Question Set D

cbse 12th Standard Maths Continuity and Differentiability Model Questions Papers Study Material - QB365

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Questions:

Multiple Choice Question

Derivative of cot x° with respect to x is

(a)

cosec x°

(b)

cosec x° cot x°

(c)

-1° cosec2 x°

(d)

-1° cosec x° cot x°

If y = sin^-1 \(\left( \frac { 3x }{ 2 } -\frac { { x }^{ 3 } }{ 2 } \right) \), then \(\frac { dy }{ dx } \) is

(a)

\(\frac { 3 }{ \sqrt { 4-{ x }^{ 2 } } } \)

(b)

\(\frac { -3 }{ \sqrt { 4-{ x }^{ 2 } } } \)

(c)

\(\frac { 1 }{ \sqrt { 4-{ x }^{ 2 } } } \)

(d)

\(\frac {- 1 }{ \sqrt { 4-{ x }^{ 2 } } } \)

If y = Ae^5x,+ Be^-5x x then \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } \) is equal to

(a)

25y

(b)

(c)

-25y

(d)

10y

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

(a)

10^–1

(b)

\({ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)

(c)

\({ \left( \frac { 9 }{ 2 } \right) }^{ 5 }\)

(d)

\(\frac { 9 }{ 10 } \)

Which of the following functions are not continuous.

(a)

[x]

(b)

|x|

(c)

e^x

(d)

\(\frac1x\), x ≠ 0

Let \(\begin{cases} 4x \ >2 \\ ax \ 0\le x\le 2 \\ b \ x<0 \end{cases}\) For what values of a and b, f is a continuous function.

(a)

a = 2, b = 0

(b)

a = 1, b = 0

(c)

a = 0, b = 2

(d)

a = 0, b = 0

What is the point of discontinuity for signum function?

(a)

x = 1

(b)

x = -1

(c)

x = 0

(d)

function is continuous on R

The linear inequalities or equations or restrictions on the variables of a linear programming problem are called …… The conditions x ≥ 0 , y ≥ 0 are called …….

(a)

Objective functions, optimal value

(b)

Constraints, non-negative restrictions

(c)

Objective functions, non-negative restrictions

(d)

Constraints, negative restrictions

Let Z = ax + by is a linear objective function. Variables x and y are called ……… variables.

(a)

Independent

(b)

Continuous

(c)

Decision

(d)

Dependent

10.

How many of the following points satisfy the inequality 2x – 3y > -5?
(1, 1), (-1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (-1, 2) and (-2, -1)

(a)

(b)

(c)

(d)

11.

If \(x^{y}=y^{x}, \text { then } x(x-y \log x) \frac{d y}{d x}\) is equal to

(a)

y(y-xlogy)

(b)

y(y + xlogy)

(c)

x(x + ylogx)

(d)

x(y -xlogy)

12.

If \(y=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}\),then \(\frac{d y}{d x}\) is equal to

(a)

\(\frac{y \tan x}{y \log \cos x-1}\)

(b)

\(\frac{y^{2} \tan x}{y \log \cos x-1}\)

(c)

\(\frac{y \tan x}{1+y \log \cos x}\)

(d)

None ofthese

13.

If \(y=x \cos x, \text { then } \frac{d^{2} y}{d x^{2}}\) is

(a)

\(-x \cos x-2 \sin x\)

(b)

xcosx + 2sinx

(c)

xsinx + cosx

(d)

None of these

14.

The maximum value of z= 4x + 3y, if the feasible region for an LPP is as shown below, is

(a)

112

(b)

100

(c)

(d)

110

15.

Solution of LPP
To maximise Z = 4x + 8y
subject to constraints: \(2 x+y \leq 30, x+2 y \leq 24, x \geq 3,\)\(y \leq 9, y \geq 0 \text { is }\)

(a)

x = 12, y = 6

(b)

x = 6, y = 12

(c)

x = 9, y = 6

(d)

none of these

16.

A bag A contains 3 white, 2 red balls and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red, then the probability that it was drawn from bag B is

(a)

\(\frac{27}{43}\)

(b)

\(\frac{20}{43}\)

(c)

\(\frac{25}{43}\)

(d)

none of these

17.

The corner points of the feasible region in the graphical representation of a linear programming problem are (2, 72), (15, 20) and (40,15). If Z = 18x + 9y be the objective function, then

(a)

Z is maximum at (2, 72), minimum at (15, 20).

(b)

Z is maximum at (15, 20), minimum at (40, 15).

(c)

Z is maximum at (40, 15), minimum at (15, 20).

(d)

Z is maximum at (40,15), minimum at (2, 72).

18.

For any two events A and B, if \(P(\bar{A})=\frac{1}{2}, P(\bar{B})=\frac{2}{3}\) and \(P(A \cap B)=\frac{1}{4}\), then \(P\left(\frac{\bar{A}}{\bar{B}}\right)\)equals to

(a)

\(\frac{3}{8}\)

(b)

\(\frac{5}{8}\)

(c)

\(\frac{1}{8}\)

(d)

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

19.

A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4},\) respectively. If the events of their solving the problem are independent,then the probability that the problem will be solved, is

(a)

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

(b)

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

(c)

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

(d)

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

20.

If A and B are two events such that P(A) =0.2, P(B) =0.4 and P(A U B) = 0.5, then value of P(A /B) is?

(a)

0.1

(b)

0.25

(c)

0.5

(d)

0.08

Case Study Questions

21.

(a) A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at every point in this interval.
(b) A function f(x) is said to be continuous in the closed interval [a, b], if f(x) is continuous in (a, b) and \(\begin{equation} \lim _{h \rightarrow 0} f(a+h)=f(a) \text { and } \lim _{h \rightarrow 0} f(b-h)=f(b) \end{equation}\)
If function \(\begin{equation} f(x)=\left\{\begin{array}{ll} \frac{\sin (a+1) x+\sin x}{x} & , x<0 \\ c & , x=0 \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}} & , x>0 \end{array}\right. \end{equation}\) is continuous at x = 0, then answer the following questions.
(i) The value of a is

(a) -3/2

(b) 0

(c) 1/2

(d) -1/2

(ii) The value of b is

(a) 1

(b) -1

(c) 0

(d) any real number

(iii) The value of c is

(a) 1

(b) 1/2

(c) -1

(d) -1/2

(iv) The value of a + c is

(a) 1

(b) 0

(c) -1

(d) -2

(v) The value oi c - a is

(a) 1

(b) 0

(c) -1

(d) 2

22.

Deepa rides her car at 25 km/hr, She has to spend Rs. 2 per km on diesel and if she rides it at a faster speed of 40 km/hr, the diesel cost increases to Rs. 5 per km. She has Rs. 100 to spend on diesel. Let she travels x kms with speed 25 km/hr and y kms with speed 40 km/hr. The feasible region for the LPP is shown below:
Based on the above information, answer the following questions

Based on the above information, answer the following questions.
(i) What is the point of intersection of line l₁ and l₂,

\(\text { (a) }\left(\frac{40}{3}, \frac{50}{3}\right)\)

\(\text { (b) }\left(\frac{50}{3}, \frac{40}{3}\right)\)

\(\text { (c) }\left(\frac{-50}{3}, \frac{40}{3}\right)\)

\(\text { (d) }\left(\frac{-50}{3}, \frac{-40}{3}\right)\)

(ii) The corner points of the feasible region shown in above graph are

\(\text { (a) }(0,25),(20,0),\left(\frac{40}{3}, \frac{50}{3}\right)\)

\(\text { (b) }(0,0),(25,0),(0,20)\)

\(\text { (c) }(0,0),\left(\frac{40}{3}, \frac{50}{3}\right),(0,20)\)

\(\text { (d) }(0,0),(25,0),\left(\frac{50}{3}, \frac{40}{3}\right),(0,20)\)

(iii) If Z = x + y be the objective function and max Z = 30. The maximum value occurs at point

\(\text { (a) }\left(\frac{50}{3}, \frac{40}{3}\right)\)

(b) (0, 0)

(c) (25, 0)

(d) (0, 20)

(iv) If Z = 6x - 9y be the objective function, then maximum value of Z is

(a) -20

(b) 150

(c) 180

(d) 20

(v) If Z = 6x + 3y be the objective function, then what is the minimum value of Z?

(a) 120

(b) 130

(c) 0

(d) 150

23.

In a play zone, Aastha is playing crane game. It has 12 blue balls, 8 red balls, 10 yellow balls and 5 green balls. If Aastha draws two balls one after the other without replacement, then answer the following questions.

(i) What is the probability that the first ball is blue and the second ball is green?

\((a) \ \frac{5}{119}\)

\((b) \ \frac{12}{119}\)

\((c) \ \frac{6}{119}\)

\((d) \ \frac{5}{119}\)

(ii) What is the probability that the first ball is yellow and the second ball is red?

\((a) \ \frac{6}{119}\)

\((b) \ \frac{8}{119}\)

\((c) \ \frac{24}{119}\)

(d) None of these

(iii) What is the probability that both the balls are red?

\((a) \ \frac{4}{85}\)

\((b) \ \frac{24}{595}\)

\((c) \ \frac{12}{119}\)

\((c) \ \frac{64}{119}\)

(iv) What is the probability that the first ball is green and the second ball is not yellow?

\((a) \ \frac{10}{119}\)

\((b) \ \frac{6}{85}\)

\((c) \ \frac{12}{119}\)

(d) None of these

(v) What is the probability that both the balls are not blue?

\((a) \ \frac{6}{595}\)

\((b) \ \frac{12}{85}\)

\((c) \ \frac{15}{17}\)

\((d) \ \frac{253}{595}\)

24.

A card is lost from a pack of 52 cards. From the remaining cards two cards are drawn at random.

Based on the above information, answer the following questions.
(i) The probability of drawing two diamonds, given that a card of diamond is missing, is

a) \(\frac{21}{425}\)

(b) \(\frac{22}{425}\)

(c) \(\frac{23}{425}\)

(d) None of these

(ii) The probability of drawing two diamonds, given that a card of heart is missing, is

a) \(\frac{26}{425}\)

(b) \(\frac{22}{425}\)

(c) \(\frac{19}{425}\)

(d) \(\frac{23}{425}\)

(iii) Let A be the event of drawing two diamonds from remaining 51 cards and E₁, E₂, E₃ and E₄ be the events that lost card is of diamond, club, spade and heart respectively, then the approximate value of \(\sum_{i=1}^{4} P\left(A \mid E_{i}\right) \text { is }\)

a) 0.17

(b) 0.24

(c) 0.25

(d) 0.18

(iv) All of a sudden, missing card is found and, then two cards are drawn simultaneously without replacement. Probability that both drawn cards are king is

a) \(\frac{1}{52}\)

(b) \(\frac{1}{221}\)

(c) \(\frac{1}{121}\)

(d) \(\frac{2}{221}\)

(v) If two cards are drawn from a well shuffled pack of 52 cards, one by one with replacement, then probability of getting not a king in 1^st and 2^nd draw is

a) \(\frac{144}{169}\)

(b) \(\frac{12}{169}\)

(c) \(\frac{64}{169}\)

(d) None of these

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Answers Key:

Multiple Choice Question

As x^o = \(\frac { \pi }{ 180 } { x }^{ c }\)
\(\therefore \frac { d }{ dx } (cot{ x }^{ o })=\)\(\frac { d }{ dx } \left( cot\frac { \pi }{ 180 } x \right) \)
\(=-\frac { \pi }{ 180 } { cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }{ x }^{ 0 }\)

As y = sin^-1 \(y={ sin }^{ 2 }\left\{ 3.\frac { x }{ 2 } -4.{ \left( \frac { x }{ 2 } \right) }^{ 3 } \right\} \)
= 3 sin^-1 \(\frac { x }{ 2 } \)
\(\therefore \ \frac { dy }{ dx } =3.\frac { 1 }{ \sqrt { 1-\frac { { x }^{ 2 } }{ 4 } } } .\frac { 1 }{ 2 } \)
\(=\frac { 3 }{ \sqrt { 4-{ x }^{ 2 } } } \)

As y' = 5Ae^5x - 5Be^-5x
and y'' = 25Ae^5x + 25Be^-5x
= 25y

(a)
[x]

(a)
a = 2, b = 0

(b)
Constraints, non-negative restrictions

10.

(d)
5

11.

Since \(x^{y}=y^{x} \Rightarrow y \log x=x \log y\)
On differentiating w.r.t. x, we get
\( \Rightarrow y \cdot \frac{1}{x}+\log x \frac{d y}{d x}=\frac{x}{y} \frac{d y}{d x}+\log y \)
\(\Rightarrow \frac{d y}{d x}\left(\frac{x}{y}-\log x\right)=\frac{y}{x}-\log y \)
\(\Rightarrow x(x-y \log x) \frac{d y}{d x}=y(-x \log y+y) \)

12.

Given that,\(y=(\cos x)^{(\cos x)^{(\cos x)} \cdots^{\infty}}\)
\(\Rightarrow \quad y=(\cos x)^{y}\)
Taking log on both sides, we get
log y = y log(cos x)
Now, differentiating w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}\)
\(\Rightarrow \frac{d y}{d x}=y\left\{-y \tan x+\log \cos x \cdot \frac{d y}{d x}\right\} \)
\(\Rightarrow(1-y \log \cos x) \frac{d y}{d x}=-y^{2} \tan x \)
\(\Rightarrow \frac{d y}{d x}=\frac{y^{2} \tan x}{(y \log \cos x-1)} \)

13.

(a)
\(-x \cos x-2 \sin x\)

14.

(a)
112

15.

(a)
x = 12, y = 6

16.

17.

18.

(b)
\(\frac{5}{8}\)

19.

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

20.

(b)
0.25

Case Study Questions

21.

L.H.L (at x = 0) = \(\begin{equation} \lim _{x \rightarrow 0} \frac{\sin (a+1) x+\sin x}{x}\left(\frac{0}{0} \text { form }\right) \end{equation}\)
Using L' Hospital rule, we get
L.H.L.(at x = 0)
\(\begin{equation} =\lim _{x \rightarrow 0}(a+1) \cos (a+1) x+\cos x=a+2 \end{equation}\)
R.H.L \(\begin{equation} \text { (at } x=0)=\lim _{x \rightarrow 0} \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}}=\lim _{x \rightarrow 0} \frac{\sqrt{1+b x}-1}{b x} \end{equation}\)
\(\begin{equation} =\lim _{x \rightarrow 0} \frac{1}{\sqrt{1+b x}+1}=\frac{1}{2} \end{equation}\)
Since,f(x) is continuous at x = 0.
\(\therefore\) From (i) and (ii), we get
\(\begin{equation} a+2=c=\frac{1}{2} \Rightarrow a=-\frac{3}{2}, c=\frac{1}{2} \end{equation}\)
Also, value of b does not affect the continuity of f(x), so b can be any real number.
(i) (a)
(ii) (d)
(iii) (b)
(iv) (c) : \(\begin{equation} a+c=-\frac{3}{2}+\frac{1}{2}=-1 \end{equation}\)
(v) (d) : \(\begin{equation} c-a=\frac{1}{2}+\frac{3}{2}=2 \end{equation}\)

22.

(i) (b): Let B(x, y) be the point of intersection of the given lines
2x + 5y = 100 ....(i)
and \(\frac{x}{25}+\frac{y}{40}=1 \Rightarrow 8 x+5 y=20\)...(ii)
Solving (i) and (ii), we get
\(x=\frac{50}{3}, y=\frac{40}{3}\)
ஃ The point of intersection \(B(x, y)=\left(\frac{50}{3}, \frac{40}{3}\right)\)
(ii) (d): The corner points of the feasible region shown in the given graph are
\((0,0), A(25,0), B\left(\frac{50}{3}, \frac{40}{3}\right), C(0,20)\)
(iii) (a): Here Z = x + y

Corner Points	Value of Z = x + y
(0,0)	0
(25,0)	25
\(\left(\frac{50}{3}, \frac{40}{3}\right)\)	30 ⇠ Maximum
(0,20)	20

Thus, max Z = 30 occurs at point \(\left(\frac{50}{3}, \frac{40}{3}\right)\)
(iv) (b):

Corner Points	Value of Z = 6x - 9y
(0,0)	0
(25,0)	150 ⇠ Maximum
\(\left(\frac{50}{3}, \frac{40}{3}\right)\)	-20
(0,20)	-180

(v) (c):

Corner Points	Value of Z = 6x + 3y
(0,0)	0 ⇠ Maximum
(25,0)	150
\(\left(\frac{50}{3}, \frac{40}{3}\right)\)	140
(0,20)	60

23.

Let B, R, Y and G denote the events that ball drawn is blue, red, yellow and green respectively.
\(\therefore P(B)=\frac{12}{35}, P(R)=\frac{8}{35}, P(Y)=\frac{10}{35} \text { and } P(G)=\frac{5}{35}\)
\((i) \ (c): P(G \cap B)=P(B) \cdot P(G \mid B)=\frac{12}{35} \cdot \frac{5}{34}=\frac{6}{119}\)
\((ii) \ (\mathbf{b}): P(R \cap Y)=P(Y) \cdot P(R \mid Y)=\frac{10}{35} \cdot \frac{8}{34}=\frac{8}{119}\)
(iii) (a): Let E = event of drawing a first red ball and
F = event of drawing a second red ball
Here, \(P(E)=\frac{8}{35} \text { and } P(E)=\frac{7}{34}\)
\(\therefore \P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{8}{35} \cdot \frac{7}{34}=\frac{4}{85}\)
\(\text {(iv) }(c): P\left(Y^{\prime} \cap G\right)=P(G) \cdot\left(Y^{\prime} \mid G\right)=\frac{5}{35} \cdot \frac{24}{34}=\frac{12}{119}\)
(v) (d): Let E = event of drawing a first non-blue ball and F = event of drawing a second non-blue ball
Here, \(P(E)=\frac{23}{35} \text { and } P(F)=\frac{22}{34}\)
\(\therefore \ P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{23}{35} \cdot \frac{22}{34}=\frac{253}{595}\)

24.

(i) (b): Required probability = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}\)
\(=\frac{12 \times 11}{51 \times 50}=\frac{22}{425}\)
(ii) (a): Required probability = \(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 \times 12}{51 \times 50}=\frac{26}{425}\)
(iii) (b): Clearly, P(A l E₁) = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{22}{425}\)
\(P\left(A \mid E_{2}\right)=\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{26}{425} \)
\(P\left(A \mid E_{3}\right)=P\left(A \mid E_{4}\right)=\frac{26}{425}\)
\(\therefore \sum_{i=1}^{4} P\left(A \mid E_{i}\right)=\frac{22}{425}+\frac{26}{425}+\frac{26}{425}+\frac{26}{425}=\frac{100}{425}=0.24\)
(iv) (b): P(getting both king) = \(\frac{{ }^{4} C_{2}}{{ }^{52} C_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
(v) (a): P(drawing a king) = \(\frac{4}{52}=\frac{1}{13}\)
\( \therefore \quad P(\text { not drawing a king })=1-\frac{1}{13}=\frac{12}{13} \)
\(\therefore \quad \text { Required probability }=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}\)