JEE MATHS

DAILY MATHS TEST QUESTIONS

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1 Fundamentals of Mathematics

Nine-fifth of the temperature in centrigrade (C) of a body plus 32^o is equal to the temperature of the same body in Fahrenheit (F). Find F, when C=55^o

1) 131^o

2) 1201^o

3) 140^o

4) 166^o

Option 1 Solution:

Solution

According to the given condition,

$$F= \dfrac { 9 }{ 5 } C +32$$

$$\therefore F = \dfrac { 9 }{ 5 } (55) +32$$

$$= 99+32 $$ $$=131^o$$

Option 2

Option 3

Option 4

2 Trigonometric Equations

The value of cot (cosec^-1 5/3 + tan^-1 2/3) is

1) 4/17
2) 5/17
3) 6/17
4) 3/17

Option 1

Option 2

Option 3 Solution:

SOLUTION

Option 4

3 Limits Continuity and Differentiability

Which one of the following is not correct ?

1) If f(x) is not continuous at x = a, then it is not differentiable at x = a
2) If f(x) is continuous at x = a, then it is differentiable at x = a
3) If f(x) and g(x) are differentiable at x = a, then f(x) + g(x) is also differentiable at x = a
4) If f(x) is continuous at x = a, then exists

Option 1

Option 2 Solution:

SOLUTION

Option 3

Option 4

4 Probability

For a biased die the probabilities for different faces to turn up are given below: The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is .....

Face	1	2	3	4	5	6
Probability	0.1	0.32	0.21	0.15	0.05	0.17.

1) $$\displaystyle \frac{5}{42}$$
2) $$\displaystyle \frac{16}{21}$$
3) $$\displaystyle \frac{5}{21}$$
4) $$\displaystyle \frac{16}{42}$$

Option 1

Option 2

Option 3 Solution:

Solution

Let $$A$$ be the event that face 1 turns up
$$ B$$ be the event that face 2 turns up.
Then $$\displaystyle P\left ( A \right )= 0.10$$ and $$\displaystyle P\left ( B \right )= 0.32,$$
Since A, B are mutually exclusive, we have
$$\displaystyle P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )= 0.10+0.32= 0.42.$$
We are to find $$\displaystyle P\left ( A|A\cup B \right )$$
But $$\displaystyle P\left ( A|A\cup B \right )= \frac{P\left [ A\cap \left ( A\cup B \right ) \right ]}{P\left ( A\cup B \right )}$$

$$= \dfrac{P\left ( A \right )}{P\left ( A\cup B \right )}$$

Hence $$\displaystyle P\left ( A|A\cup B \right )= \frac{0.10}{0.42}= \frac{5}{21}$$

Option 4

5 Probability

In a family the husband tells a lie in $$30\%$$ and the wife in $$35\%$$ cases. Find the probability that both contradict each other.

1) $$ = 34\% $$
2) $$ = 44\% $$
3) $$ = 25\% $$
4) $$ = 35\% $$

Option 1

Option 2 Solution:

Solution

The husband tells lie in $$30\% = 0.3$$ cases.

The wife tells lie in $$35\% = 0.35$$ cases.

The probability that they both will contradict is,

$$P = P\left( {{\rm{husband}}\;{\rm{lies}}\;{\rm{but}}\;{\rm{wife}}\;{\rm{don't}}} \right) + P\left( {{\rm{wife}}\;{\rm{lies}}\;{\rm{but}}\;{\rm{husband}}\;{\rm{don't}}} \right)$$

$$ = 0.3\left( {1 - 0.35} \right) + \left( {0.35} \right)\left( {1 - 0.3} \right)$$

$$ = 0.195 + 0.245$$

$$ = 0.44$$

$$ = 44\% $$

Option 3

Option 4

6 Binomial Theorem

The coefficient of three consecutive of (1 + χ)ⁿ⁺⁵ are in the ratio 5 : 10 : 14. Then, n is equal to

1) 1
2) 3
3) 4
4) 6

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

7 Sets relations and functions

Find the range of the function f (x) = sin ^-¹ $\left[ {{x}^{2}}+\cfrac{1}{2} \right]$ + cos ^-¹ $\left[ {{x}^{2}}-\cfrac{1}{2} \right]$$\left[ {{x}^{2}}-\cfrac{1}{2} \right]$, where [ . ] is the greatest integer function.

1)$\left\{ 0,-\cfrac{1}{2} \right\}$
2) $\left\{ \cfrac{\pi }{2},\pi \right\}$
3) $\left( 0,\cfrac{\pi }{2} \right)$
4) {π}

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

$ f (x) = sin -1 \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+$cos -1$\left[ {{x}^{2}}-\cfrac{1}{2} \right]$   = sin -1 $\left[ {{x}^{2}}+\cfrac{1}{2} \right]+$ cos -1 $\left[ {{x}^{2}}+\cfrac{1}{2}-1 \right]$   = sin -1 $\left[ {{x}^{2}}+\cfrac{1}{2} \right]+$ cos -1 $\left( \left[ {{x}^{2}}+\cfrac{1}{2} \right]-1 \right)$   Since ${{x}^{2}}+\cfrac{1}{2}$ $\ge \cfrac{1}{2}$, $\left[ {{x}^{2}}+\cfrac{1}{2} \right]$ = 0 or 1 as Sin -1$\left[ {{x}^{2}}+\cfrac{1}{2} \right]$ is defined only for these two values   Hence $\left[ {{x}^{2}}+\cfrac{1}{2} \right]=$ 0   ⇒f (x) = sin -1 0 + cos -1 ( - 1) = $\pi \left[ {{x}^{2}}+\cfrac{1}{2} \right]=$ 1 ⇒ f (x) = sin -1 (1) + cos -10 = π Therefore range of f (x) = {$\pi $}\)

8 Probability

In a college, 25% boys and 10% girls offer mathematics. There are 60% girls in the college. If a Mathematics student is chosen at random, then the probability that the student is a girl, will be

1) 1/6
2) 3/8
3) 5/8
4) 5/6

Option 1

Option 2 Solution:

SOLUTION

Option 3

Option 4

9 Probability

In a hurdle race, a runner has probability $$p$$ of jumping over a specific hurdle. Given that in $$5$$ trials, the rummer succeeded $$3$$ times, the conditional probability that the runner had succeeded in the first trial is

1) $$\dfrac {3}{5}$$
2) $$\dfrac {2}{5}$$
3) $$\dfrac {1}{5}$$
4) $$\dfrac {4}{5}$$

Option 1 Solution:

Solution

Let $$A$$ denote the event that the runner succeeds exactly $$3$$ times out of five and $$B$$ denote the event that the number succeeds on the first trial.
$$\displaystyle P(B|A)=\frac {P(B\cap A)}{P(A)}$$
But $$P(B\cap A)=P$$ (succeeding in the first trial and exactly once in two other trials)
$$=p(^4C_2p^2(1-p)^2)=6p^3(1-p)^2$$
and $$P(A)=^5C_3p^3(1-p)^2=10p^3(1-p)^2$$
Thus, $$\displaystyle P(B|A)=\frac {6p^3(1-p)^2}{10p^3(1-p)^2}=\frac {3}{5}$$.

Option 2

Option 3

Option 4

10 Differentiation

If the functions f(x)=sin(x+a) and g(x)=bsinx+ccosx satisfy f(0)=g(0) and f′(0)=g′(0) then

1) b=π / 2?
2) b=cosa
3) c=sina
4) c=cosa

Option 1

Option 2 Solution:

SOLUTION

Given, f(x)=sin(x+a)

and g(x)=bsinx+ccosx
f(0)=g(0)⇒sina=c
f′(x)=cos(x+a)
g′(x)=bcosx−csinx
f′(0)=g′(0)⇒cosa=b

Option 3

Option 4

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