Mathematics Test

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Mathematics Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

∫secx /secx+tanxdx is equal to

A
tanx−secx+c

B
loge|1+secx|+c

C
secx+tanx+c

D
log|sinx|+log|cosx|+c

Solution

Let \(I=\int \frac{\sec x}{\sec x+\tan x} d x\)
\(=\int \frac{\sec x(\sec x-\tan x)}{\sec ^{2} x-\tan ^{2} x} d x\)
\(=\int\left(\sec ^{2} x-\sec x \tan x\right) d x\)
\(=\tan x-\sec x+c\)

KEY CONCEPTS

Pythagorean Identities
\(\sin ^{2} x+\cos ^{2} x=1\)
\(1+\tan ^{2} x=\sec ^{2} x\)
\(1+\cot ^{2} x=\csc ^{2} x\)

\(\int \frac{d x}{\cos ^{2} x}=\int \sec ^{2} x d x=\tan x+C\)

\(\int \sec x \tan x d x=\sec x+C\)
\(\int \csc x \cot x d x=-\csc x+C\)
Question 2
1 / -0

Let p be the statement 'Ravi races' and let qbe the statement 'Ravi wins'. Then, the verbal translation of ∼(p∨(∼q))

A
Ravi does not race and Ravi does not win

B
It is not true that Ravi races and that Ravi does not win

C
Ravi races and Ravi wins

D
It is not true that Ravi races or that Ravi does not win

Solution

Given, p: Ravi races, q: Ravi wins

∴ The statement of given proposition ∼(p∨(∼q)) is

"It is not true that Ravi races or that Ravi does not win."

Negation of a statement in mathematical reasoning

The denial of a statement p is called its negation, written as ~ p~ p. Negation of any statement p is formed by writing "It is not the case that ..... "or " It is false that......." before p or, if possible by inserting in p the word "not".

Negation is called a connective although it does not combine two or more statements. In fact, it only modifies a statement.

Negation of compound statements:

We have learnt about negation of a simple statement. Writing the negation of compound statements having conjunction, disjunctions, implication, equivalence, etc, is not very simple. So, let us discuss the negation of compound statement.

(i) Negation of conjunction:If p and q are two statements, then ∼ (p∧q )≡(∼p∨∼q)∼ p∧q ≡(∼p∨∼q)

(ii) Negation of disjunction:If p and q are two statements, then ∼ (p∨q )≡(∼p∧ ∼q)∼ p∨q ≡(∼p∧ ∼q)

(iii) Negation of implication:If p and q are two statements, then ∼(p⇒q)≡∼p⇒q≡∼(∼p∨q)≡∼(∼p∨q)≡(p ∧ ∼q)p ∧ ∼q

(iv) Negation of biconditional statement:If p and q are two statements, then∼(p ⇔q)≡(p ∧ ∼q) ∨(q ∧ ∼p) ≡ ∼p⇔q≡p⇔∼q
Question 3
1 / -0

The mean and variance of a random variable XX having a binomial distribution are 44 and 22 respectively. Then, P(X>6) is equal to

A
1/256

B
3/256

C
9/256

D
7/256

Solution

Given, \(n p=4, n p q=2\)
\(\Rightarrow p=q=\frac{1}{2}\)
\(\therefore n=4 \times 2=8\)
\(\therefore P(X>6)=^{8} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)+^{8} C_{8}\left(\frac{1}{2}\right)^{8}\)
\(=\frac{8}{256}+\frac{1}{256}=\frac{9}{256}\)

KEY CONCEPTS

We have \(P(X=0)+P(X=1)+\ldots+P(X=n)\)
\(=^{n} C_{0} p^{0} q^{n-0}+^{n} C_{1} p^{1} q^{n-1}+\ldots+^{n} C_{n} p^{n} q^{n-n}=(q+p)^{n}=1^{n}=1\)
Now probability of
(a) Occurrence of the event exactly r times \(P(X=r)=^{n} C_{r} q^{n-r} p^{r}\)
Question 4
1 / -0

The locus of the point PP which is equidistant from 3x+4y+5=03x+4y+5=0 and 9x+12y+7=09x+12y+7=0 is

A
A hyperbola

B
An ellipse

C
A parabola

D
A straight line

Solution

\(\left|\frac{3 x+4 y+5}{5}\right|=\left|\frac{9 x+12 y+7}{25}\right|\)
\(15 x+20 y+25-9 x-12 y-7=0\)
\(6 x+8 y+18=0\)
\(3 x+4 y+9=0\)
\(\therefore\) locus is a straight line
KEY CONCEPTS
Distance of a Point from a Line The length p of the perpendicular from the point \(\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c=0\) is given by \(p=\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)
. Length of perpendicular from origin to the line \(a x+b y+c=0\) is \(\left|\frac{c}{\sqrt{a^{2}+b^{2}}}\right|\)
- Length of perpendicular from the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) to the line \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) is \(\left|\mathrm{x}_{1} \cos \alpha+\mathrm{y}_{1} \sin \alpha-\mathrm{p}\right|\)
Question 5
1 / -0

The distance travelled by a motor car in t seconds after the brakes are applied is s feet, where s=22t−12t2. The distance travelled by the car before it stops, is

A
10.08 ft

B
10 ft

C
11 ft

D
11.5 ft

Solution

Given, s=22t−12t2

∴ v=dsdt=22−24t

When the car stop, v=0,

⇒22−24t=0

⇒t=1112

∴ s=22(1112)−12(1112)2=10.08 ft

KEY CONCEPTS

Rate Measure in Calculus The rate of change of \(y=f(x)\) at a point \(x=a\) is defined as the slope of the tangent to the curve at \(x\) \(=\) al i.e. rate \(=d y d x a t x=a\)
If a variable quantity y is some function of time t i.e., \(y=f(t),\) then small change in time \(\Delta t\) have a corresponding change \(\Delta y\) in \(y .\) Thus, average rate of change \(=\Delta y \Delta t\)
When limit \(\Delta t \rightarrow 0\) is applied, the rate of change becomes instantaneous and we get the rate of change with respect to \(t\) at the instant \(t\)
¡i.e. \(\lim \Delta \mathrm{t} \rightarrow 0 \Delta \mathrm{y} \Delta \mathrm{t}=\mathrm{dydt}\)
Hence, it is clear that the rate of change of any variable with respect to some other variable is derivative of first variable with respect to other variable.
Question 6
1 / -0

The minimum value of linear objective function z=2x+2y, under linear constraints 3x+2y≥12,, x+3y≥11 and x, y≥0x,, is

A
10

B
12

C
6

D
5

Solution

The graph of feasible region for the given L.P.P. is

The values of objective function at the corner point of the feasible region are
\(z=[2 x+2 y]_{\mathrm{B}(0,6)}=12\)
\(z=[2 x+2 y]_{\mathrm{E}(2,3)}=10\)
\(z=[2 x+2 y]_{\mathrm{C}(11,0)}=22\)
So, the minimum value of the objective function is 10

KEY CONCEPTS

Graphical solution of two variable linear programming problem (Iso-profit or Iso-cost method)
Iso-profit or 1 so -cost method : Various steps of the method are as follows:
(1) Find the feasible region of the LP.P.
(ii) Assign a constant value \(Z_{1}\) to \(Z\) and draw the corresponding line of the objective function.
(iii) Assign another value \(Z_{2}\) to \(Z\) and draw the corresponding line of the objective function.
(iv) if \(Z_{1}\) is less than \(Z_{2}\) then in case of maximization (minimization) move the line \(P_{1} Q_{1}\) corresponding to \(Z_{1}\) to the line \(P_{2} Q_{2}\) corresponding to \(Z_{2}\) paraliel to itself as far as possible, until the farthest point within the feasible region is touched by this line. The coordinates of the point give maximum (minimum) value of the objective function.
(v) The problem with more equations/inequations can be handled easily by this method.
(vi) In case of unbounded region, it either finds an optimal solution or declares an unbounded solution. Unbounded solutions are not considered optimal solution. In real worid problems, unlimited profit or loss is not possible.
Question 7
1 / -0

The equation y=±√3x, y=1 are the sides of

A
An equilateral triangle

B
A right angled triangle

C
An isosceles triangle

D
An obtuse triangle

Solution
Question 8
1 / -0

In the expansion of (1+x)⁵⁰ the sum of the coefficients of the odd powers of x is

A
0

B
2⁴⁹

C
2⁵⁰

D
2⁵¹

Solution

\((1+x)^{50}=1+^{50} C_{1} x^{1}+^{50} C_{2} x^{2}+^{50} C_{3} x^{3}+\ldots . . . +^{50} C_{49} x^{49}+\)\(^{50} \mathrm{C}_{50} \times^{50}\)
the sum of the coefficients of odd powers of \(x\) is:
\(=^{50} \mathrm{C}_{1}+^{50} \mathrm{C}_{3}+\ldots \ldots \ldots \ldots \ldots . .+^{50} \mathrm{C}_{49}\)
\((1-x)^{50}=1+^{50} C_{1}(-x)^{1}+^{50} C_{2}(-x)^{2}+^{50} C_{3}(-x)^{3}+\ldots \ldots \ldots \ldots . .+^{50} C_{49}(-x)^{49}+\)\(^{50} \mathrm{C}_{50}(-x)^{50}\)
\(=1-^{50} \mathrm{C}_{1} \mathrm{x}^{1}+^{50} \mathrm{C}_{2} \mathrm{x}^{2}-^{50} \mathrm{C}_{3} \mathrm{x}^{3}+\ldots \ldots \ldots \ldots \ldots . .^{50} \mathrm{C}_{49} \mathrm{x}^{49}+^{50} \mathrm{C}_{50} \mathrm{x}^{50}\)
\((1+x)^{50}-(1-x)^{50}=2\left(^{50} C_{1} x^{1}+^{50} C_{3} x^{3}+\ldots . . . ^{50} C_{49} x^{49}\right)\)
putting \(x=1\)
\(\Rightarrow 2^{50}-0=2\left(^{50} \mathrm{C}_{1}+^{50} \mathrm{C}_{3}+\ldots \ldots . . . .+^{\mathrm{so}} \mathrm{C}_{49}\right)\)
\(\Rightarrow 2^{50}=2\left(^{50} \mathrm{C}_{1}+^{50} \mathrm{C}_{3}+\ldots . . . . .+^{50} \mathrm{C}_{49}\right)\)
\(\Rightarrow^{50} \mathrm{C}_{1}+^{50} \mathrm{C}_{3}+\ldots \ldots \ldots . . . ..^{50} \mathrm{C}_{49}=2^{50} / 2\)
\(\Rightarrow^{50} \mathrm{C}_{1}+^{50} \mathrm{C}_{3}+\ldots \ldots \ldots . . .+^{50} \mathrm{C}_{49}=2^{49}\)
Question 9
1 / -0

The number of ways in which 7 persons can address a meeting so that out of the three persons A, B and C, A will speak before B and B before C, is

A
1680

B
840

C
120

D
32

Solution

We can solve this problem using fundamental principle of counting.

Fundamental Principle of Multiplication:

If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then two jobs in succession can be completed in m × n ways.

By using this definition we can find the answer.

Let the 7 people be A, B, C, D, E, F and G. There are 7 positions to be filled D, E, F and G can occupy any of the 7 positions.

Thus :

Number of options for D = 7

(Any of the 7 position)

Number of options for E = 6

(Any of the 6 remaining positions)

Number of options for F = 5

(Any of the 5 remaining positions)

Number of options for G = 4

(Any of the 4 remaining positions)

Since, A must speak before B and B must speak before C, the 3 remaining positions must be occupied as follows : A - B - C

Thus,

number of options for A = 1

(A must occupy the leftmost remaining position)

Number of options for C = 1

(C must occupy the right most remaining positions)

Number of options for B = 1

(Only 1 position left)

To combine all of these options, we multiply :

7 × 6 × 5 × 4 × 1 × 1 × 1 = 840

Hence, the number of ways in which 7 persons can address a meeting so that out of the three persons A, B and C, A will speak before B and B before C, is 840.

Another method :

The same problem can be done by permutation also. If you do it by permutation we get

We have

Let the 7 people be A, B, C, E, F and G. Out of the three persons A, B and C. A will speak before B and B before C. Then arrange the remaining 4 peoples in 7 places. This is done in7P4ways. Then arrange ABC is

7P4= 7 × 6 × 5 × 4 × 1 × 1 × 1 = 840
Question 10
1 / -0

Solution of dy / dx = xlogx²+x / sin y+ycos y is

A
y sin y= x²logx+c

B
y sin y= x²+c

C
y sin y =x²+log x

D
y sin y = x log x + c

Solution

Solution of differential equations

Solution of differential equation in many type form of equation.

(1) Differential equation of the first order and first degree.

(2) Equation in which the variables are separable.

(3) Differential equation reducible to the variable separable type.

(4) Homogeneous differential equation.

(5) First order liner differential equation.

(6) Exact differential equation.

(7) Bernoulli’s equation.

Variable Separable form of Differential

Equations

(A) Equations in variable separable form: If the differential equation of the form f(x)dx=g(y)dy .....(i)

Where f(x) and g(y) functions of x and y only. Then we say that the variables are separable in the differential equation.

Thus, integrating both sides of (i), we get its solution as, where c is an arbitrary constant.

There is no need of introducing arbitrary constants to both sides as they can be combined together to give just one.

Exact Differential Equation

If M and N are functions of x and y, the equation Mdx + Ndy = 0 is called exact when it can be derived by direct differentiation of an equation of the form f(x,y)=C

∂M∂y=∂N∂x

An exact differential equation can always be derived from its general solution directly by differentiation without any subsequent multiplication, elimination etc.

If an equation of the form Mdx + Ndy = 0 is not exact, it can always be made exact by multiplying by some function of x and y. Such a multiplier is called an integrating factor.

(a) xdy+ydx=d(xy)

(b) xdy−ydxx2=d(yx)

(c) xdy−ydxxydln(yx)

(d) xdy+ydxxy=d(lnxy)

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Mathematics Test - 35

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Mathematics Test - 35

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Question 1

tanx−secx+c

loge|1+secx|+c

secx+tanx+c

log|sinx|+log|cosx|+c

Solution

Question 2

Ravi does not race and Ravi does not win

It is not true that Ravi races and that Ravi does not win

Ravi races and Ravi wins

It is not true that Ravi races or that Ravi does not win

Solution

Question 3

1/256

3/256

9/256

7/256

Solution

Question 4

A hyperbola

An ellipse

A parabola

A straight line

Solution

Question 5

10.08 ft

10 ft

11 ft

11.5 ft

Solution

Question 6

10

12

6

5

Solution

Question 7

An equilateral triangle

A right angled triangle

An isosceles triangle

An obtuse triangle

Solution

Question 8

0

249

250

251

Solution

Question 9

1680

840

120

32

Solution

Question 10

y sin y= x2logx+c

y sin y= x2+c

y sin y =x2+log x

y sin y = x log x + c

Solution

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2⁴⁹

2⁵⁰

2⁵¹

y sin y= x²logx+c

y sin y= x²+c

y sin y =x²+log x