12th Standard English Medium Maths Reduced Syllabus Annual Exam Model Question Paper - 2021

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Questions:

Generic 5

12 x 5 = 60

Solve :\(x\frac { dy }{ dx } sin\left( \frac { y }{ x } \right) +x-ysin\left( y\frac { y }{ x } \right) =,y(1)=\frac { \pi }{ 2 } \)

Solve :x²dy+y(x+y)dx=0 given that y=1 when x=1.

Find the area bounded by the curve y=xe^x and y=xe^-xand the line x=1.

Find the area bounded by x=at²,y=2at between the ordinates corresponding to t = 1 and t = 2.

A manufacturer can sell x items at a price of rupees \(\left( 5-\frac { x }{ 100 } \right) \) each. The cost price of x items is Rs.\(\left( \frac { x }{ 5 } +500 \right) \) .Find the numbers of items he should sell to earn maximum profit.

Solve: \(\frac { dy }{ dx } \) = (3x+2y+1)²

The slope of the tangent at p(x,y) on the curve is -\(\left( \frac { y+3 }{ x+2 } \right) \). If the curve passes through the origin, find the equation of the curve.

Show that the area under the curve y = sin x and y = sin 2x between x = 0 and x = \(\frac { \pi }{ 3 } \) and x axis are as 2:3

If \(\left| \overset { \rightarrow }{ A } \right| =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and \(\overset { \wedge }{ i } =\overset { \wedge }{ j } -\overset { \wedge }{ k } \) are two given vector, then find a vector B satisfying the equations \(\overset { \rightarrow }{ A } \times \overset { \rightarrow }{ B } \)= \(\overset { \rightarrow }{ C } \) and \(\overset { \rightarrow }{ A } \).\(\overset { \rightarrow }{ B } \) = 3

10.

If A = \(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \), show that A^-1 = \(\frac {1}{2}\) (A² - 3I).

11.

Find the domain of f(x) = sin^-1 \((\frac{|x|-2}{3})+ \) cos^-1\((\frac{1-|x|}{4})\)

12.

If p and q are the roots of the equation I x²+ nx + n = 0, show that \(\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{i}}\) = 0

Generic 3

6 x 3 = 18

13.

Find the domain of
g(x) = sin⁻¹x + cos⁻¹x

14.

Find the adjoint of the following:
\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)

15.

A room 34m long is constructed to be a whispering gallery. The room has an elliptical ceiling, as shown in Figure. If the maximum height of the ceiling is 8 m, determine where the foci are located.

16.

Find the equation of the ellipse with foci (±2, 0), vertices (±3, 0)

17.

If A = \(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \), verify that (AB)^-1 = B^-1A^-1

18.

The complex numbers u, v, and w are related by \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \) If v = 3−4i and w = 4+3i, find u in rectangular form.

Multiple Choice Question

20 x 1 = 20

19.

p	q	(p ∧ q) ⟶ ¬q
T	T	(a)
T	F	(b)
F	T	(c)
F	F	(d)

Which one of the following is correct for the truth value of (p ∧ q)⟶ ¬p?

(a)

(a)	(b)	(c)	(d)
T	T	T	T

(b)

(a)	(b)	(c)	(d)
F	T	T	T

(c)

(a)	(b)	(c)	(d)
F	F	T	T

(d)

(a)	(b)	(c)	(d)
T	T	T	F

20.

If \(\frac{\Gamma(n+2)}{\Gamma(n)}=90\) then n is

(a)

(b)

(c)

(d)

21.

The approximate change in the volume V of a cube of side x metres caused by increasing the side by 1% is

(a)

0.3xdx m³

(b)

0.03x m³

(c)

0.03x² m³

(d)

0.03x³m³

22.

The change in the surface area S = 6x² of a cube when the edge length varies from x_o to x_o+ dx is

(a)

12 x_o+dx

(b)

12x_o dx

(c)

6x_o dx

(d)

6x_o+ dx

23.

If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is

(a)

0.4 cu.cm

(b)

0.45 cu.cm

(c)

2 cu.cm

(d)

4.8 cu.cm

24.

Let X have a Bernoulli distribution with mean 0.4, then the variance of (2X - 3) is

(a)

0.24

(b)

0.48

(c)

0.6

(d)

0.96

25.

The solution of \(\frac { dy }{ dx } ={ 2 }^{ y-x }\) is

(a)

2^x+ 2^y= C

(b)

2^x- 2^y= C

(c)

\(\frac { 1 }{ { 2 }^{ x } } -\frac { 1 }{ { 2 }^{ y } } =C\)

(d)

x + y = C

26.

The order and degree of the differential equation \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 1/3 }+{ x }^{ 1/4 }=0\) are respectively

(a)

2, 3

(b)

3, 3

(c)

2, 6

(d)

2, 4

27.

The minimum value of the function |3 - x| + 9 is

(a)

(b)

(c)

(d)

28.

The number given by the Mean value theorem for the function \(\frac { 1 }{ x } \), x ∈ [1, 9] is

(a)

(b)

2.5

(c)

(d)

3.5

29.

A balloon rises straight up at 10 m/s. An observer is 40 m away from the spot where the balloon left the ground. The rate of change of the balloon's angle of elevation in radian per second when the balloon is 30 metres above the ground.

(a)

\(\frac{3}{25} \text { radians } / \mathrm{sec}\)

(b)

\(\frac{4}{25} \text { radians } / \mathrm{sec}\)

(c)

\(\frac{1}{5} \text { radians } / \mathrm{sec}\)

(d)

\(\frac{1}{3} \text { radians } / \mathrm{sec}\)

30.

If A^T is the transpose of a square matrix A, then ___________

(a)

|A| ≠ |A^T|

(b)

|A| = |A^T|

(c)

|A| + |A^T| =0

(d)

|A| = |A^T| only

31.

If A, B and C are invertible matrices of some order, then which one of the following is not true?

(a)

adj A = |A|A^-1

(b)

adj(AB) = (adj A)(adj B)

(c)

det A^-1 = (det A)^-1

(d)

(ABC)^-1 = C^-1B^-1A^-1

32.

If A = \(\left[ \begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix} \right] \), then 9I₂ - A =

(a)

A^-1

(b)

\(\frac { { A }^{ -1 } }{ 2 } \)

(c)

3A^-1

(d)

2A^-1

33.

If \(\vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } \times \vec { b } )\times \vec { c } \) where \(\vec { a } ,\vec { b } ,\vec { c } \) are any three vectors such that \(\vec{b} \cdot \vec{c} \neq 0 \text { and } \vec{a} \cdot \vec{b} \neq 0\), then \(\vec { a } \) and \(\vec { c } \) are

(a)

perpendicular

(b)

parallel

(c)

inclined at an angle \(\frac{\pi}{3}\)

(d)

inclined at an angle \(\frac{\pi}{6}\)

34.

An ellipse has OB as semi minor axes, F and F′ its foci and the angle FBF′ is a right angle. Then the eccentricity of the ellipse is

(a)

\(\frac { 1 }{ \sqrt { 2 } } \)

(b)

\(\frac { 1 }{ 2 } \)

(c)

\(\frac { 1 }{ 4 } \)

(d)

\(\frac { 1 }{ \sqrt { 3 } } \)

35.

If cot^-1 2 and cot^-1 3 are two angles of a triangle, then the third angle is

(a)

\(\frac{\pi}{4}\)

(b)

\(\frac{3\pi}{4}\)

(c)

\(\frac{\pi}{6}\)

(d)

\(\frac{\pi}{3}\)

36.

If \(\omega \neq 1\) is a cubic root of unity and \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & { -\omega }^{ 2 }-1 & { \omega }^{ 2 } \\ 1 & { \omega }^{ 2 } & { \omega }^{ 7 } \end{matrix} \right| \) = 3k, then k is equal to

(a)

(b)

-1

(c)

\(\sqrt { 3i } \)

(d)

\(-\sqrt { 3i } \)

37.

The product of all four values of \(\left( cos\cfrac { \pi }{ 3 } +isin\cfrac { \pi }{ 3 } \right) ^{ \frac { 3 }{ 4 } }\) is

(a)

-2

(b)

-1

(c)

(d)

38.

If \(z=\cfrac { \left( \sqrt { 3 } +i \right) ^{ 3 }\left( 3i+4 \right) ^{ 2 } }{ \left( 8+6i \right) ^{ 2 } } \) , then |z| is equal to

(a)

(b)

(c)

(d)

Generic 2

10 x 2 = 20

39.

solve: x dy + y dx = xy dx

40.

Solve : \(\frac { dy }{ dx } =\frac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } \)

41.

Find the volume of the solid y=x³,x=0,y=1 is revolved about the y-axis.

42.

If \(w={ e }^{ { x }^{ 2 }+{ y }^{ 2 } }\) ,x=cosθ,y=sinθ, find \(\frac { dw }{ d\theta } \)

43.

If w=e^xy,x=at²,y=2at, find \(\frac { dw }{ dt } \)

44.

Find the point on the parabola y²=18x at which the ordinate increases at twice the rate of the abscissa.

45.

If of f(x, y) = x² + y³ + 2xy² find f_xx, f_yy, f_xy and f_yx.

46.

Use differentials to find \(\sqrt{25.2}\)

47.

Find the intervals of increasing and decreasing function for f(x) = x³ + 2x² - 1.

48.

If z = 2−2i, find the rotation of z by θ radians in the counter clockwise direction about the origin when \(\theta =\frac { \pi }{ 3 } \).

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Answers Key:

Generic 5

12 x 5 = 60

\(log|x|=cos\left( \frac { y }{ x } \right) ,x\neq 0\)

\(y=\frac { 2x }{ { 2x }^{ 2-1 } } ,x\neq \pm \frac { 1 }{ \sqrt { 2 } } \)

\(\frac { 2 }{ e } \)

\(\frac { { 56a }^{ 2 } }{ 3 } \)

240

Given \(\frac { dy }{ dx } \) =(3x+2y+1)²
Let z =3x+2y+1
\(\frac { dz }{ dx } =3+2\frac { dy }{ dx } \)
⇒ \(\frac { dz }{ dx } -3=2\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { dz }{ dx } -3 \right) \)
(1) becomes,
\(\frac { 1 }{ 2 } \left( \frac { dz }{ dx } -3 \right) \) =z²
⇒ \(\frac { dz }{ dx } \)-3 =2z²
⇒ \(\frac { dz }{ dx } \)=2x²+3
⇒ \(\frac { dz }{ 2{ z }^{ 2 }+3 } \) =dx
⇒ \(\frac { 1 }{ 2 } \int { \frac { dz }{ { z }^{ 2 }+\frac { 3 }{ 2 } } } =\int { dx } \)
⇒ \(\frac { 1 }{ 2 } \int { \frac { dz }{ { z }^{ 2 }+\left( \sqrt { \frac { 3 }{ 2 } } \right) ^{ 2 } } } \) =x+c
⇒ \(\frac { 1 }{ 2 } .\frac { 1 }{ \frac { \sqrt { 3 } }{ \sqrt { 2 } } } tan^{ -1 }\left( \frac { 1 }{ \frac { \sqrt { 3 } }{ \sqrt { 2 } } } \right) \) =x+c
⇒ \(\frac { 1 }{ \sqrt { 6 } } tan^{ -1 }\left( \frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \)= x+c
⇒ \(tan^{ -1 }\left( \frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) =\sqrt { 6 } \) x+c
⇒ \(\frac { \sqrt { 2 } }{ \sqrt { 3 } } .z=tan(\sqrt { 6 } x+c)\)
⇒ \(\frac { \sqrt { 2 } }{ \sqrt { 3 } } (3x+2y+1)=tan(\sqrt { 6 } x+c)\).

Given \(\frac { dy }{ dx } =-\left( \frac { y+3 }{ x+2 } \right) \)
⇒ \(\frac { dy }{ y+3 } =-\frac { dx }{ x+2 } \)
⇒ \(\int { \frac { dy }{ y+3 } } =-\int { \frac { dx }{ x+2 } } \)
⇒ log(y + 3) = -log (x + 2) + log c
⇒ log(y + 3) + log (x + 2) = log c
⇒ log(x + 2) (y + 3) = log c
⇒ (x+2)(y+3) = c ..(1)
Since the curve passes through (0, 0)
(0 + 2)(0 + 3) = c ⇒ c = 6
(1) becomes,
(x + 2)(y + 3) = 6
⇒ xy + 3x + 2y + 6 = 6
⇒ xy + 3x + 2y = 0

Area under the curve y = sin x between x = 0 and x = \(\frac { \pi }{ 3 } \) is
\({ A }_{ 1 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ ydx } =\int _{ 0 }^{ \frac { \pi }{ 3 } }{ sinxdx } =-{ \left[ cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)
\(=-(cos\frac { \pi }{ 3 } -cos0)=-\left( \frac { 1 }{ 2 } -1 \right) \)
\(=-\left( -\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } \)
Area under the curve y = sin 2x between x = 0 and \(\frac { \pi }{ 3 } \) is
\({ A }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ { sin2 \ x \ dx=-\left[ \frac { cos2 \ x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 3 } } } \)
\(=-\frac { 1 }{ 2 } [cos2\frac { \pi }{ 3 } -cos0]-\frac { 1 }{ 2 } \left[ -\frac { 1 }{ 2 } -1 \right] =-\frac { 1 }{ 2 } \left( -\frac { 3 }{ 2 } \right) =\frac { 3 }{ 4 } \)
\(\therefore \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 2 } \times \frac { 4 }{ 3 } =\frac { 2 }{ 3 } \)
∴ A₁:A₂ =2 : 3

Let \(\overset { \rightarrow }{ B } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ A } \times \overset { \rightarrow }{ B } =\overset { \rightarrow }{ C } \Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } \\ 1 \\ x \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 1 \\ y \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ z \end{matrix} \right| =\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\Rightarrow \overset { \wedge }{ i } (z-y)-\overset { \wedge }{ j } (z-x)+\overset { \wedge }{ k } (y-x)\quad \overset { \wedge }{ j } -\overset { \wedge }{ k } \)
Equating the like components on both sides, we get
z - y = 0 .....(1)
x - y = 1 .....(2)
y - x = -1 .....(3)
Also, \(\overset { \rightarrow }{ A } .\overset { \rightarrow }{ B } =3\Rightarrow \left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) =3\)
⇒ x + y + z = 3 ....(4)
Solving (1), (2), (3) and (4), we get \(x=\frac { 5 }{ 3 } ,y=\frac { 2 }{ 3 } \)and \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

10.

Given A =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
|A| = 0-1\(\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \)
= -1(0-1) + 1(1-0) = 1 + 1 = 2
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} (0-1) & -(0-1) & +(1-0) \\ -(0-1) & +(0-1) & -(0-1) \\ +(1+0) & -(0-1) & +(0-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \)
∴ A^-1= \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) ................(1)
Now A²=\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] \)
A²- 3I =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2-3 & 1-0 & 1-0 \\ 1-0 & 2-3 & 1-0 \\ 1-0 & 1-0 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) .............(2)
From (1) and (2), it is proved that A^-1= \(\frac{1}{2}\) [A²- 3I]

11.

f(x) = sin^-1 \((\frac{|x|-2}{3})+cos^-1(\frac{1-|x|}{4})\)
From the definition of sin^-1
\(-1\le \frac { \left| x \right| -2 }{ 3 } \le 1\)
\(\Rightarrow -3\le \left| x \right| -2\le 3\)
\(\Rightarrow -3+2\le \left| x \right| \le \left| x \right| \le 3+2\)
\(\Rightarrow -1\le \left| x \right| \le 5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(\Rightarrow 0\le \left| x \right| and\left| x \right| \le 5\)
\(\Rightarrow \left| x \right| \ge 0and-5\le x\le 5\)
From the definition of cos^-1x.
\(-1\le \frac { 1-\left| x \right| }{ 4 } \le 1\)
\(\Rightarrow -4\le 1-\left| x \right| \le 4\)
\(\Rightarrow -4-1\le \left| x \right| \le 4-1\)
\(\Rightarrow -5\le -\left| x \right| \le 3\)
\(\Rightarrow -3\le \left| x \right| >5\)
It reduces to
\(0\le \left| x \right| \le 5\)
\(-5\le |x|\le 5\)
From (1) & (2),
Domain is [-5, 5]

12.

Given p, q are the roots of lx²+ nx + n = 0
\(p+q=\frac { -b }{ a } =\frac { -n }{ l } ...(1)\)
\(pq=\frac { c }{ a } =\frac { n }{ l } ...(2)\)
L.H.S
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}} =\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} \sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\( =\frac{-\frac{n}{l}}{\sqrt{p q}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\frac{n}{l}}{\frac{\sqrt{n}}{\sqrt{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\(=\frac{-\frac{n}{l}+\frac{n}{l}}{\sqrt{\frac{n}{l}}}=0\) R.H.S

Generic 3

6 x 3 = 18

13.

Given g(x) = sin^-1 x + cos^-1x
From the definition of sin^-1x.
\(-1\le x\le 1\) ...(1)
Also from the definition of cos^-1x
\(-1\le x\le 1\) .........(2)
\(\therefore \) From (1) & (2),
Domain ofg(x) = [-1, 1] U [-1, 1]
= [-1, 1]
Hence the domain of g(x) is [-1, 1].

14.

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left( \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right) \)
adj A =\(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) \)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)-(14-9) \\ +(3-4)-(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

15.

The length a of the semi major axis of the elliptical ceiling is17 m. The height b of the semi minor axis is 8 m. Thus c² = a² -b² = 17²- 8²
then c =\(\sqrt { 289-64 } =\sqrt { 225 } =15\)
For the elliptical ceiling the foci are located on either side about 15 m from the centre, along its major axis.

16.

SS′ = 2c and 2c = 4; A'A = 2a = 6
c = 2 and a = 3,
b²= a²−c²= 9−4 = 5.
Major axis is along x-axis, since a > b.
Centre (0, 0) and Foci are (±2, 0)
Therefore, equation of the ellipse is \(\frac { { x }^{ 2 } }{ 9 } \frac { { y }^{ 2 } }{ 5 } =1\)

17.

Given A =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] =\left[ \begin{matrix} -3+10 & -9+4 \\ -7+25 & -21+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 & -5 \\ 18 & -11 \end{matrix} \right] \)
|AB| = -77+90 = 13 ≠ 0 ⇒ (AB)^-1 exists
|A| = 15-14 = 1 ≠ 0 ⇒ A^-1 exists
|B| = -2+15 = 13 ≠ 0 ⇒ B^-1 exists
(AB)^-1 = \(\frac { 1 }{ |AB| } adj(AB)=\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ...............(1)
B^-1= \(\frac { 1 }{ |B| } adj(B)=\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \)
A^-1= \(\frac { 1 }{ |A| } \)(adj A)
= \(\frac { 1 }{ 1 } \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) =\left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
∴ B^-1A^-1 = \(\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} 10-21 & -4+9 \\ -25+7 & 10-3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ..............(2)
From (1) or (2) it is prove that
(AB)^-1 = B^-1 A^-1

18.

Given v = 3-4i, w = 4+3i and \(\frac { 1 }{ u } =\frac { 1 }{ v } +\frac { 1 }{ w } \)
∴ \(\frac { 1 }{ u } =\frac { 1 }{ 3-4i } +\frac { 1 }{ 4+3i } \)
= \(\frac { 3+4i }{ (3-4i)(3+4i) } +\frac { 4-3i }{ (4+3i)(4-3i) } \)
= \(\\ \frac { 3+4i }{ 9-(4i)^{ 2 } } +\frac { 4-3i }{ 16-(3i)^{ 2 } } =\frac { 3+4i }{ 9+16 } +\frac { 4-3i }{ 16+9 } \)
= \(\frac { 3+4i }{ 25 } +\frac { 4-3i }{ 25 } =\frac { 3+4i+4-3i }{ 25 } \)
\(\frac { 1 }{ u } =\frac { 7+i }{ 25 } \)
∴ u = \(\frac { 25 }{ 7+i } \times \frac { 7-i }{ 7-i } =\frac { 25(7-i }{ 7^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { 25(7-i) }{ 49+1 } =\frac { 25(7-i) }{ 50 } =\frac { 1 }{ 2 } \)(7-i)
∴ u = \(\frac { 1 }{ 2 } \)(7-i) or \(\frac { 7 }{ 2 } \) - \(\frac { i }{ 2 } \)

Multiple Choice Question

20 x 1 = 20

19.

(b)

(a) (b) (c) (d)

F T T T

20.

(d)
9

21.

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(1)
\({ f }_{ xy }=\frac { \partial }{ \partial y } ({ f }_{ x })\)
\({ f }_{ xy }=\frac { ({ x }^{ 2 }+{ y }^{ 2 })(1)-y(2y) }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-2{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \) ...(2)
∴ From (1) and (2), f_xy = f_yz

22.

(b)
12x_o dx

23.

(d)
4.8 cu.cm

24.

(d)
0.96

25.

(b)
2^x- 2^y= C

26.

(a)
2, 3

27.

(d)
9

28.

29.

(b)
\(\frac{4}{25} \text { radians } / \mathrm{sec}\)

30.

(b)
|A| = |A^T|

31.

(b)
adj(AB) = (adj A)(adj B)

32.

(d)
2A^-1

33.

(b)
parallel

34.

(a)
\(\frac { 1 }{ \sqrt { 2 } } \)

35.

36.

(d)
\(-\sqrt { 3i } \)

37.

(b)
-1

38.

Generic 2

10 x 2 = 20

39.

|xy| = ce^x

40.

y = log |e^x+e^-x|+c

41.

\(\frac { 3\pi }{ 5 } \)

42.

\(\frac { dw }{ d\theta } =0\)

43.

\(\frac { dw }{ dt } ={ 6a }^{ 2 }{ t }^{ 2 }e^{ 2{ a }^{ 2 } }{ t }^{ 3 }\)

44.

\(\left( \frac { 9 }{ 8 } ,\frac { 9 }{ 2 } \right) \)

45.

Given f(x,y) = x² + y³ + 2xy²
f_x = 3x² + 2y²
f_xx = 6x
f_y = 0+ 3y² + 4xy
= 3y² + 4xy
f_yy = 6y + 4x
f_xy = 4y
f_yx = 4y

46.

Let y = f(x) = \(\sqrt x\)
Let x_o = 25, dx = 25.2 - 25 = 0.2
y = \(\sqrt x\)
dy = \(\frac{1}{2\sqrt{x}}\) dx
dy = \(\frac{1}{2\sqrt{x}}\) (0.2) = 0.02
∴\(\sqrt{25.2}\) = f(x₀) + f'(x₀) dx
= \(\sqrt{25}\) + 0.02
= 5 + 0.02 = 5.02

47.

f(x) = x³+ 2x²-1
f'(x) = 3x² + 4x = 0
⇒ x (3x + 4) = 0
⇒ x = 0 or \(\frac { 4 }{ 3 } \)
The possible intervals are \(\left( -\infty ,-\frac { 4 }{ 3 } \right) \left( -\frac { 4 }{ 3 } ,0 \right) \) and (0, ∞).

Interval	\(\left( -\infty ,-\frac { 4 }{ 3 } \right) \)	\(\left( -\frac { 4 }{ 3 } ,0 \right) \)	(0, ∞)
Sign of f'(x)	Say x = -2 3(-2)²+4(-2) = 4 +ve	say x = -1 3(-1)²+4(-1) = -1 -ve	say x = 1 3(1)²+4(1) = 7 +ve
Monotonicity	Strictly increasing	Strictly decreasing	Strictly increasing

48.

\(\theta =\frac { \pi }{ 3 } \)
Given z = 2-2i
θ = \(\frac { \pi }{ 3 } \)
Let z = 2-2i = r(cos θ + i sin θ)
r =\(\sqrt { { 2 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
α = \(\\ tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -2 }{ 2 } \right| \)
= tan^-1(1) = \(\frac { \pi }{ 4 } \)
The complex number 2-2i lie in the IV quadrant
∴ θ = -α = - \(\frac { \pi }{ 4 } \) [∵ x is +ve, y is -ve]
∴ 2-2i = 2\(\sqrt{2}\)\(\left[ cos\left( -\frac { \pi }{ 4 } \right) +isin\left( -\frac { \pi }{ 4 } \right) \right] \)
Z = \(2\sqrt { 2 } { e }^{ -i\frac { \pi }{ 4 } }\) ......... (1) [By uler'e formula]
Th rotation of z by θ radians in the counter clockwise direction about the origin in ze^iθ
∴ Rotaton of z is \(z{ e }^{ i\frac { \pi }{ 3 } }\)
= \(2\sqrt { 2 } { e }^{ -i\frac { \pi }{ 4 } }.{ e }^{ i\frac { \pi }{ 3 } }\) [using (1)]
= \(2\sqrt { 2 } \left[ { e }^{ \left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) } \right] =2\sqrt { 2 } e^{ i\frac { \pi }{ 12 } }\)