∪ E)′ = 1 – P (H ∪ E)\\= 1 – [P (H) + P (E) – P (H ∩ E)]\\=1-\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\\=1-\frac{4}{5}=\frac{1}{5}\\\text{(ii) Probability that a randomly chosen student reads English newspaper if she reads a Hindi newspaper is}\\P (E / H) =\frac{P (H ∩ E)}{P(H)}=\frac{\frac{1}{5}}{\frac{3}{5}}=\frac{1}{3}\\\text{(iii) Probability that a randomly chosen student reads Hindi newspaper if she reads a English newspaper is}\\P (E / H) =\frac{P (H ∩ E)}{P(H)}=\frac{\frac{1}{5}}{\frac{2}{5}}=\frac{1}{2}$$

33. Since,
$$f (x) =\frac{2x+1}{2x^3+3x^2+x}> 0\\⇒ f (x) =\frac{2x+1}{x(2x^2+3x+1)}> 0\\⇒ f (x) =\frac{2x+1}{x(2x+1)(x+1)}> 0\\⇒ f (x) =\frac{1}{x(x+1)}> 0,x≠-\frac{1}{2}\\\text{[(2x + 1) cancels out in the numerator and denominator]}\\\text{By using the sign rule (or otherwise) we can now find out the region on the number line where}\frac{1}{x(1+x)}\text{> 0 will be true.}\\\text{The region is – ∞ to – 1 and 0 to ∞, where neither – 1 nor 0 are included in the solution set.}\\\text{ Hence, we write the solution as : f(x) ∈ (– ∞, – 1) ∪ (0, ∞)}$$

OR

Given, R = {(T1, T2) : T1 is congruent to T2}
Reflexivity: Let (T, T ) ∈ R, " T ∈ A
Since every triangle is congruent to itself
i.e., R is reflexive.
Symmetric : Let (T1, T2) ∈ R
then, T1 is congruent to T2
⇒ T2 is congruent to T1
⇒ (T2, T1) ∈ R
So, R is symmetric.
Transitivity : Let T1, T2, T3 ∈ A such that (T1, T2) ∈ R and (T2, T3) ∈ R. Then,
⇒ T1 is congruent to T2 and T2 is congruent to T3
⇒ T1 is congruent to T3
⇒ (T1, T3) ∈ R
So, R is transitive.
Hence, R is an equivalence relation on set A.

34. Given y = | x + 3 |
$$\text{Required area} =\int^0_{-6}|x+3|dx\\=\int^{-3}_{-6}-(x+3)dx+\int^{0}_{-3}(x+3)dx$$

$$= −[\frac{x^2}{2}+3x]^{-3}_{-6}+[\frac{x^2}{2}+3x]^{-0}_{-3}\\=-[(\frac{9}{2}-9)-(\frac{36}{2}-18)]+[(0+0)-(\frac{9}{2}-9)]\\=-(\frac{9}{2}-0)+(0-(\frac{-9}{0}))\\=\frac{9}{2}+\frac{9}{2}=\frac{18}{2}=9 \text{sq. units.}$$

35. Given that
al + bm + cn = 0 …(i)
and fmn + gnl + hlm = 0 …(ii)
From (i), al + bm = – cn
$$⇒\space n=-\frac{(al+bm)}{C}…(iii)\\\text{Substituting this value of n in (ii), we get}\\(fm + gl)(-\frac{al+bm}{C})+ hlm = 0\\⇒ – (fm + gl) (al + bm) + hlmc = 0\\⇒ – aflm – agl^2 – bfm^2 – bglm + chlm = 0\\⇒ agl^2 + (af + bg – ch)lm + bfm^2 = 0 …(iv)\\\text{We note that both l and m cannot be zero.}\\\text{For, if l = m = 0}\\\text{Then from (iii), we get n = 0}\\\text{which is wrong as}\space l^2 + m^2 + n^2 = 1.\\\text{So, without any loss of generality we may take m }≠ 0.\\\text{Dividing both sides of (iv) by}\space m^2, we get\\ag(\frac{1}{m})^2+(af+bg-ch)(\frac{l}{m})+bf=0…(v)\\\text{The above equation is quadratic in}(\frac{l}{m})\text{giving two values of}(\frac{l}{m})\\If (l_1, m_1, n_1) and (l_2, m_2, n_2)\\\text{are the direction cosines of the two given lines (i) and (ii), then}\frac{l_1}{m_1},\frac{l_2}{m_2}\\\text{are roots of equation (v).}\\\text{Therefore, product of root =}\frac{bf}{ag}\\⇒(\frac{l_1}{m_1})(\frac{l_2}{m_2})=\frac{bf}{ag}\\⇒\frac{l_1l_2}{m_1m_2}=\frac{f/a}{g/b}\\⇒\frac{l_1l_2}{f/a}=\frac{m_1m_2}{g/b}\\⇒\frac{l_1l_2}{f/a}=\frac{m_1m_2}{g/b}=\frac{n_1n_2}{h/c}\space\text{[By symmetry of result]}\\\text{Let}\frac{l_1l_2}{f/a}=\frac{m_1m_2}{g/b}=\frac{n_1n_2}{h/c}=\lambda\\\text{where l is a non-zero real number.}\\l_1l_2 = \lambda·\frac{f}{a},m_1m_2=\lambda.\frac{g}{b},n_1n_2=\lambda.\frac{h}{c}\\\text{Now, these lines whose direction cosines are given by (i) and (ii) are at right angles.}\\If l_1l_2 + m_1m_2 + n_1n_2 = 0\\i.e., if\space \lambda\frac{f}{a}+\lambda\frac{g}{b}+\lambda\frac{h}{c}=0\\i.e., if\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0\\\text{Also, the two lines will be parallel if,}\\l_1 = l_2, m_1= m_2, n_1 = n_2\\i.e., if\frac{l_1}{m_1}=\frac{l_2}{m_2}\\\text{i.e., if equation (v) has equal roots.}\\\text{i.e., if discriminant of (v) is zero.}\\i.e., if (af + bg – ch)^2 – 4agbf = 0\\i.e., if a^2f^2 + b^2g^2 + c2h2 + 2abfg – 2bgch – 2afch – 4agbf = 0\\i.e., if a^2f^2 + b^2g^2 + c^2h^2 – 2(bcgh + cahf + abfg) = 0.$$

OR

Here, T is the image of point P(1, 6, 3). Q is the foot of perpendicular PQ on the line AB.
First, we find Q.
Equation of line AB is given by
$$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\\Let\space\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda(say)\\⇒ x = l, y – 1 = 2l, z – 2 = 3l\\⇒ x = l, y = 2l + 1, z = 3l + 2\\\text{Let coordinates of}\\Q = (l, 2l + 1, 3l + 2) …(ii)$$

$$\text{Now, direction ratios of line}\\PQ = (l – 1, 2\lambda + 1 – 6, 3\lambda + 2 – 3)\\\text{Direction ratios of} PQ = (l – 1, 2\lambda – 5, 3\lambda – 1)\\Now, line PQ AB\\\ a_1a_2 + b_1b_2 + c_1c_2 = 0\\Where\space a_1 = \lambda – 1, b_1 = 2\lambda – 5, c_1 = 3\lambda – 1\\\text{and} a_2 = 1, b_2 = 2, c_2 = 3\\\ 1(\lambda – 1) + 2(2\lambda – 5) + 3(3\lambda – 1) = 0\\⇒ l – 1 + 4\lambda – 10 + 9\lambda – 3 = 0\\⇒ 14\lambda – 14 = 0\\⇒ \lambda = 1\\\text{Putting} \lambda = 1 in equation (ii), \text{we get}\\Q(1, 2 + 1, 3 + 2) = (1, 3, 5)\\\text{Now, as discussed earlier Q is the mid-point of PT.}\\\text{Let coordinates of T = (x, y, z)}\\\text{\ Using by mid-point formula,}\\\text{Q = Mid-point of P(1, 6, 3) and T(x, y, z)}\\=(\frac{x+1}{2},\frac{y+6}{2},\frac{z+3}{2},)\\But\space Q = (1, 3, 5)\\(\frac{x+1}{2},\frac{y+6}{2},\frac{z+3}{2},)= (1, 3, 5)\\⇒\frac{x+1}{2}=1,\frac{y+6}{2}=3,\frac{z+3}{2}=5\\⇒ x = 2 – 1, y = 6 – 6, z = 10 – 3\\⇒ x = 1, y = 0, z = 7\\\text{Coordinates of T} = (x, y, z) = (1, 0, 7)\\\text{Hence, coordinates of image of point}\\P(1, 6, 3) = T(1, 0, 7).$$

## Section-E

36. (i) Direction ratios of Z-axis = ( 0, 0, a).
(ii) Cartesian equation of side BN where B(0, a, 0), N(0, a, a) is
$$=\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\\=\frac{x-0}{0-0}=\frac{y-0}{0-0}=\frac{z-0}{0-0}\\=\frac{x}{0}=\frac{y}{0}=\frac{z}{0}\\\text{(iii) Directions ratios of diagonal}\space \vec{AN}=\vec{ON}-\vec{OA}\\\vec{ON}=0\hat{i}+a\hat{j}+a\hat{k}\\\vec{OA}=a\hat{i}+0\hat{j}+0\hat{k}\\\vec{AN}=(x_2 − x_1)\hat{i}+(y_2 − y_1)\hat{j}+(z_2 − z_1)\hat{k}\\=(0 − a)\hat{i}+(a − 0)\hat{j}+(a − 0)\hat{k}\\-a\hat{i}+a\hat{j}+a\hat{k}\\d.r.’s = 〈 – a, a, a 〉\\d.c.’s =\frac{-a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}},\frac{a}{\sqrt{a^2+a^2+a^2}}\\=(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$$

OR

Coordinates of diagonal MC
M = (a, a, 0), C = (0, 0, a)
$$d.r.’s of MC=(0 − a)\hat{i}+(a − 0)\hat{j}+(a − 0)\hat{k}\\-a\hat{i}+a\hat{j}+a\hat{k}\\d.c.’s of MC =\frac{-a}{\sqrt{(-a^2)+(-a^2)+a^2}},\frac{-a}{\sqrt{(-a^2)+(-a^2)+a^2}},\frac{a}{\sqrt{(-a^2)+(-a^2)+a^2}}\\=(\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}})$$

37. (i)
$$\vec{OA}=2\hat{i}-\hat{j}+\hat{k}\\|\vec{OA}|=\sqrt{2^2+(-1)^2+1^2}\\=\sqrt{4+1+1}=\sqrt{6}\space\text{unit}$$

$$(ii)\space\vec{OB}=\hat{i}-3\hat{j}-5\hat{k}\\\text{Distance covered by Rahul is}\\|\vec{OB}|=\sqrt{1^2+(-3)^2+(-5)^2}\\=\sqrt{1+9+25}=\sqrt{35}\space\text{unit}\\\text{(iii) Distance between Raja’s office and Rahul’s client office will be}|\vec{AB}|\\\vec{AB}=\vec{OB}-\vec{OA}=(\hat{i}-3\hat{j}-5\hat{k})-(2\hat{i}-\hat{j}+\hat{k})\\|\vec{AB}|=\sqrt{(-1)+(-2)^2+(-6)^2}\\=\sqrt{1+4+36}=\sqrt{41}\text{unit}.$$

OR

$$\vec{OC}=(x\hat{i}+y\hat{j}+z\hat{k})\\\text{Which divides BA in the ratio 2 : 1. By section formula, coordinates of C will be :}\\x=\frac{2(2)+1}{3},y=\frac{2(-1)+1(-3)}{2+1},x=\frac{2(1)+1(5)}{2+1}\\x=\frac{4+1}{3}=\frac{5}{3},y=\frac{-2-3}{3}=\frac{-5}{3}\\\frac{2-5}{3}=\frac{-3}{3}=1\\\text{Coordinates of C}(\frac{5}{3},\frac{-5}{3},-1)$$

38. (i) The total requirement of each material is given by A'B.

$$A′=\begin{bmatrix}2&4&2\\3&2&4\\1&5&2\end{bmatrix}$$
$$A′B=\begin{bmatrix}2&4&2\\3&2&4\\1&5&2\end{bmatrix}\begin{bmatrix}100\\100\\100\end{bmatrix}=\begin{bmatrix}800\\900\\800\end{bmatrix}$$
$$(ii)AC=\begin{bmatrix}2&3&1\\4&2&5\\2&4&2\end{bmatrix}\begin{bmatrix}5\\10\\5\end{bmatrix}=\begin{bmatrix}45\\65\\60\end{bmatrix}$$

#### CBSE Practice Paper Mathematics Class 12

All Practice Paper for Class 12 Exam 2024

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