Oswal Practice Papers CBSE Class 12 Mathematics Solutions (Practice Paper - 10)

Section-A

1. (a) 20

Explanation :

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

2. (a) – 80

Explanation :

Δ = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃
= a₁₁M₁₁ – a₁₂M₁₂ + a₁₃ M₁₃
= (1) (– 40) – 3(–10) + (–2) (35)
= – 40 + 30 – 70 = – 80

3. (a) – cot y cosec²y

Explanation :

Given, y = cos^–1 x
⇒ x = cos y
On differentiating w.r.t. y, we get
$$\frac{dx}{dy}=-sin\space y\\or\space\frac{dy}{dx}= – cosec\space y…(i)\\\text{On again differentiating w.r.t. x, we get}\\\frac{d^2y}{dx^2}=\frac{d}{dx}(– cosec\space y) = – (– cosec\space y\space cot\space y)\frac{dy}{dx}\\=cosec\space y\space cot\space y\space (– cosec\space y) = – cot\space y\space cosec^2\space y\space [from (i)]$$

4. (c)
$$\hat{i}.\hat{k}=0$$

Explanation :

$$As\space\hat{i}.\hat{k}=|\hat{i}|.|\hat{k}|cos\frac{\pi}{2}= 1 × 1× 0 = 0$$

5. (d)
$$\frac{tan4x}{8}+C$$

Explanation :

$$I=\int\frac{1}{1+cos8x}dx=\int\frac{1}{1-\frac{(1-tan^24x)}{(1+tan^24x)}}dx\\=\int\frac{(1+tan^24x)}{1+tan^24x+1-tan^24x}dx\\=\int\frac{1+tan^24x}{2}dx\\=\frac{1}{2}\int sec^24xdx=\frac{tan4x}{8}+C$$

6. (d) infinite

Explanation :

Since, the objective functon has the same maximum value on two corner points, therefore every point on the line segment joining these two points also gives the maximum value. Hence, the number of points at which Zmax occurs is infinite.

7. (a)
$$2|\vec{a}|^2-9\vec{a}.\vec{b}-5|\vec{b}|^2$$

Explanation :

$$(\vec{a}-5\vec{b}).(2\vec{a}+\vec{b})=\vec{a}.2+\vec{a}.\vec{b}-5\vec{b}.2\vec{a}-5\vec{b}.\vec{b}\\=2\vec{a}.\vec{a}+\vec{a}.\vec{b}-10\vec{a}.\vec{b}-5\vec{b}.\vec{b}\\=2|\vec{a}|^2-9\vec{a}.\vec{b}-5|\vec{b}|^2$$

8. (b) 20

Explanation :

$$\int^{4}_{-4}|x+2|dx=\int^{-2}_{-4}-(x+2)dx+\int^{4}_{-2}(x+2)dx\\=\begin{bmatrix}\frac{-x^2}{2}-2x\end{bmatrix}^{-2}_{-4}+\begin{bmatrix}\frac{x^2}{2}+2x\end{bmatrix}^{-2}_{-4}\\=\begin{bmatrix}\frac{-4}{2}+4+\frac{16}{2}-8\end{bmatrix}+\begin{bmatrix}\frac{16}{2}+8-\frac{4}{2}+8\end{bmatrix}\\= 2 + 18 = 20$$

9. (b) – 2

Explanation :

$$[x –1]\begin{bmatrix}1&0\\-2&0\end{bmatrix}=0\\⇒{[x +2] = 0}\\⇒ x + 2 = 0\\⇒ x = – 2$$

10. (d) Feasible region

Explanation :

The region of feasible solution in the graphical method of LPP is called feasible region.

11. (d) – 7000

Explanation :

|AB| = |A| |B|
= (400 – 500) (150 – 80) = – 100 × 70 = – 7000

12. (d) c(a² – b²)

Explanation :

$$M_{31} =\begin{bmatrix}a&bc\\b&ca\end{bmatrix}=ca^2 – b^2c = c (a^2 – b^2)$$

13. (a) 0

Explanation :

Number of arbitary constants in particular solution of differential equaton is 0.

14. (a) x = 0

Explanation :

$$\lim\limits_{x\rightarrow0^{-}}f(x)=1=\lim\limits_{x\rightarrow 0^{+}}f(x)=1≠f(0)=2\\\text{\f(x)is not continuous at x = 0}$$

15. (b)3/4

Explanation :

$$\text{Let A = Numbers multiple of 2}\\n(A) = 30\\P(A) =\frac{30}{60}\\\text{and B = Numbers multiple of 5}\\n(B) = 12\\P(B) =\frac{12}{60}\\P(A∪B) = P(A) + P(B) – P(A ∩ B)\\\frac{30}{60}+\frac{12}{60}-\frac{6}{60}=\frac{36}{60}=\frac{3}{5}$$

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

Explanation :

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

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

Explanation :

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

18. (b) 2

Explanation :

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

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

Explanation :

$$\text{Period of sin x is}\frac{2\pi}{1}\text{and period of cos ax =}\frac{2\pi}{a}\\\text{⇒ Hence, period of f(x) =}(\frac{2\pi}{1},\frac{2\pi}{a})=\frac{LCM of (2\pi,2\pi)}{HCF of(1,a)}=\frac{2\pi}{k}\\\text{where, k is the HCF of 1 and a.}\\⇒\frac{1}{k}= integer = q (say) (≠ 0)\\and\frac{1}{k}= integer = p (say)\\\frac{a/k}{1/k}=\frac{p}{q}\\⇒ a =\frac{p}{q}\\⇒\text{a is rational}$$

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

Explanation :

$$\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AC}-\vec{AC}=0$$

Section-B

21. Here, f : A → B is defined as {(1, 4) (2, 5) (3, 6)}.
Since, the image of distinct elements of A under f are distinct as:
f (1) = 4, f(2) = 5 and f (3) = 6
From above it is evident that x1 ≠ x2 and f(x1) ≠ (x2)
f : A → B is one one.

$$\text{Here,}\space f (x) =\frac{|x-1|}{x-1}, x ≠ 1\\⇒ f(x) =\frac{(x-1)}{(x-1)}\text{∀ then x > 1}\\and\space f(x) =\frac{-(x-1)}{x-1}\text{∀ then x > 1}\\\text{i.e., The range of f(x) is {1, – 1}.}$$

22.We have, f(x) = – |x – 1| + 5 for x ∈ R
Clearly, |x – 1| ≥ 0 for all x ∈ R
Taking minus sign both sides, we get
– |x – 1| ≤ 0 for all x∈R
– |x – 1| + 5 ≤ 5 for all x∈R
⇒ f (x) ≤ 5 for all x∈R

So, 5 is the maximum value of f (x)
Now, f (x) = 5
⇒
– |x – 1| + 5 = 5
⇒ |x – 1| = 0 ⇒ x = 1
Thus, f (x) attains the maximum value 5 at x = 1.
Since, f(x) can be made as small as possible.Therefore, the minimum value of f(1) does not exist.

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

Let l, m, n be the direction cosines of the line then
l = cos α, m = cos β, n = cos γ
Since, l² + m² + n² = 1
⇒ cos²α + cos²β + cos² γ = 1
⇒ 1 – sin²α+ 1 – sin²β + 1– sin² γ = 1
⇒ 3 – (sin²α + sin²β + sin² γ) = 1
⇒ sin²α + sin²β + sin² γ = 3 – 1
⇒ sin²α + sin²β+ sin² γ = 2

24. Given, differential equation is:
$$\frac{dy}{dx}=(\frac{1-cosx}{1+cosx})\\⇒{dy}=(\frac{1-cosx}{1+cosx}).{dx}\\⇒dy =(\frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}})dx\\⇒ dy =(tan^2\frac{x}{2})dx\\⇒ dy =(sec^2\frac{x}{2}-1)dx\\\text{On integrating both sides, we get}\\y=\int(sec^2\frac{x}{2}-1)dx+C\\\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+C\\⇒ y =2tan\frac{x}{2}-x+C$$

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

Section-C

26. Let

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

27.

x + y = 50

x	50	0
y	0	50

3x + y = 90

x	0	30
y	90	0

x = 10 is a line which is parallel to y-axis.

Corner Points	z = 4x + y
A (30,0)	z = 120 + 0 = 120 → Max.
B (20,30)	z = 80 + 30 = 110
C (10,40)	z = 40 + 40 = 80
D (10,0)	z = 40 + 0 = 40

Maximum value is 120 at (30, 0).

28. Clearly, f (x) = x³ – x = x (x – 1) (x + 1). The sign of f (x) for different values of x are given in figure.

We observe that,
f (x) > 0 for all x ∈ (–1, 0) ∪ (1, 2)
f (x) < 0 for all x ∈ (0, 1)
$$\int^{2}_{-1}|x^3-x|dx=\int^{0}_{-1}(x^3-x)dx+\int^{1}_{0}-(x^3-x)dx+\int^{2}_{1}(x^3-x)dx\\=[\frac{x^4}{4}-\frac{x^2}{2}]^{0}_{-1}+[\frac{-x^4}{4}+\frac{x^2}{2}]^{1}_{0}+[\frac{x^4}{4}+\frac{x^2}{2}]^{2}_{1}\\=[(0-0)-(\frac{1}{4}-\frac{1}{2})]+[(\frac{-1}{4}+\frac{1}{2})-0]+\frac{1}{2})]+[(\frac{16}{4}-\frac{4}{2})-(\frac{1}{4}-\frac{4}{2})]\\=-\frac{1}{4}+\frac{1}{2}-\frac{1}{4}+\frac{1}{2}+4-2-\frac{1}{4}+\frac{1}{2}\\=\frac{-1+2-1+2+8-1+2}{4}\\=\frac{11}{4}$$

$$Let\space I=\int^{\pi/4}_{0}\frac{dx}{cos^3x\sqrt{sin2x}}\\=\int^{\pi/4}_{0}\frac{dx}{cos^4x\sqrt{\frac{2sin2x}{cos^2x}}}\\=\int^{\pi/4}_{0}\frac{sec^4xdx}{\sqrt{\frac{2sin2x}{cos^2x}}}\\=\int^{\pi/4}_{0}\frac{(1+tan^2x)sec^2xdx}{2\sqrt{\frac{sinx\space cosx}{cos^2x}}}\\{[... sin 2x = 2 sin x.cos x]}\\=\int^{\pi/4}_{0}\frac{(1+tan^2x)sec^2xdx}{2\sqrt{tanx}}dx\\\text{Put tan x = t}\\sec^2x dx = dt\\\text{When x = 0, t = tan 0 = 0}\\When x =\frac{\pi}{4},t=\frac{\pi}{4}=1\\I=\int^{1}_{0}\frac{(1+t^2)dt}{2\sqrt{t}}\\=\frac{1}{2}\int^{1}_{0}(\frac{1}{\sqrt{t}}+t^{3/2})dt\\=\frac{1}{2}[2\sqrt{t}+\frac{2}{5}t^{5/2}]^{1}_{0}\\=[\sqrt{t}+\frac{2}{5}t^{5/2}]^{1}_{0}\\=[\sqrt{1}+\frac{2}{5}(1)^{5/2}]-{0}\\=\frac{6}{5}$$

29. Given, differential equation is:
$$xlog\space x\frac{dy}{dx}+y=2logx\\⇒\frac{dy}{dx}+\frac{dy}{xlogx}=\frac{2}{x}\\Here,P=\frac{1}{xlogx},Q=\frac{2}{x}\\I.F.=e^{\int px}=e^{\frac{1}{xlogx}dx}\\e^{\int}(\frac{\frac{1}{x}}{logx})dx\\=e^{log(log x)}=log x\\\text{Solution of given equation will be given by:}\\y.I.F. =\int Q × I.F. dx + C\\⇒ylog\space x =2\int logx.\frac{1}{x}+C\\⇒ y\space logx = 2\frac{(log\space x)^2}{2}+C\\y=log\space x+\frac{C}{log\space x}$$

$$\text{We have}\space y+x\frac{dy}{dx}=x-y\frac{dy}{dx}\\⇒x\frac{dy}{dx}+y\frac{dy}{dx}=x-y\\⇒\frac{dy}{dx}=\frac{x-y}{x+y}…(i)\\\text{Which is a homogeneous equation}\\\text{put y = vx ⇒}\frac{dy}{dx}=v+x\frac{dv}{dx}\text{in (i), we get}\\v + x\frac{dv}{dx}=\frac{x-vx}{x+vx}\\⇒ x\frac{dv}{dx}=\frac{1-v}{1+v}-v\\=\frac{1-v-v-v^2}{1+v}\\⇒ x\frac{dv}{dx}=\frac{1-2v-v^2}{1+v}\\⇒\frac{1+v}{1-2v-v^2}dv =\frac{dx}{x}, x ≠ 0…(ii)\\\text{Putting}\space t=1–2v–v^2\\dt=(–2-2v)dv\\\frac{-1}{2}dt=(1+v)dv\\\text{Now equation (i) becomes}\\\frac{-1}{2}dt=\frac{dx}{x}\\\text{On integrating above equation, we get}\\\frac{-1}{2}\int\frac{1}{t}dt=\int\frac{1}{x}dx\\⇒\frac{-1}{2}log|1–2v-v^2|=log (Cx)\\⇒log {C|x|}+log|1–2v–v^2|^{1/2}=0\\⇒C|x|(1-\frac{2y}{x}-\frac{y^2}{x^2})^{1/2}=0\\⇒C(x^2–2xy–y^2)^{1/2}=0$$

30. Let E₁ and E₂ be the events that the set-up is correct and incorrect respectively. Let E be the event that the items produces are good one.
$$\text{Given that,}\space\frac{\text{Setting is correct}}{\text{Setting is incorrect}}=\frac{7}{3}\\P(E_1) =\frac{7}{10}\\and P(E_2) =\frac{3}{10}\\10\%\text{defective items means}\space90\% \text{good items}\\P(\frac{E}{E_1})=\frac{90}{100}=\frac{9}{10}\\and\space P(\frac{E}{E_2})=\frac{10}{100}=\frac{1}{10}\\\text{\ By Bayes’ theorem,}\\P(\frac{E_1}{E})=\frac{P(E_1).P(\frac{E}{E_1})}{P(E_1).P(\frac{E}{E_1})+P(E_2).P(\frac{E}{E_2})}\\\frac{\frac{7}{9}×\frac{9}{10}}{\frac{7}{10}×\frac{9}{10}+\frac{3}{10}×\frac{1}{10}}\\=\frac{\frac{63}{100}}{\frac{66}{100}}=\frac{63}{66}=\frac{21}{22}$$

Let M = Event that a person chosen at random is a male.
F = Event that a person chosen at random is a female.
C = Event that a person chosen at random is colourblind.
$$\text{Required probability =}p(\frac{M}{C})\text{given that P(M) = P(F) =}\frac{1}{2}\text{(equal proportion)}\\Now,P(\frac{C}{M})=\frac{5}{100},P(\frac{C}{F})=\frac{25}{1000}\\\text{By Baye’s theorem}\\P(\frac{M}{C})=\frac{P(\frac{c/f}{M})P(M)}{P(\frac{C}{M})P(M)+P(\frac{C}{F})P(F)}\\=\frac{\frac{5}{100}×\frac{1}{2}}{\frac{5}{100}×\frac{1}{2}+\frac{25}{10000}×\frac{1}{2}}\\=\frac{\frac{5}{200}}{\frac{5}{200}+\frac{5}{200}}=\frac{2}{3}$$

31. Let.
$$I=\int^{\pi}_{0}\frac{x}{1+sinx}dx\\⇒I=\int^{\pi}_{0}\frac{\pi-x}{1+sin{(\pi-x})}dx\space[...\int^{a}_{0}f(a-x)dx]\\⇒I=\int^{\pi}_{0}\frac{\pi}{1+sinx}dx-\int^{\pi}_{0}\frac{x}{1+sinx}dx\\⇒\int^{\pi}_{0}\frac{\pi}{1+sinx}dx-I\\⇒ 2I =\pi\int^{\pi}_{0}\frac{1}{1+sinx}×\frac{1-sinx}{1-sinx}dx\\=\pi[\int^{\pi}_{0}sec^2x-x\int secx\space tanx]d=\pi[tanx]^{\pi}_{0}-[secx]^{\pi}_{0}\\=\pi[(tan\pi–tan\theta)–(sec\pi–sec\theta)]\\⇒I=\frac{\pi}{2}[(0–0)–(– 1– 1)]\\=\frac{\pi}{2}[0–(-2)]=\frac{\pi}{2}×2=\pi$$

Section-D

32. Given, x = y² is a parabola symmetric to positive x-axis with vertex (0, 0).

Equation of the side AB is
$$\text{We have given that,}\space I=\int^{a}_{0}\sqrt{x}dx=\int^{4}_{a}\sqrt{x}dx\\\begin{bmatrix}\frac{x^{3/2}}{\frac{3}{2}}\end{bmatrix}^{a}_{0}=\begin{bmatrix}\frac{x^{3/2}}{\frac{3}{2}}\end{bmatrix}^{4}_{a}\\⇒\frac{2}{3}a^{3/2}=\frac{2}{3}(4^{3/2}-a^{3/2})\\⇒\frac{2}{3}a^{3/2}+\frac{2}{3}a^{3/2}=\frac{2}{3}×8\\⇒\frac{4}{3}a^{3/2}=\frac{8×2}{3}\\⇒a\sqrt{a}=\frac{16}{3}×\frac{3}{4}\\⇒ a^{3/2} = 4\\⇒a = 4^{2/3} =(2^2)^{2/3}\\= 2^{4/3} =\sqrt[3]{16}$$

33.Give, R = {(P₁, P₂) : P1 and P2 have same number of sides}
Reflexivity : Let P be any polygon in A
Then P and P have same number of sides
⇒ (P,P) ∈ R
Then (P, P) ∈ R ∀ P ∈ A
So, R is reflexive.
Symmetric : Let P₁ and P₂ be two polygons A in such that (P₁, P₂) ∈R
Now, P₁ and P₂ have same number of sides
⇒ P₂ and P₁ have same number of sides
⇒ (P₂ , P₁) ∈ R
So, R is symmetric on A
Transitivity: Let P₁, P₂ and P₃ be three polygons in A such that (P₁, P₂) ∈ R and (P₂, P₃) ∈ R.
Then, (P₁, P₂) ∈ R
⇒ P₁ and P₂ have same number of sides
(P₂, P₃) ∈ R
⇒ P₂ and P₃have same number of sides
∴ P₁
and P₃ have same number of sides
⇒ (P₁, P₃) ∈ R
Thus, if (P₁, P₂) ∈ R and (P₂, P₃) ∈ R
⇒ (P₁, P₃) ∈ R
So, R is transitive.
Hence, R is an equivalence relation on A.
Let P be a polygon in A such that (P, T)∈R. Then, polygon P and triangle T have same number of sides.
Thus, P is any triangle in A with sides 3, 4 and 5.

$$Let\space\text{x = cos}^{–1}\frac{4}{5}\text{and cos y =}\space\frac{12}{13}\\\sin x=\sqrt{1-cos^2x}\\=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\\\text{and sin y =}\sqrt{1-cos^2y}\\=\sqrt{1-\frac{144}{169}}\\=\frac{5}{13}\\\text{Now, using the identity}\\\text{cos (x + y) = cos x cos y – sin x sin y}\\\text{We get, cos} (x + y) =(\frac{4}{5}×\frac{12}{13})-(\frac{3}{5}×\frac{5}{13})\\⇒ cos (x + y) =\frac{48}{65}-\frac{15}{65}\\⇒ cos (x + y) =\frac{33}{65}\\or x + y = cos^{−1}\frac{33}{65}\\cos^{-1}\frac{4}{5}+cos^{-1}\frac{12}{13}=cos^{-1}\frac{33}{65}[...x=cos^{-1}\frac{4}{5}\space and\space y =cos^{-1}\frac{12}{13}]$$

34. Given
$$A=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\\A^2=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}\\A^2–5A–14I =\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-5\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}\\=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}+\begin{bmatrix}-15&25\\20&-10\end{bmatrix}+\begin{bmatrix}-14&0\\0&-14\end{bmatrix}\\=\begin{bmatrix}29-15-14&-25+25+0\\-20+20+0&24-10-14\end{bmatrix}\\=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0\\A^2 – 5A – 14I = 0\\\text{On premultiplying by} A^{–1},\text{we get}\\A^{–1}(A^2) – 5A^{–1}A –14A^{–1} I = 0\\⇒ A – 5I – 14A^{–1} = 0\\⇒ A^{–1} =\frac{1}{14}(A-5I)\\=\frac{1}{14}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-5\begin{bmatrix}1&0\\00&1\end{bmatrix}\\=\frac{1}{14}\begin{bmatrix}-2&-5\\-4&-3\end{bmatrix}$$

35. Given lines are,
l – 5m + 3n = 0 …(i)
7l² + 5m² – 3n² = 0 …(ii)
Putting l = (5m – 3n) from (i) into (ii), we get
7 (5m – 3n)² + 5m² – 3n² = 0
⇒ 175 m² + 63n² – 210 mn + 5m² – 3n² = 0
⇒ 180m² + 60n² – 210 mn = 0
6m² – 7mn + 2n² = 0
$$⇒(\frac{m}{n})^2-7(\frac{m}{n})+2=0\\\text{Let,}p=\frac{m}{n}\\⇒ 6P^2–7P+2=0\\or(3P–2)(2P–1)=0\\P=\frac{2}{3},\frac{1}{2}\\So,\frac{m}{n}=\frac{2}{3}or\frac{m}{n}=\frac{1}{2}(since,P=\frac{m}{n})\\\text{Now,}\frac{m}{n}=\frac{2}{3}\\⇒\frac{m}{n}=\frac{n}{3}\\⇒\frac{5m-3n}{5×2-3×3}=\frac{l}{1}\\Thus,\frac{l}{1}=\frac{m}{2}=\frac{n}{2}=\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{1^2+2^2+3^2}}\\=\frac{3}{\sqrt{14}}\\Again,\frac{m}{n}=\frac{1}{2}⇒\frac{m}{1}=\frac{n}{2}\\⇒\frac{5m-3n}{5×1-3×2}=\frac{l}{-1}\\\frac{l}{-1}=\frac{m}{1}=\frac{n}{2}=\frac{\sqrt{l^2+m^2+n^2}}{(-1)^2+1^2+2^2}=\frac{1}{\sqrt{6}}\\So, l =\frac{-1}{\sqrt{6}},m=\frac{1}{\sqrt{6}},n=\frac{2}{\sqrt{6}}\\\text{Hence, d.c.’s of the lines are}(\frac{1}{\sqrt{6}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}})\space and\space(\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$$

Cartesian equation of the line passing through (1, 2, –4) is:
$$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z+4}{c}…(i)\\\text{Given lines are}\\\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}…(ii)\\\text{and}\space\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}…(iii)\\Let\space\vec{b_1},\vec{b_2},\vec{b_3}\text{are parallel vectors of (i), (ii), (iii) respectively,}\\\vec{b_1}=a\hat{i}+b\hat{j}+c\hat{k}\\\vec{b_2}=3\hat{i}-16\hat{j}+7\hat{k}\\\vec{b_3}=3\hat{i}+8\hat{j}-5\hat{k}\\\text{Given that (i) is perpendicular to both (ii) and (iii)}\\\vec{b_1}.\vec{b_2}=0\\⇒ 3a – 16b + 7c = 0 …(iv)\\and\space\vec{b_1}.\vec{b_3}=0\\⇒ 3a + 8b – 5c = 0 …(v)\\\text{From (iv) and (v)}…(v)\\\frac{a}{80-56}=\frac{b}{21+15}=\frac{c}{24+48}…(i)\\⇒\frac{a}{24}=\frac{b}{36}=\frac{c}{72}\\⇒\frac{a}{2}=\frac{b}{3}=\frac{c}{6}=\lambda\\⇒ a=2\lambda,b=3\lambda,c=6\lambda\\\text{Putting in equation (i)}\frac{x-1}{2\lambda}=\frac{y-2}{3\lambda}=\frac{z+4}{6\lambda}\\⇒\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\\\text{Which is the required cartesian equation of line and vector equation of this line is:}\\\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda(2\hat{i}+3\hat{j}+6\hat{k})$$

Section-E

36. (i) Let f(x) = x³ sin⁴⁰ x
f(x) = (– x)³sin⁴⁰ (– x)
= – x³ (– sin x)⁴⁰
= – x³ sin⁴⁰ x
= – f(x)
⇒ f (x) is an odd function.
$$(ii) Let I =\int^{\pi/4}_{-\pi/4}sec^2xdx\\\text{Consider}\\f(x) = sec^2x\\f(–x) = sec^2(–x)\\= sec^2x = f(x)\\\text{⇒ f(x) is an even function.}\\(iii) Let f(x) = x^3 + x cos x + tan^5 x\\Then, f (–x) = (–x)^3 + (–x) cos (–x) + tan^5 (–x)\\= –x^3 – xcos x – tan^5x = – f(x)\\⇒ \text{f(x) is an odd function}\\Now\space\int^{\pi/2}_{-\pi/2}(x^3+xcosx+tan^5x+1)dx=\int^{\pi/2}_{-\pi/2}(x^3+xcosx+tan^5x)dx=\int^{\pi/2}_{-\pi/2}(1).dx\\= 0 +[x]^{\pi/2}_{-\pi/2}=(\frac{\pi}{2}+\frac{\pi}{2})=\pi$$

We know that,
$$|x + 2|=\begin{cases}x+2,&\text{When x+2 ≥ 0 or x ≥ -2}\\-(x+2)&\text{When x+2 < 0 or x < -2}\end{cases}\\So, \int^{5}_{-5}|x + 2|dx=\int^{-2}_{-5}|x + 2|dx+\int^{5}_{-2}|x + 2|dx\\=\int^{-2}_{-5}(x + 2)dx+\int^{5}_{-2}(x + 2)dx\\\begin{bmatrix}\frac{x^2}{2}+2x\end{bmatrix}^{5}_{-2}-\begin{bmatrix}\frac{x^2}{2}+2x\end{bmatrix}^{-2}_{5}\\=\begin{bmatrix}\frac{25}{2}+10-2+4\end{bmatrix}-\begin{bmatrix}-2+4-\frac{25}{2}+10\end{bmatrix}=29$$

37. (i) Let C be the cost of the fuel consumed by running the train for time t.
$$Then,\space\frac{C}{t}=kv^2\\Also\space 48=k(16)^2\\3=16k\\\text{(ii) Let C' be the total cost of running the train.}\\A.T.Q.,\frac{C'}{t}= kv^2 + 1200\\\frac{C'}{t}=\frac{3}{16}v^2t+1200t[from (i)]\\\text{... Train travelled a distance of 1000 km.}\\v=\frac{1000}{t}\\or\space t=\frac{1000}{v}\\C^{'}=\frac{3}{16}v^2(\frac{1000}{v})+1200(\frac{1000}{v})\\=\frac{375}{2}v+\frac{1200000}{v}\\(iii) From (ii)\space C^′=\frac{375}{2}v+\frac{1200000}{v}\\\frac{dC'}{dv}=\frac{375}{2}+1200000(\frac{-1}{v^2})\\\text{To find economical speed,}\frac{dC'}{dv}=0\\\text{Put}\frac{375}{2}-\frac{1200000}{v^2}=0\\⇒ v^2 =\frac{2}{375}×1200000\\== 2× 3200 = 6400\\v = 80 km/hr\\\text{So, The most economical speed is 80 km/hr.}$$

$$Let y =\frac{1-x+x^2}{1-x+x^2}, x ∈R.\\Then, x^2 (y – 1) + x(y + 1) + y – 1 = 0 …(i)\\\text{Since, x is real, discriminant of equation (i) should be greater than or equal to zero.}\\⇒(y + 1)^2 – 4(y – 1) (y – 1) ≥ 0\\i.e., – 3y^2 + 10y – 3 ≥ 0\\⇒ –3(y – 3)(y-\frac{1}{3})≥0\\⇒ (y – 3)(y-\frac{1}{3})≤0\\\frac{1}{3}≤ y≤ 3\\\text{Hence, minimum-value of y is}\frac{1}{3}$$

38. (i)
$$\displaystyle\sum_{k=1}^2P(\frac{E}{E_k})P(E_k)=P(\frac{E}{E_1})p(E_1)+P(\frac{E}{E_2})P(E_2)\\=1×\frac{3}{5}+\frac{3}{5}×\frac{2}{5}=\frac{3}{5}+\frac{2}{15}\\=\frac{9+2}{15}=\frac{11}{5}\\\text{(ii) Required probability =}P(\frac{E_1}{E})\\=\frac{P(\frac{E_1}{E})×P(E_1)}{P(\frac{E_1}{E})×P(E_1)+P(\frac{E}{E_2})×P(E_2)}\\=\frac{1×\frac{3}{5}}{1×\frac{3}{5}+\frac{2}{5}×\frac{1}{5}}\\=\frac{\frac{3}{5}}{\frac{3}{5}+\frac{2}{15}}\\=\frac{\frac{3}{5}}{\frac{9+2}{15}}\\=\frac{\frac{3}{5}}{\frac{11}{15}}\\=\frac{9}{11}$$