Mathematics Test

Result

Mathematics Test - 49

/

Score
-

Rank

Time Taken: -

Question 1/10
1 / -0

A point $R$ with $x$ -co-ordinate 4 lies on the line segment joining the points $P (2, - 3, 4)$ and $Q (8, 0, 10)$ . Find the co-ordinates of thepoint R.

Correct

Wrong

Skipped
A
(4,-2,6)

Correct

Wrong

Skipped
B
(2,-4,6)

Correct

Wrong

Skipped
C
(3,-3,6)

Correct

Wrong

Skipped
D
(1,-3,4)

Solutions

Let R divide $PQ$ in the ratio $k : 1$
where $P$ is (2,-3,4) and $Q$ is (8,0,10)
$∴$ The co-ordinates of $R$ are $(\frac{8 k + 2}{k + 1}, \frac{- 3}{k + 1}, \frac{10 k + 4}{k + 1})$
By the question, $\frac{8 k + 2}{k + 1} = 4$
$\Rightarrow 8 k + 2 = 4 k + 4 \Rightarrow 4 k = 2 \Rightarrow k = \frac{1}{2}$
From $(1),$ co-ordinates of $R$ are $: (\frac{8 (\frac{1}{2}) + 2}{\frac{1}{2} + 1}, \frac{- 3}{\frac{1}{2} + 1}, \frac{10 (\frac{1}{2}) + 4}{\frac{1}{2} + 1})$
i.e. $(\frac{4 + 2}{\frac{3}{2}}, \frac{- 3}{\frac{3}{2}}, \frac{5 + 4}{\frac{3}{2}})$ i.e. (4,-2,6)
Hence, the correct option is (A)
Question 2/10
1 / -0

The points $A, B$ and $C$ represent the complex numbers $z_{1}, z_{2}, (1 - i) z_{1} + i z_{2} ($ where $i^{2} = \sqrt{- 1})$ respectively on the complex plane. The triangle $ABC$ is :

Correct

Wrong

Skipped
A
Isosceles but not right angled

Correct

Wrong

Skipped
B
Right angled but not isosceles

Correct

Wrong

Skipped
C
Isosceles and right angled

Correct

Wrong

Skipped
D
None of these

Solutions

since $A \equiv z_{1}, B \equiv z_{2}, C \equiv (1 - i) z_{1} + i z_{2}$

$∴ A B = | z_{1} - z_{2} |, B C = | (1 - i) z_{1} + i z_{2} - z_{2} |$

$= | 1 - i | | z_{1} - z_{2} | = \sqrt{2} | z_{1} - z_{2} |$

$C A = | (1 - i) z_{1} + i z_{2} - z_{1} |$

and $= | i ‖ - z_{1} + z_{2} | = | z_{1} - z_{2} |$

$∴ A B = C A$ and $(A B)^{2} + (C A)^{2} = (B C)^{2}$

Hence the triangle is isosceles and right angled.

Hence, the correct option is (C)
Question 3/10
1 / -0

Let $f (x)$ be a quadratic expression which is positive for all real $x$ . If $g (x) = f (x) - f^{'} (x) + f^{''} (x),$ then for any real $x$ :

Correct

Wrong

Skipped
A
$g (x) \leq 0$

Correct

Wrong

Skipped
B
$g (x) < 0$

Correct

Wrong

Skipped
C
$g (x) \geq 0$

Correct

Wrong

Skipped
D
$g (x) > 0$

Solutions

Let $f (x) = a x^{2} + b x + c > 0, \forall x \in R$
$\Rightarrow b^{2} - 4 a c < 0 & a > 0 \dots$ (i)
Now, $g (x) = f (x) - f^{'} (x) + f^{'}$ " $(x)$
Discriminant $= a x^{2} + (b - 2 a) x + (2 a - b + c)$
$= (b - 2 a)^{2} - 4 a (2 a - b + c)$
$= (b^{2} - 4 a c) - 4 a^{2} < 0$
$\Rightarrow g (x) > 0, \forall x \in R$
Hence, the correct option is (D)
Question 4/10
1 / -0

If the area bounded by y=ax² and x=ay², a>0, is 1, then a is equal to :

Correct

Wrong

Skipped
A
1

Correct

Wrong

Skipped
B
$\frac{1}{\sqrt{3}}$

Correct

Wrong

Skipped
C
$\frac{1}{3}$

Correct

Wrong

Skipped
D
$- \frac{1}{\sqrt{3}}$

Solutions

$y^{2} = \frac{x}{a}, y = a x^{2}$
solving simultaneously
$\Rightarrow a^{2} x^{4} = \frac{x}{a}$
$\Rightarrow x = 0$ or $x^{3} = \frac{1}{a^{3}} \Rightarrow x = \frac{1}{a} \Rightarrow y = \frac{1}{a}$
$∴ (0, 0)$ and $(\frac{1}{a}, \frac{1}{a})$ are the points of intersection
\thereforearea between the given curves
$= | \int_{0}^{1 / a} \frac{\sqrt{x}}{\sqrt{a}} - a x^{2} d x |$
$= {| \frac{x^{3 / 2}}{\sqrt{a} (\frac{3}{2})} - \frac{a x^{3}}{3} |}_{0}^{\frac{1}{a}}$
$= | [\frac{2}{3} \frac{1}{a^{2}} - \frac{1}{3 a^{2}}] |$
$= | \frac{1}{3 a^{2}} | = 1 ($ given $)$
$\Rightarrow a^{2} = \frac{1}{3} \Rightarrow a = + \frac{1}{\sqrt{3}} (∵ a > 0)$
Hence, the correct option is (B)
Question 5/10
1 / -0

Tangents to the parabola $y^{2} = 16 x$ ,which are parallel & perpendicular to the line
$2 x - y + 5 = 0.$
Then the coordinates of their points of contact are:

Correct

Wrong

Skipped
A
(16, 16), (1, -4)

Correct

Wrong

Skipped
B
(-16, -16), (-1, 4)

Correct

Wrong

Skipped
C
(-16, 16), (1, 4)

Correct

Wrong

Skipped
D
(16, -16), (1, 4)

Solutions

We know that $y = m x + \frac{a}{m} \dots \dots . (1)$ is a tangent to the parabola $y^{2} = 4 a x$ and the point of contact is $(\frac{a}{m^{2}}, \frac{2 a}{m}$
Here, $y^{2} = 16 x$
$∴ 4 a = 16 \Rightarrow a = 4$
Slope of the line $2 x - y + 5 = 0$ is 2
Any tangent parallel to it will have its slope 2 and any tangent perpendicular to it will have slope $- \frac{1}{2}$
Hence putting a $= 4$ and $, m = 2$ in equation (1) then the equation of tangent is
$y = 2 x + \frac{4}{2} \Rightarrow 2 x - y + 2 = 0$ at (1,4)
for $m = - \frac{1}{2}$ the equation of the tangent is
$y = - \frac{1}{2} x + \frac{4}{(- \frac{1}{2})}$
or $x + 2 y + 16 = 0$ at (16,-16)
Hence, the correct option is (D)
Question 6/10
1 / -0

A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel, is :

Correct

Wrong

Skipped
A
$\frac{2}{9}$

Correct

Wrong

Skipped
B
$\frac{4}{9}$

Correct

Wrong

Skipped
C
$\frac{3}{9}$

Correct

Wrong

Skipped
D
None of these

Solutions

Out of 11 letters in the word PROBABILITY
O, A, I and I are the vowels.
So probability that the letter selected is a vowel, is 4/11.

Classical Definition of Probability If a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes, out of which m are favourable to the occurrence of an event A, then the probability of occurrence of A is given by
Question 7/10
1 / -0

Find the derivative with respect to $x$ of the function

$(\log_{\cos x} \sin x) {(\log_{\sin x} \cos x)}^{- 1} + \arcsin \frac{2 x}{1 + x^{2}}$ at $x = \frac{π}{4}$

Correct

Wrong

Skipped
A
$- \frac{8}{\ln 2} + \frac{32}{16 + π^{2}}$

Correct

Wrong

Skipped
B

Correct

Wrong

Skipped
C
$- \frac{8}{ℓ n 2} + \frac{32}{16 - π^{2}}$

Correct

Wrong

Skipped
D

Solutions

$\begin{aligned} y = (\log_{\cos x} \sin x) {(\log_{\sin x} \cos x)}^{- 1} + \arcsin \frac{2 x}{1 + x^{2}} \\ = \frac{\ln \sin x}{\ln \cos x} {(\frac{\ln \cos x}{\ln \sin x})}^{- 1} + 2 \tan^{- 1} x \\ y = {(\frac{\ln \sin x}{\ln \cos x})}^{2} + 2 \tan^{- 1} x \\ \Rightarrow \frac{dy}{d x} = 2 (\frac{ℓ n \sin x}{ℓ n \cos x}) \frac{(ℓ n \cos x) \frac{\cos x}{\sin x} + (ℓ n \sin x) \frac{\sin x}{\cos x}}{(\ln \cos x)^{2}} + \frac{2}{1 + x^{2}} [Using chain rule] \\ {\Rightarrow \frac{d y}{d x} |}_{\frac{π}{4}} = 2 \frac{(1 + 1)}{(\ln \frac{1}{\sqrt{2}})} + \frac{2}{1 + {(\frac{π}{4})}^{2}} \\ = \frac{2 \times 2}{(\frac{- 1}{2}) \ln 2} + \frac{2}{1 + {(\frac{π}{4})}^{2}} \\ = - \frac{8}{\ln 2} + \frac{32}{16 + π^{2}} \end{aligned}$
Hence, the correct option is (A)
Question 8/10
1 / -0

If ABC be a triangle with $∠ B A C = 2 π / 3$ and AB = x such that (AB) (AC) = 1. If x varies, then the longest possible length of the angle bisector AD is

Correct

Wrong

Skipped
A
$\frac{1}{4}$

Correct

Wrong

Skipped
B
$\frac{1}{8}$

Correct

Wrong

Skipped
C
$\frac{1}{2}$

Correct

Wrong

Skipped
D
$\frac{1}{3}$

Solutions

$AD = y = \frac{2 bc}{b + c} \cos \frac{A}{2} = \frac{bx}{b + x} (asc = x)$

But $b x = 1$ or $b = \frac{1}{x}$

$∴ y = \frac{x}{1 + x^{2}} = \frac{1}{x + \frac{1}{x}}$

Thus $y_{max} = \frac{1}{2}$ , since the minimum value of the denominator is 2 if $x > 0$
Hence, the correct option is (C)
Question 9/10
1 / -0

Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number.

Correct

Wrong

Skipped
A
$\frac{4}{7}$

Correct

Wrong

Skipped
B
$\frac{8}{7}$

Correct

Wrong

Skipped
C
$\frac{6}{7}$

Correct

Wrong

Skipped
D
$\frac{2}{7}$

Solutions

Let the events be :

A : Number on the card drawn is even.

B : Number on the card is more than 3.

We are to find P(A/B).

We have : S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Then A = {2, 4, 6, 8, 10}

andB = {4, 5, 6, 7, 8, 9, 10}
$∴ A \cap B = \{4, 6, 8, 10\}$
$∴ P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{4}{10}}{\frac{7}{10}} = \frac{4}{7}$
Hence, the correct option is (A)

Question 10/10

1 / -0

Which of the following statement is a tautology -

Correct

Wrong

Skipped

(∼q∧p)∧q

Correct

Wrong

Skipped

(∼q∧p)∧(p∧∼p)

Correct

Wrong

Skipped

(∼q∧p)∨(p∧∼p)

Correct

Wrong

Skipped

(p∧q)∨(∼(p∧q)

Solutions

Truth Table

p	q	$~ p$	$~ q$	$~ q \land p$	$p \land ~ p$	$(~ q \land p) \land p$	$(~ q \land p) \land (p \land ~ p)$	$(- p \land q) \lor (p \lor - p)$	$(p \land q) \lor (~ (p \land q))$
T	T	F	F	F	F	F	F	F	T
T	F	F	T	F	F	F	T	T	T
F	T	T	F	F	F	F	F	F	T
F	F	T	T	F	F	F	F	F	T

Validity of statements in Mathematical Reasoning

In this Section, we will discuss when a statement is true. To answer this question, one must answer all the following questions.

What does the statement mean?

What would it mean to say that this statement is true and when this statement is not true?

The answer to these questions depend upon which of the special words and phrases and, or, and which of the implications if and only, if-then, and which of the quantifiers for every, there exists, appear in the given statement. Here, we shall discuss some techniques to find when a statement is valid. We shall list some general rules for checking whether a statement is true or not.

Rule 1 If p and q are mathematical statements, then in order to show that the statement p and q is true, the following steps are followed.

Step-1 Show that the statement p is true.

Step-2 Show that the statement q is true.

Rule 2 Statements with Or If p and q are mathematical statements, then in order to show that the statement p or q is true, one must consider the following.

Case 1 By assuming that p is false, show that q must be true.

Case 2 By assuming that q is false, show that p must be true.

Rule 3 Statements with If-then In order to prove the statement if p then q we need to show that any one of the following case is true.

Case 1 By assuming that p is true, prove that q must be true. (Direct method)

Case 2 By assuming that q is false, prove that p must be false. (Contrapositive method)

Rule 4 Statements with if and only if In order to prove the statement p if and only if q, we need to show.

(i) If p is true, then q is true and

(ii) If q is true, then p is true.

Hence, the correct option is (D)

Question 1/10

(4,-2,6)

(2,-4,6)

(3,-3,6)

(1,-3,4)

Question 2/10

Isosceles but not right angled

Right angled but not isosceles

Isosceles and right angled

None of these

Question 3/10

g(x)≤0

g(x)<0

g(x)≥0

g(x)>0

Question 4/10

1

13

13

−13

Question 5/10

(16, 16), (1, -4)

(-16, -16), (-1, 4)

(-16, 16), (1, 4)

(16, -16), (1, 4)

Question 6/10

29

49

39

None of these

Question 7/10

−8ln⁡2+3216+π2

+8ln⁡2+3216+π2

−8ℓn2+3216−π2

+8ln⁡2−3216−π2

Question 8/10

14

18

12

13

Question 9/10

47

87

67

27

Question 10/10

(∼q∧p)∧q

(∼q∧p)∧(p∧∼p)

(∼q∧p)∨(p∧∼p)

(p∧q)∨(∼(p∧q)

$g (x) \leq 0$

$g (x) < 0$

$g (x) \geq 0$

$g (x) > 0$

$\frac{1}{\sqrt{3}}$

$\frac{1}{3}$

$- \frac{1}{\sqrt{3}}$

$\frac{2}{9}$

$\frac{4}{9}$

$\frac{3}{9}$

$- \frac{8}{\ln 2} + \frac{32}{16 + π^{2}}$

$- \frac{8}{ℓ n 2} + \frac{32}{16 - π^{2}}$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{4}{7}$

$\frac{8}{7}$

$\frac{6}{7}$

$\frac{2}{7}$