Mathematics Test

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Mathematics Test - 23

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Weekly Quiz Competition

Question 1
4 / -1

Number of possible tangents to the curve $y=\cos (x+y),-3 \pi \leq x \leq 3 \pi$ that are
parallel to the line $x+2 y=0$, is

A
1

B
2

C
3

D
4

Solution

We have, $y=\cos (x+y)$
$\frac{d y}{d x}=\sin (x+y)\left(1+\frac{d y}{d x}\right)$
since, the tangents are parallel to the line $x+2 y=0$ $-\frac{1}{2}=-\sin (x+y)\left(1-\frac{1}{2}\right)$
$\Rightarrow \sin (x+y)=1$
$\Rightarrow x+y=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{3 \pi}{2}$
$-1 \leq y \leq 1$
Hence (c) is the correct answer.
Question 2
4 / -1

If the function $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1,$ where $a>0,$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2}=q$, then a equals

A
3

B
1

C
2

D
1/2

Solution

We have, $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$
$\therefore f(x)=6 x^{2}-18 a x+12 a^{2}=0$
$\Rightarrow 6\left[x^{2}-3 a x+2 a^{2}\right]=0$
$\Rightarrow x^{2}-3 a x+2 a^{2}=0 \Rightarrow x^{2}-2 a x-a x+2 a^{2}=0$
$\Rightarrow x(x-2 a)-a(x-2 a)=0$
$\Rightarrow(x-a)(x-2 a)=0 \Rightarrow x=a, x=2 a$
Now, $f^{\prime}(x)=12 x-18 a$
$\therefore f^{\prime}(a)=12 a-18 a=-6 a<0$
$\therefore f(x)$
will be maximum at $x=a$
l.e. $p=a$ Also, $f^{\prime}(2 a)=24 a-18 a=6 a$
$\therefore f(x)$
will be minimum at $x=2$ a
I.e.q $=2 \mathrm{a}$
Given, $p^{2}=q$ $\Rightarrow a^{2}=2 a \Rightarrow a=2$
Hence $(\mathrm{c})$ is the correct answer.
Question 3
4 / -1

$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}$ is equal to

A
π

B
π/2

C
π/3

D
π/4

Solution

$\quad$ Since $2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}$
$\therefore 4 \tan ^{-1} \frac{1}{5}=2\left[2 \tan ^{-1} \frac{1}{5}\right]=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\frac{1}{25}}$
$=2 \tan ^{-1} \frac{10}{24}=\tan ^{-1} \frac{\frac{20}{24}}{1-\frac{100}{576}}=\tan ^{-1} \frac{120}{119}$
So, $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239}$
$=\tan ^{-1} \frac{\frac{120}{119} \frac{1}{239}}{1+\frac{120}{119} \frac{1}{239}}=\tan ^{-1} \frac{(120 \times 239)-119}{(119 \times 239)+120}$
$\Rightarrow \tan ^{-1} 1=\frac{\pi}{4}$
Question 4
4 / -1

If one root of the equation ax²+ bx + c = 0 be n times the other root, then

A
na²= bc (n + 1)²

B
nb² = ac (n + 1)²

C
nc² = ab (n + 1)²

D
nb³ = ac (n + 1)²

Solution

Let the roots be $\alpha$ and $n \alpha$
Sum of roots, $a+n a=-\frac{b}{a} \Rightarrow a=-\frac{b}{a(n+1)} \quad \ldots . .(i)$
and product, a.n. $a=\frac{c}{a} \Rightarrow \alpha^{2}=\frac{c}{n a}$
From (i) and (ii), we get
$\Rightarrow\left[-\frac{b}{a(n+1)}\right]^{2}=\frac{c}{n a} \Rightarrow \frac{b^{2}}{a^{2}(n+1)^{2}}=\frac{c}{n a}$
$\Rightarrow n b^{2}=\operatorname{ac}(n+1)^{2}$
Question 5
4 / -1

If $A$ and $B$ are two given sets, then $A \cap(A \cap B)$ eis equal to

A
A

B
B

C
∅

D
A∩( $B^{c}$ ).

Solution

$\begin{aligned}\left.\mathrm{A} \cap(A \cap B)^{c}\right)=& A \cap\left(A^{c} \cup B^{c}\right) \\ &=\left(A \cap\left(A^{c}\right) \cup\left(A \cap\left(B^{c}\right)\right.\right.\\ &=\emptyset \cup\left(A \cap\left(B^{c}\right)=A \cap\left(B^{c}\right)\right.\end{aligned}$
Question 6
4 / -1

The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by

A
{(1, 4, (2, 5), (3, 6),.....}

B
{(4, 1), (5, 2), (6, 3),.....}

C
{(1, 3), (2, 6), (3, 9),..}

D
None of these

Solution

$R=\{(a, b): a, b \in N, a-b=3\}=\{((n+3), n): n \in N\}$

$=[(4,1),(5,2),(6,3), \ldots \ldots\}$
Question 7
4 / -1

The value of log₃ 4log₄5log₅6log₆7log₇8log₈9 is

A
1

B
2

C
3

D
4

Solution
Question 8
4 / -1

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^{3}}$ is equal to

A
1

B
π/2

C
1/16

D
0

Solution

\[
\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^{3}}
\]
Let $x=\frac{\pi}{2}+t$
If $x \rightarrow \frac{\pi}{2}, t \rightarrow 0$
\[
\begin{array}{l}
\lim _{t \rightarrow 0} \frac{\sin t-\tan t}{-8 t^{3}} \\
=\lim _{t \rightarrow 0} \frac{\left(t-\frac{t^{3}}{3 !}+\frac{t^{5}}{5 !}+\ldots\right)-\left(t+\frac{t^{3}}{3}+\frac{2 t^{5}}{15}+\ldots\right)}{-8 t^{3}}
\end{array}
\]
$=\frac{1}{16}$
We can put $x=\frac{\pi}{2}-t$ and we'll get L.H.L also same. since L.H.L = R.H.L the limit exists and is $=\frac{1}{16}$
Question 9
4 / -1

The number of values of x where the function f(x) = 2 (cos 3x + cos√3xattains its maximum, is

A
1

B
2

C
0

D
Infinite

Solution

We have,
\[
\begin{array}{l}
f(x)=2(\cos 3 x+\cos \sqrt{3} x) \\
=4 \cos \left(\frac{3+\sqrt{3}}{2}\right) x \cos \left(\frac{3-\sqrt{3}}{2}\right) x \leqslant 4
\end{array}
\]
and it is equal to 4 when both $\cos \left(\frac{3+\sqrt{3}}{2}\right) x$ and $\cos \left(\frac{3-\sqrt{3}}{2}\right)$
Are equal to 1 for a value of $x$. This is possible only when $x=0$.
Hence (a) is the correct answer.
Question 10
4 / -1

$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=$

A
-3/16

B
5/16

C
3/16

D
-5/16

Solution

$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$
$=\frac{1}{2} \sin 20^{\circ} \sin 60^{\circ}\left(2 \sin 40^{\circ} \sin 80^{\circ}\right)$
$=\frac{1}{2} \sin 20^{\circ} \sin 60^{\circ}\left(\cos 40^{\circ}-\cos 120^{\circ}\right)$
$=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \sin 20^{\circ}\left(1-2 \sin ^{2} 20^{\circ}+\frac{1}{2}\right)$
$=\frac{\sqrt{3}}{4} \sin 20^{\circ}\left(\frac{3}{2}-2 \sin ^{2} 20^{\circ}\right)$
$=\frac{\sqrt{3}}{8}\left(3 \sin 20^{\circ}-4 \sin ^{3} 20^{\circ}\right)$
$=\frac{\sqrt{3}}{8} \sin 60^{\circ}=\frac{\sqrt{3}}{8} \cdot \frac{\sqrt{3}}{2}=\frac{3}{16}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 23

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Mathematics Test - 23

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SHARING IS CARING

Question 1

1

2

3

4

Solution

Question 2

3

1

2

1/2

Solution

Question 3

π

π/2

π/3

π/4

Solution

Question 4

na2 = bc (n + 1)2

nb2 = ac (n + 1)2

nc2 = ab (n + 1)2

nb3 = ac (n + 1)2

Solution

Question 5

A

B

∅

A∩(Bc).

Solution

Question 6

{(1, 4, (2, 5), (3, 6),.....}

{(4, 1), (5, 2), (6, 3),.....}

{(1, 3), (2, 6), (3, 9),..}

None of these

Solution

Question 7

1

2

3

4

Solution

Question 8

1

π/2

1/16

0

Solution

Question 9

1

2

0

Infinite

Solution

Question 10

-3/16

5/16

3/16

-5/16

Solution

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na²= bc (n + 1)²

nb² = ac (n + 1)²

nc² = ab (n + 1)²

nb³ = ac (n + 1)²

A∩( $B^{c}$ ).