Mathematics Test

Result

Mathematics Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Let f(x) = [x], then f′(1) = ?

A
1

B
0

C
Does not exist

D
None of these

Solution

Given: f(x) = [x]

$\operatorname{Rf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$=\lim _{h \rightarrow 0} \frac{[1+h]-[1]}{h}=\lim _{h \rightarrow 0} \frac{(1-1)}{h}=0$

$L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{[1-h]-[1]}{-h}$

$=\lim _{h \rightarrow 0} \frac{0-1}{-h}=\infty$

$R f^{\prime}(1)^{1} L f^{\prime}(1) .$

So, $f^{\prime}(1)$ does not exist.

Hence, the correct option is (D).
Question 2
1 / -0

Let f(x) = x^3/2, then f′(0) =

A
3/2

B
1/2

C
Does not exist

D
None of these

Solution
Question 3
1 / -0

If f(x) = | x |, then f′(2) =

A
2

B
1

C
Does not exist

D
None of these

Solution
Question 4
1 / -0

Let f(x + y) = f(x)f(y), for all x, y ∈ R. If f'(0) = 2 and f(4) = 4, then f'(4) is equal to-

A
4

B
1

C
3

D
8

Solution

We have $f(x+y)=f(x)$ f $(y)$, for all $x, y$ Î $R$.

Putting $x=y=0$, we get

f(0)=f(0)f(0)

Now f'(0)=2 (Given)

We know,

$\lim _{x \rightarrow 0} \frac{f(0+h)-f(0)}{h}=2$

$\Rightarrow \lim _{h \rightarrow 0} \frac{f(0) f(h)-f(0)}{h}=2$

$\Rightarrow \lim _{h\rightarrow 0} \frac{f(h)-1^{n}}{h}=2$ [using $\left.f(0)=1\right]$

Again, $f(4)=\frac{f(4+h)-f(4)}{h}$

$f'(x)=\lim _{1 \rightarrow 0} \frac{((4)+(h)-f(4)}{h}$

$=\left(\lim _{n} \frac{f(h)-1}{h}\right) f(4)=2 f(4)$

$\therefore\ f'(x)= 2×4=8$ (since f(4)=4)
Question 5
1 / -0

A
{1}

B

C
{-1, 1}

D
None of these

Solution

Hence , options (B ) is the correct answer.
Question 6
1 / -0

A
g(x)

B
1 + g(x)

C
1 + xg(x)

D
None of these

Solution

Here the correct option is (C).
Question 7
1 / -0

A
0

B
1 +π /4

C
1

D
3/2

Solution
Question 8
1 / -0

A
24

B
36

C
48

D
12

Solution
Question 9
1 / -0

A
π

B
π/3

C
π/2

D
π/4

Solution
Question 10
1 / -0

A
π/4

B
π/3

C
π/2

D
none of these

Solution
Question 11
1 / -0

A
2

B
1

C
√2

D
2√2

Solution
Question 12
1 / -0

A
2

B
2√2

C
4

D
None of these

Solution
Question 13
1 / -0

If A, B and C are subsets of a given set, then which one of the following relations is not correct?

A
A ∪ (A ∩ B) = A ∪ B

B
A ∩ (A ∪ B) = A

C
(A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

D
(A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Solution

We know that, If A, B and C are subsets of a set X. Then,

I. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

II. A ∪ A = A, A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A and A ∩ A = A

III. (A ∩ B) ∪ C = (A ∩ C) ∪ (B ∩ C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Now, checking the options,

(A): A ∪ (A ∩ B)

= (A ∪ A) ∩ (A ∪ B) ---- (Using property I)

= A ∩ (A ∪ B)

= A ---- (Using property II)

So, option (A) is not correct.

(B): A ∩ (A ∪ B)

= (A ∩ A) ∪ (A ∩ B) ---- (Using property I)

= A ∪ (A ∩ B)

= A ---- (Using property II)

So, option (B) is correct.

(C): (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) ---- (Using property III)

So, option (C) is correct.

(D): (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) ---- (Using property IV)

So, option (D) is correct.
Question 14
1 / -0

A parallelepiped is formed by planes drawn through the points P(6,8,10) and Q(3,4,8) parallel to the coordinate planes. Find the length of edges and diagonals of the parallelepiped.

A
$\sqrt{29}$

B
$3 \sqrt{28}$

C
√2

D
√3

Solution

$\mathrm{P}=(6,8,10)$

$\mathrm{Q}=(3,4,8)$

$\mathrm{P}^{1}=\mathrm{x}=6 \quad \mathrm{P}_{1}^{1} \mathrm{x}=3$

$\mathrm{P}^{2}=\mathrm{y}=8 \quad \mathrm{P}_{2}^{1} \mathrm{y}=4$

$\mathrm{P}^{3}=\mathrm{z}=10 \quad \mathrm{P}_{3}^{1}=\mathrm{z}=8$

$\mathrm{S} 0$

$\mathrm{I}=6-3=3$

$\mathrm{b}=8-4=4$

$\mathrm{h}=10-8=2$

diagonal $=\sqrt{1^{2}+\mathrm{b}^{2}+\mathrm{b}^{2}}$

$=\sqrt{3^{2}+4^{2}+2^{2}}$

$=\sqrt{9+16+4}$

$=\sqrt{29}$
Question 15
1 / -0

What is the value of $\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{\tan 3 x^{\circ}}$ ?

A
$\frac{1}{4}$

B
$\frac{1}{3}$

C
$\frac{1}{2}$

D
1

Solution

We know that,

$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ ---- (1)

and $\lim _{x \rightarrow 0} \frac{\tan x}{x}=1$ ---- (2)

We have to find the value of $\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{\tan 3 x^{\circ}}$.

We also know that $180^{\circ}=\pi$ radian

So, if $ 1^{\circ}=\frac{\pi}{360}$

$ \therefore x^{\circ}=\frac{\pi x}{180}$

Therefore, the given expression will become,

$\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{\tan 3 x^{\circ}}=\lim _{x \rightarrow 0} \frac{\sin \frac{\pi x}{180}}{\tan \frac{3 \pi x}{180}}$

$=\lim _{x \rightarrow 0} \frac{\frac{\sin \frac{\pi x}{180}}{\frac{\pi x}{180}} \times \frac{\pi x}{180}}{\frac{\tan \frac{3 \pi x}{180}}{\frac{3 \pi x}{18}} \times \frac{3 \pi x}{180}}$

$=\lim _{x \rightarrow 0} \frac{\frac{\pi x}{180}}{\frac{3 \pi x}{180}}$ [From equation (1) and (2)]

$=\frac{1}{3}$