Question Set C

TN 12th Standard Maths Differentials and Partial Derivatives Previous Year Question Papers Study Material - QB365

Download Tamil Nadu 12th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Multiple Choice Question

20 x 1 = 20

If \(\rho\) (A) = \(\rho\)([A| B]), then the system AX = B of linear equations is

(a)

consistent and has a unique solution

(b)

consistent

(c)

consistent and has infinitely many solution

(d)

inconsistent

If x^ay^b = e^m, x^cy^d = eⁿ, Δ₁ = \(\left| \begin{matrix} m & b \\ n & d \end{matrix} \right| \), Δ₂ = \(\left| \begin{matrix} a & m \\ c & n \end{matrix} \right| \), Δ₃ = \(\left| \begin{matrix} a & b \\ c & d \end{matrix} \right| \), then the values of x and y are respectively,

(a)

e^(Δ₂^{/ Δ}₁⁾, e^(Δ₃^{/ Δ}₁⁾

(b)

log (Δ₁/ Δ₃), log (Δ₂/ Δ₃)

(c)

log (Δ₂/ Δ₁), log(Δ₃/ Δ₁)

(d)

e^(Δ₁^{/ Δ}₃⁾,e^(Δ₂^{/ Δ}₃⁾

If A = \(\left[ \begin{matrix} 2 & 0 \\ 1 & 5 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 1 & 4 \\ 2 & 0 \end{matrix} \right] \) then |adj (AB)| =

(a)

-40

(b)

-80

(c)

-60

(d)

-20

If A is a 3 \(\times\) 3 non-singular matrix such that AA^T = A^TA and B = A^-1A^T, then BB^T =

(a)

(b)

(c)

I₃

(d)

B^T

If the direction cosines of a line are \(\frac { 1 }{ c } ,\frac { 1 }{ c } ,\frac { 1 }{ c } \), then

(a)

\(c=\pm 3\)

(b)

\(c=\pm \sqrt { 3 } \)

(c)

c > 0

(d)

0 < c < 1

If \(\vec { a } =2\hat { i } +3\hat { j } -\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -5\hat { k } ,\vec { c } =3\hat { i } +5\hat { j } -\hat { k } ,\) then a vector perpendicular to \(\vec { a } \) and lies in the plane containing \(\vec { b } \) and \(\vec { c } \) is

(a)

\(-17\hat { i } +21\hat { j } -97\hat { k } \)

(b)

\(17\hat { i } +21\hat { j } -123\hat { k } \)

(c)

\(-17\hat { i } -21\hat { j } +97\hat { k } \)

(d)

\(-17\hat { i } -21\hat { j } -97\hat { k } \)

If \(\vec { a } ,\vec { b } ,\vec { c } \) are three unit vectors such that \(\vec { a } \) is perpendicular to \(\vec { b } \) and is parallel to \(\vec { c } \) then \(\vec { a } \times (\vec { b } \times \vec { c } )\) is equal to

(a)

\(\vec { a } \)

(b)

\(\vec { b} \)

(c)

\(\vec { c } \)

(d)

\(\vec { 0 } \)

If a vector \(\vec { \alpha } \) lies in the plane of \(\vec { \beta } \) and \(\vec { \gamma } \), then

(a)

\([\vec { \alpha } ,\vec { \beta } ,\vec { \gamma } ]\) = 1

(b)

\([\vec { \alpha } ,\vec { \beta } ,\vec { \gamma } ]\) = -1

(c)

\([\vec { \alpha } ,\vec { \beta } ,\vec { \gamma } ]\) = 0

(d)

\([\vec { \alpha } ,\vec { \beta } ,\vec { \gamma } ]\) = 2

Consider an ellipse whose centre is of the origin and its major axis is along x-axis. If its eccentrcity is \(\frac { 3 }{ 5 } \) and the distance between its foci is 6, then the area of the quadrilateral inscribed in the ellipse with diagonals as major and minor axis of the ellipse is

(a)

(b)

(c)

(d)

10.

The centre of the circle inscribed in a square formed by the lines x²− 8x − 12 = 0 and y²− 14y + 45 = 0 is

(a)

(4, 7)

(b)

(7, 4)

(c)

(9, 4)

(d)

(4, 9)

11.

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is

(a)

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

(b)

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

(c)

\(\frac { 2 }{ \sqrt { 3 } } \)

(d)

\(\frac { 3 }{ 2 } \)

12.

If \(\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}\), then x is equal to

(a)

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

(b)

\(\frac{1}{\sqrt{5}}\)

(c)

\(\frac{2}{\sqrt{5}}\)

(d)

\(\frac{\sqrt3}{2}\)

13.

If the function f(x) = sin^-1(x²- 3), then x belongs to

(a)

[-1, 1]

(b)

[\(\sqrt2\), 2]

(c)

\(\\ \\ \\ \left[ -2,-\sqrt { 2 } \right] \cup \left[ \sqrt { 2 } ,2 \right] \)

(d)

\([-2,-\sqrt{2}]\)

14.

If sin⁻¹x = 2sin⁻¹\(\alpha\) has a solution, then

(a)

\(|\alpha |\le \frac { 1 }{ \sqrt { 2 } } \)

(b)

\(|\alpha |\ge \frac { 1 }{ \sqrt { 2 } } \)

(c)

\(|\alpha |<\frac { 1 }{ \sqrt { 2 } } \)

(d)

\(|\alpha |>\frac { 1 }{ \sqrt { 2 } } \)

15.

If \(\omega \neq 1\) is a cubic root of unity and \(\left( 1+\omega \right) ^{ 7 }=A+B\omega \), then (A, B) equals

(a)

(1, 0)

(b)

(−1, 1)

(c)

(0, 1)

(d)

(1, 1)

16.

The number of positive zeros of the polynomial \(\overset { n }{ \underset { j=0 }{ \Sigma } } { n }_{ C_{ r } }\)(-1)^rx^r is

(a)

(b)

(c)

< n

(d)

17.

The number of real numbers in [0, 2π] satisfying sin⁴x - 2sin²x + 1 is

(a)

(b)

(c)

(d)

∞

18.

A polynomial equation in x of degree n always has

(a)

n distinct roots

(b)

n real roots

(c)

n complex roots

(d)

at most one root

19.

If |z₁| = 1, |z₂| = 2, |z₃| = 3 and |9z₁z₂+ 4z₁z₃+ z₂z₃| = 12, then the value of |z₁+z₂+z₃| is

(a)

(b)

(c)

(d)

20.

The value of \(\sum_{n=1}^{13}\left(i^{n}+i^{n-1}\right)\) is

(a)

1+ i

(b)

(c)

(d)

Generic 5

8 x 5 = 40

21.

If u = sin^-1 \(\left( \frac { x+y }{ \sqrt { x } +\sqrt { y } } \right) \), Show that \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tanu\)

22.

Find the area of the region bounded by the parabola y² = x and the line y = x − 2

23.

Find the area of the region bounded by the curve 2+x−x²+y = 0 , x-axis, x = −3 and x = 3.

24.

Water at temperature 100^oC cools in 10 minutes to 80^oC in a room temperature of 25^oC.
Find
(i) The temperature of water after 20 minutes
(ii) The time when the temperature is 40^oC
\(\left[ { log }_{ e }\frac { 11 }{ 15 } =-0.3101;{ log }_{ e }5=1.6094 \right] \)

25.

The equation of electromotive force for an electric circuit containing resistance and self inductance is E = Ri + L\(\frac{di}{dt},\) Where E is the electromotive force is given to the circuit, R the resistance and L, the coefficient of induction. Find the current i at time t when E = 0.

26.

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?

27.

A ladder 17 metre long is leaning against the wall. The base of the ladder is pulled away from the wall at a rate of 5 m/s. When the base of the ladder is 8 metres from the wall.
(i) How fast is the top of the ladder moving down the wall?
(ii) At what rate, the area of the triangle formed by the ladder, wall and the floor is changing?

28.

A particle moves along a line according to the law s(t) = 2t³ − 9t² +12t − 4, where t ≥ 0.
(i) At what times the particle changes direction?
(ii) Find the total distance travelled by the particle in the first 4 seconds.
(iii) Find the particle’s acceleration each time the velocity is zero.

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Multiple Choice Question

20 x 1 = 20

(b)
consistent

(a)
e^(Δ₂^{/ Δ}₁⁾, e^(Δ₃^{/ Δ}₁⁾

(b)
-80

(b)
\(c=\pm \sqrt { 3 } \)

(d)
\(-17\hat { i } -21\hat { j } -97\hat { k } \)

(d)
\(\vec { 0 } \)

(d)
40

10.

(a)
(4, 7)

11.

12.

(b)
\(\frac{1}{\sqrt{5}}\)

13.

14.

(a)
\(|\alpha |\le \frac { 1 }{ \sqrt { 2 } } \)

15.

(b)
(−1, 1)

16.

(a)
0

17.

(a)
2

18.

19.

(b)
2

20.

(a)
1+ i

Generic 5

8 x 5 = 40

21.

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\) ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

22.

Given equation of the parabola is y² = x and the
\(\therefore\) y²= y + 2 \(\Rightarrow\) y²- y - 2 = 0
\(\Rightarrow\) (y-2) (y+1) = 0
\(\Rightarrow\) y = -1, 2
y = x - 2
x 0 2 y -2 0 \(\therefore \) Required area = \(\int _{ -1 }^{ 2 }{ ({ x }_{ 1 }-{ x }_{ 2 }) } dy\)

\(=\int _{ -1 }^{ 2 }{ (y+2)-{ y }^{ 2 }dy } ={ \left[ \frac { { y }^{ 2 } }{ 2 } +2y-\frac { { y }^{ 3 } }{ 3 } \right] }_{ -1 }^{ 2 }\)
\(=\left( 2+4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 2 } -2+\frac { 1 }{ 3 } \right) \)
\(=\left( \frac { 5+12-8 }{ 3 } \right) -\left( \frac { 3-12+2 }{ 6 } \right) \)
\(=\left( \frac { 10 }{ 3 } \right) -\left( \frac { -7 }{ 6 } \right) =\frac { 10 }{ 3 } +\frac { 7 }{ 6 } =\frac { 20+7 }{ 6 } =\frac { 27 }{ 6 } \)
Area \(=\frac{9}{2}\) sq.units

23.

Equation of the given curve is 2+x-x²+y = 0

x	0	2	-1
y	-2	0	0

\(\Rightarrow\)y = x²-x-2

\(\therefore\) Required area= \(\int _{ -3 }^{ -1 }{ ydx } +\int _{ -1 }^{ 2 }{ -y } dx+\int _{ 2 }^{ 3 }{ ydx } \)
\(\\ =\int _{ -3 }^{ -1 }{ \left( { x }^{ 2 }-x-2 \right) } dx\int _{ -1 }^{ 2 }{ (2+x-{ x }^{ 2 })dx } +\int _{ 2 }^{ 3 }{ ({ x }^{ 2 }-x-2) } dx\)
\(={ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ -3 }^{ -1 }+{ \left( 2x+\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 2 }+{ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ 2 }^{ 3 }\)
\(=\left( -\frac { 1 }{ 3 } -\frac { 1 }{ 2 } +2 \right) -\left( -9-\frac { 9 }{ 2 } +6 \right) +\left( 4+2-\frac { 8 }{ 3 } \right) -\left( -2+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } \right) +\left( 9-\frac { 9 }{ 2 } -6 \right) -\left( \frac { 8 }{ 3 } -2-4 \right) \)
\(\\ =\left( \frac { -2-3+6 }{ 6 } \right) -\left( \frac { -6-9 }{ 2 } \right) +\left( \frac { 18-8 }{ 3 } \right) -\left( \frac { -12+3+2 }{ 6 } \right) +\left( \frac { 6-9 }{ 2 } \right) -\left( \frac { 8-24 }{ 3 } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 15 }{ 2 } +\frac { 10 }{ 3 } +\frac { 7 }{ 6 } -\frac { 3 }{ 2 } +\frac { 16 }{ 3 } \)
\(=\frac { 1+45+20+7-9+32 }{ 6 } =\frac { 90 }{ 6 } =15 \)

24.

Let T be the temperature of water at any time t.
Then, by Newton's law of cooling,
\(\frac { dT }{ dt } \infty (T-{ 25 }^{ o })\)
\(\Rightarrow \frac { dT }{ dt } =-\lambda (T-25)\)
\(\Rightarrow \int { \frac { dT }{ T-25 } } =-\lambda \int { dt } \)
\(\Rightarrow log(T-25)=-\lambda t+C\) ...(1)
At t = 0, T = 100⁰e in (1) we get
\(\therefore(1)\Rightarrow\)log75 = 0+C
\(\Rightarrow\)C = log75
\(\therefore\)(1) becomes, log (T-25) = -\(\lambda\)t + log 75
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =-\lambda t...(2)\)
When t = 10, T = 80^oC
\(\therefore log\left( \frac { 80-25 }{ 75 } \right) =-10|\)
\(\Rightarrow log\left( \frac { 11 }{ 15 } \right) =-10\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \)
Substituting \(\lambda\) in (2) we get
\(log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t ...(3)\)
\(\Rightarrow log\left( \frac { T-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) \times 20\)
\(={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow \frac { T-25 }{ 75 } ={ \left( \frac { 11 }{ 15 } \right) }^{ 2 }\)
\(\Rightarrow T-25=\frac { 121 }{ 225 } \times 75=\frac { 121 }{ 3 } =40.33\)
\(\Rightarrow T=40.33+25\)
\(\\ =65{ .33 }^{ 0 }C\)
So the temperature of water after 20 minutes is 65.33°C
(ii) putting T = 40^oC in (3) we get
\(log\left( \frac { 40-25 }{ 75 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(\Rightarrow log\left( \frac { 1 }{ 5 } \right) =\frac { 1 }{ 10 } log\left( \frac { 11 }{ 15 } \right) t\)
\(t=\frac { 10log\left( \frac { 1 }{ 5 } \right) }{ log\left( \frac { 11 }{ 15 } \right) } =\frac { -10log5 }{ log\left( \frac { 11 }{ 15 } \right) } \)
\(=\frac { -10\times 1.6094 }{ -0.3101 } \)
\(\therefore\) t = 53.46 minutes

25.

Given E = Ri + L \(\frac{di}{dt}\)
\(\frac { E }{ L } =\frac { Ri }{ L } +\frac { di }{ dt } \)
\(\Rightarrow \frac { Ri }{ L } +\frac { di }{ dt } =\frac { E }{ L } \)
This is a linear differential equation
\(Here\quad P=\frac { R }{ L } and\quad Q=\frac { E }{ L } \)
\(\therefore \int { pdt } =\int { \frac { R }{ L } dt } =\frac { R }{ L } t\)
\(\therefore I.F={ e }^{ \int { pdt } }={ e }^{ \frac { Rt }{ L } }\)
\(\therefore\) Solution is i\({ e }^{ \int { pdt } }=\int { Q{ e }^{ \int { pdt } }dt+C } \)
\(\Rightarrow i{ e }^{ \frac { Rt }{ L } }=\int { \frac { E }{ L } . } { e }^{ \frac { Rt }{ L } }dt+C\)
\(\therefore i{ e }^{ \frac { Rt }{ L } }=\frac { E }{ L } \frac { { e }^{ \frac { Rt }{ L } } }{ \frac { R }{ L } } dt+C\)
\(i=\frac { E }{ R } { e }^{ \frac { Rt }{ L } }+C\)
\(i=\frac { E }{ R } +c{ e }^{ -\frac { Rt }{ L } }\)
When E = 0,
\(i=0+c{ e }^{ -\frac { Rt }{ L } }\)
\(\Rightarrow i=c{ e }^{ -\frac { Rt }{ L } }\)

26.

Let A be the number of bacteria at any time t
Given \(\frac { dA }{ A } \alpha A\)
\(\Rightarrow \frac { dA }{ dt } =kA\)
\(\Rightarrow \frac { dA }{ dt } =kdt\)
\(\Rightarrow \int { \frac { dA }{ dt } } =k\int { \frac { dt }{ t } } \)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow log\left( \frac { A }{ c } \right) =kt\)
\(\Rightarrow \frac { A }{ c } ={ e }^{ kt }\)
\(\Rightarrow A=c.{ e }^{ kt }\) ...(1)
Initially when t = 0, assume that A = A₀
\(\therefore\) becomes, = ce₀ \(\Rightarrow\) c = A₀
\(\therefore\) A = A0ekt ...(2)
Given when t = 5, A = 3A₀
3A₀= A₀e^5k \(\Rightarrow\) 3 = 35k
when t = 10, 2 becomes,
A = A₀e^10k
= A₀(e^5k)2 = A₀(3)²
\(\Rightarrow\) A = 9A₀
Hence, the amount of bacteria will be present after 10 hours is 9 times the initial amount.

27.

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB²= AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
⇒ y² = 289 - 64 = 225
⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

28.

Given s (t) = 2t³ − 9t² + 12t ≥ 0
On differentiating we get
V(t) = 6t²-18t+ 12 ... (1)
= 6 (t² - 3t+ 2)
= 6 (t - 1) (t - 2)
Now V(t) = 0
⇒ 6 (t-1)(t- 2) = 0
⇒ t = 1, 2
The particle changes direction when V(t) changes its sign.
If 0 ≤ t < 1 then both (t - 1) and (t - 2) < 0
⇒ V(t) > 0
If 1 < t < 2 then (t -1) > 0 and (t - 2) < 0
⇒ V(t) < 0
If t > 2 then both (t - 1) and (t - 2) > 0
⇒ V(t) > 0
∴ The particle changes direction when t = 1 and t = 2 sec.
(ii) Total distance travelled by the particle in the first 4 seconds is |s(0)- s (1)| + |s (1) - s (2)| + |s (2) -s (4)|
s(0) = -4
s(1) = 2(1)3 - 9(1)2 + 12 (1) - 4
= 2 - 9 + 12 - 4 = 1
s (2) = 2 \(\times\) 2³ - 9 \(\times\) 2² + 12 \(\times\) 2 - 4
= 16 - 36 + 24 - 4
= 0
s (4) = 2(4)³ - 9(4)² + 12 (4) - 4
= 128 - 144 + 48 - 4 = 28
∴ Is (0) -s (1)|+ Is (1) -s (2)|+ Is (2) -s(4)|
= |-4 - 1| + |1 - 0| + |0 - 28|
= |-5| + |1| + |0 - 28|
= 5 + 1 + 28 = 34 m
(iii) Given s (t) = 2t³ − 9t² + 12t ≥ 0
[acceleration = \(\frac { dv }{ dt } \)]
When t = 1,
Acceleration = 12 (1) - 18 = -6 m/sec²
When t = 2
Acceleration = 12 (2) - 18 = 6 m/sec²