Mathematics Test

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Mathematics Test - 32

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Weekly Quiz Competition

Question 1
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 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

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 4
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 5
4 / -1

If the tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ at any point on it cuts the axes OX and OY at P and Q respectivelv. then $\mathrm{OP}+\mathrm{OQ}$ is

A
a/2

B
$\sqrt{a}$

C
2a

D
4a

Solution

$\sqrt{x}+\sqrt{y}=\sqrt{a} \ldots \ldots$

$\Rightarrow \frac{1}{2 \sqrt{x}}+\frac{1 d y}{2 \sqrt{y} d x}=0$

$\therefore \frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}$
Equation of tangent at $\left(x_{1} y_{1}\right)$ isy $-y_{1}=-\frac{\sqrt{y_{1}}}{\sqrt{x_{1}}}\left(x-x_{1}\right)$
$\Rightarrow \frac{x}{\sqrt{x_{1}}}+\frac{y}{\sqrt{y_{1}}}=\sqrt{a} ; \Rightarrow o p=\sqrt{a} \sqrt{x_{1}}, O Q=\sqrt{a} \sqrt{y_{1}}$
$\therefore O P+O Q=\sqrt{a}$
Question 6
4 / -1

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

A
3100

B
3300

C
2900

D
1400

Solution

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n(A ∩ B) = 5% of 10,000 = 500

n(B ∩ C) = 3% of 10,000 = 300

n(C ∩ A) = 4% of 10,000 = 400

n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]

=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
Question 7
4 / -1

$3 \cos ^{-1} x-\pi x-\frac{\pi}{2}=0$ has :

A
One solution

B
Infinite solutions

C
No solution

D
None of these

Solution

$3 \cos ^{-1} x-\pi x+\frac{\pi}{2}$ Clearly graphs of $y=3 \cos ^{-1} x$ and $y=\pi x+\frac{\pi}{2}$ in the domain of $\cos ^{-1} x$ i.e., in [-1,1] intersect only once, therefore there is only one solution of the given equation.
Question 8
4 / -1

If the sum of two of the roots of x³+ px² + qx + r = 0is zero, then pq =

A
- r

B
r

C
2 r

D
- 2 r

Solution

Given that, $a+\beta=0$
$\alpha+\beta+y=-p \Rightarrow y=-p$
Substituting $\mathrm{Y}=-\mathrm{p}$ in the given equation
$\Rightarrow-p^{3}+p^{3}-p q+r=0 \Rightarrow p q=r$
Question 9
4 / -1

If $3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}+2 \tan ^{-1} \frac{2 x}{1+x^{2}}=\frac{\pi}{3}$ then $x=$

A
√3

B
1/√3

C
1

D
None of these

Solution

$$
3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}+2 \tan ^{-1} \frac{2 x}{1+x^{2}}=\frac{\pi}{3}
$$
Putting $x=\tan \theta$
$\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)-4 \cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
$+2 \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)=\frac{\pi}{3}$
$\Rightarrow 3 \sin ^{-1}(\sin 2 \theta)-4 \cos ^{-1}(\cos 2 \theta)$
$+2 \tan ^{-1}(\tan 2 \theta)=\frac{\pi}{3}$
$\Rightarrow 3(2 \theta)-4(2 \theta)+2(2 \theta)=\frac{\pi}{3} \Rightarrow 6 \theta-8 \theta+4 \theta=\frac{\pi}{3}$
$\Rightarrow \theta=\frac{\pi}{6} \Rightarrow \tan ^{-1} x=\frac{\pi}{6} \Rightarrow x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$
Question 10
4 / -1

Let $f(x)=\left\{\begin{array}{l}1+\sin x, x<0 \\ x^{2}-x+1, x \geq 0\end{array}\right.$. Then

A
f has a local maximum at x = 0

B
f has a local minimum at x = 0

C
f is increasing every where

D
f is decreasing everywhere

Solution

f is continuous at ‘0’ and f'(0-) $>$ 0 and f'( 0 +) $<$ 0 . Thus f has a local maximum at ‘0’.

Hence option A is correct.

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Mathematics Test - 32

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Mathematics Test - 32

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Question 1

A

B

∅

A∩(Bc).

Solution

Question 2

3

1

2

1/2

Solution

Question 3

na2 = bc (n + 1)2

nb2 = ac (n + 1)2

nc2 = ab (n + 1)2

nb3 = ac (n + 1)2

Solution

Question 4

1

2

0

Infinite

Solution

Question 5

a/2

a

2a

4a

Solution

Question 6

3100

3300

2900

1400

Solution

Question 7

One solution

Infinite solutions

No solution

None of these

Solution

Question 8

- r

r

2 r

- 2 r

Solution

Question 9

√3

1/√3

1

None of these

Solution

Question 10

f has a local maximum at x = 0

f has a local minimum at x = 0

f is increasing every where

f is decreasing everywhere

Solution

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A∩( $B^{c}$ ).

na²= bc (n + 1)²

nb² = ac (n + 1)²

nc² = ab (n + 1)²

nb³ = ac (n + 1)²

$\sqrt{a}$