JEE MATHS

DAILY MATHS TEST QUESTIONS

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1 Differentiation

If f(x) = √ (1+√x?) , x > 0, then f(x) ⋅ f′(x) is equal to

1) ?1 / 2√x? .
2) 1 / 2
3) 1 / 4√x
4) (2√x?+1) / 4√x
.

Option 1

Option 2

Option 3 Solution:

Solution

f(x) = √ (1+√x?)

then, y²=1 +√x?

Differentiating w.r.t. x, we get,

Option 4

2 Complex Numbers

If the imaginary part of the expression $$\cfrac{z-1}{{{e}^{\theta t}}}+\cfrac{{{e}^{\theta t}}}{z-1}$$ be zero, then the locus of z can be

1) A parabola
2) A circle of radius 1
3) A straight line parallel to x - axis
4) A rectangle hyperbola

Option 1

Option 2 Solution:

SOLUTION

Let $$u=\cfrac{z-1}{{{e}^{\theta i}}}\Rightarrow \cfrac{{{e}^{\theta i}}}{z-1}=\cfrac{1}{u},$$ ,
Now $$\left( u+\cfrac{1}{u} \right)-\left( \bar{u}+\cfrac{1}{{\bar{u}}} \right)=0$$

$$\Rightarrow \left( u-\bar{u} \right)\left( 1-\cfrac{1}{u\bar{u}} \right)=0$$

If u is not purely real, then

$$u\bar{u}=1\Rightarrow \left| \cfrac{z-1}{{{e}^{\theta i}}} \right|=1\Rightarrow \left| z-1 \right|=1$$

Option 3

Option 4

3 Probability

On a busy road in a city the number of persons sitting in the cars passing by were observed during a particular interval of time. Data of $$60$$ such cars is given in the following table.

No. of persons in the car	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$
No of Cars	$$22$$	$$16$$	$$12$$	$$6$$	$$4$$

Suppose another car passes by after this time interval. Find the probability that it has at least $$4$$ persons in it

1) $$\dfrac {1}{6}$$
2) $$\dfrac {1}{12}$$
3) $$\dfrac {1}{10}$$
4) None of these

Option 1 Solution:

Solution

$$P$$(at least $$4$$) $$=\dfrac{(4+5)}{Total}$$

$$=\dfrac{6+4}{60}=\dfrac{1}{6}$$

Option 2

Option 3

Option 4

4 Trigonometric Equations

The number of solutions of the equation 6cos⁵θ – 6cos⁴θ – 5cos³θ + 5cos²θ + cosθ – 1 = 0, where 0 < q < 360° is-

1) 2
2) 4
3) 6
4) 8

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

5 Matrices

Let A, B, C, D be (not necessarily square) real matrices such that A^T=BCD;B^T=CDA;C^T=DAB and D^T=ABC
for the matrix S=ABCD, consider the two statements.
I S³=S
II S²=S4

1) II is true but not I
2) I is true but not II
3) Both I and II are true
4) Both I and II are false

Option 1

Option 2 Solution:

Solution

$${ A }^{ T }=BCD$$
$${ B }^{ T }=CDA$$
$${ C }^{ T }=DAB$$
$${ D }^{ T }=ABC$$
$$S=ABCD$$
$${ S }^{ T }={ \left[ ABCD \right] }^{ T }$$
$$={ D }^{ T }{ C }^{ T }{ B }^{ T }{ A }^{ T }$$
$$=(ABC)(DAB)(CDA)(BCD)$$
$$=A(BCD)A(BCD)A(BCD)$$
$$=\left( A{ A }^{ T } \right) \left( A{ A }^{ T } \right) \left( A{ A }^{ T } \right) $$
$$={ \left( A{ A }^{ T } \right) }^{ 3 }$$
$$={ (ABCD) }^{ 3 }$$
$$={ S }^{ 3 }$$
$$\therefore { S }^{ 3 }={ S }^{ T }\Rightarrow \therefore $$ I is true
$${ S }^{ 2 }={ (ABCD) }^{ 2 }$$
$$={ (A{ A }^{ T } })^{ 2 }$$
$${ S }^{ 4 }={ (ABCD) }^{ 4 }$$
$$={ \left( A{ A }^{ T } \right) }^{ 4 }$$
$${ S }^{ 2 }\neq { S }^{ 4 }\Rightarrow \therefore $$ II is not true
$$\therefore $$ I is true but not II

Option 3

Option 4

6 Differentiation

f(x) = ?x−1?+?x−3? then f′(2)=

1) ?−2? .
2) 2
3) 0
4) 1
.

Option 1

Option 2

Option 3 Solution:

Solution

Hence f′(2)=0

Option 4

7 Differentiation

$$\dfrac{dy}{dx}=e^{-2y},x=5\Rightarrow y=0,$$ then  $$y=3,find\ the\ value\ of\ 2x$$.

1) e⁵+ 9
2) e⁶+ 9
3) e⁸+ 9
4) e⁴+ 9
.

Option 1

Option 2 Solution:

Solution

$$\displaystyle\Rightarrow \int e^{2y}dy=\int dx +c$$
$$\Rightarrow \dfrac{e^{2y}}{2}=x+c$$
$$\Rightarrow e^{2y}=2x+c$$
put $$(x,y)=(5,0)$$
$$\Rightarrow 1=10+c$$ $$\Rightarrow c=-9$$
$$\Rightarrow e^{2y}=2x-9$$
$$\Rightarrow 2x=e^{2y}+9$$
$$\Rightarrow$$ $$ for\ y=3$$
$$2x=e^{6}+9$$

Option 3

Option 4

8 Probability

A room has three electric lamps. From a collection of 10 electric bulbs of which 6 are good 3 are selected at random & put in the lamps. Find the probability that the room is lighted

1)$$\displaystyle \frac {29}{30}$$

2)$$\displaystyle \frac {1}{30}$$

3)$$\displaystyle \frac {23}{30}$$

4)$$\displaystyle \frac {7}{30}$$

Option 1 Solution:

Solution

A room has $$3$$ electric lamps.

Out of $$10$$ bulbs present, $$6$$ bulbs are good.
$$\Rightarrow$$ $$4$$ bulbs are defective.
The probability that selected $$3$$ bulbs are good $$=1-$$ (Probability that selected $$3$$ bulbs are defective)
$$=1-\displaystyle\dfrac{{}^{4}C_{3}}{{}^{10}C_{3}}$$
$$=1-\displaystyle\dfrac{(4\times 3\times 2)}{(10\times 9\times 8)}=1-\dfrac{1}{30}$$
$$=\displaystyle\frac{29}{30}$$

Option 2

Option 3

Option 4

9 Matrices

For all values of $$\Theta \in \left( 0 , \dfrac { \pi } { 2 } \right)$$ the determinants of the matrix $$\left[ \begin{array} { c c c } { - 2 } & { \tan \theta + \sec ^ { 2 } \theta } & { 3 } \\ { - \sin \theta } & { \cos \theta } & { \sin \theta } \\ { - 3 } & { 4 } & { 3 } \end{array} \right]$$ always lies in the interval:

1) (-4,3)
2) (1,2)
3) (4,3)
4) (4, 5)

Option 1 Solution:

Solution

We have,

$$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ -2 } & { \tan \theta +{ { \sec }^{ 2 } }\theta } & 3 \\ { -\sin \theta } & { \cos \theta } & { \sin \theta } \\ { -3 } & 4 & 3 \end{array} } \right| =-2\left( { 3\cos \theta -4\sin \theta } \right) -\left( { { { \sec }^{ 2 } }\theta +\tan \theta } \right) \left( 0 \right) +3\left( { -4\sin \theta +3\cos \theta } \right) \\ =-6\cos \theta +8\sin \theta -12\sin \theta +9\cos \theta \\ =-4\sin \theta +3\cos \theta \\ f\left( \theta \right) =3\cos \theta -4\sin \theta \\ which\, \, always\, lies\, between\, \left( { -4,\, 3 } \right) \end{array}$$

Option 2

Option 3

Option 4

10 Probability

In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. The parameter p of the distribution is

1) $$1/3$$
2) $$1/4$$
3) $$1/5$$
4) None of these

Option 1

Option 2

Option 3 Solution:

Solution

In a binomial distribution consisting of 5 independent trials, n = 5
The probability of 1 and 2 successes are 0.4096 and 0.2048 respectively.
$$ ^{5}C_{1}{ p }^{ 1 }{ (1-p) }^{5-1 } = 0.4096$$ $$ ^{5}C_{2}{ p }^{2 }{ (1-p) }^{5-2 } = 0.2048 $$
$$ 5 { p }{ (1-p) }^{4} = 0.4096$$ _____ (1) $$ 10 { p }^{ 2 }{ (1-p) }^{3} = 0.2048 $$ ______ (2)
Divide equation (1) by (2), we get
$$ \dfrac {(1-p)}{2p} = 2 $$

$$ 5p = 1 $$

$$ p = \dfrac {1}{5} $$

Option 4

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