Mathematics Test

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Mathematics Test - 21

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Weekly Quiz Competition

Question 1
4 / -1

The approximate value of square root of 25.2 is

A
5.01

B
5.02

C
5.03

D
5.04

Solution

Let $f(x)=\sqrt{x}$
Now, $f(x+\delta x)-f(x)=f^{\prime}(x) \cdot \delta x=\frac{\delta x}{2 \sqrt{x}}$
We may write, $25.2=25+0.2$
Taking $x=25$ and $\delta x=0.2$ We have $f(25.2)-f(25)=\frac{0.2}{2 \sqrt{25}}=0.02$
$\therefore f(25.2)=f(25)+0.02$
$=\sqrt{25}+0.02=5.02$
$\Rightarrow \sqrt{(25.2)}=5.02$
Question 2
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.
Question 3
4 / -1

If one root of the quadratic equation $a x^{2}+b x+c=0$ is equal to the $n^{\text {th }}$ power of the other root, then the value of $\left(a c^{n}\right)^{\frac{1}{n+1}}+\left(a^{n} c\right)^{\frac{1}{n+1}}$

A
b

B
-b

C
$b^{\frac{1}{n+1}}$

D
$-b^{\frac{1}{n+1}}$

Solution

Let $\alpha, \alpha^{n}$ be two roots,
Then $\alpha+\alpha^{n}=-\frac{b}{a^{\prime}} \alpha \alpha^{n}=\frac{c}{a}$
Eliminating $\alpha,$ we get $\left(\frac{9}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=-\frac{b}{a}$
$\Rightarrow a \cdot a^{-\frac{1}{n+1}}, c^{\frac{1}{n+1}}+a \cdot a^{-\frac{n}{n+1}}, c^{\frac{n}{n+1}}=-b$
or $\left(a^{n} c\right)^{\frac{1}{n+1}}+\left(a c^{n}\right)^{\frac{1}{n+1}}=-b$
Question 4
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
Question 5
4 / -1

If $\lim _{x \rightarrow 0}(1+a \sin x)^{\text {cosec x}}=3,$ then a is

A
In 2

B
In 3

C
In 4

D
e³

Solution

$\quad=\lim _{x \rightarrow 0}(1+a \sin x)^{\operatorname{cosec} x}$
$\left[1^{\infty}\right.$ form $] \Rightarrow \lim _{x \rightarrow 0} e^{\operatorname{cosec} x \cdot a \sin x}=e^{a}$
$\therefore e^{a}=3 \Rightarrow a=\log _{e} 3=\ln 3$
Question 6
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 7
4 / -1

$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=$

A
1/√10

B
-1/√10

C
1/10

D
-1/10

Solution

Let $0 \leq x \leq \pi$
So, $\cos ^{-1} \frac{4}{5}=x \Rightarrow \cos x=\frac{4}{5} \quad \ldots .$ (i)
Now $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \left(\frac{x}{2}\right)$
$\ldots .$ (ii)
From (i), $\cos x=\frac{4}{5} \Rightarrow 1-2 \sin ^{2} \frac{x}{2}=\frac{4}{5}$
$\Rightarrow 2 \sin ^{2} \frac{x}{2}=1-\frac{4}{5}=\frac{1}{5} \Rightarrow \sin \frac{x}{2}=\sqrt{\frac{1}{10}}$
Question 8
4 / -1

With reference to a universal set, the inclusion of a subset in another, is relation, which is

A
Symmetric only

B
Equivalence relation

C
Reflexive only

D
None of these

Solution

since $A \subseteq A \ldots$ Relation $^{\prime} \subseteq^{\prime}$ is relfexive
since $A \subseteq B, B \subseteq C \Rightarrow A \subseteq C$
$\therefore$ Relation ' $\subseteq$ ' is transitive.
But $A^{\prime} \subseteq^{\prime} B, \Rightarrow B \subseteq A \ldots$ relation is not symmetric.
Question 9
4 / -1

The value of $\sqrt{\left(\log _{0.5}^{2} 4\right)}$ is

A
-2

B
√(-4)

C
2

D
None of these

Solution

$\sqrt{\left(\log _{0.5}^{2} 4\right)}=\sqrt{\left\{\log _{0.5}(0.5)^{-2}\right\}^{2}}=\sqrt{(-2)^{2}}=2$
Hence option C is correct.
Question 10
4 / -1

Let $g(x)$ be an antiderivative for $f(x) .$ Then $\ln \left(1+(g(x))^{2}\right)$ is an antiderivate for

A
$\frac{2 f(x) g(x)}{1+(f(x))^{2}}$

B
$\frac{2 f(x) g(x)}{1+(g(x))^{2}}$

C
$\frac{2 f(x)}{1+(f(x))^{2}}$

D
$\frac{2 f(x)}{1+(f(x))^{2}}$

Solution

Given $\int f(x) d x=g(x)$
$g^{\prime}(x)=f(x)$
Now
$\frac{d}{d x}\left(\operatorname{In}\left(1+g^{2}(x)\right)=\frac{2 g(x) g^{\prime}(x)}{1+g^{2}(x)}\right.$
$\frac{2 f(x) g(x)}{1+g^{2}(x)}$

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Mathematics Test - 21

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Mathematics Test - 21

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Question 1

5.01

5.02

5.03

5.04

Solution

Question 2

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

Question 3

b

-b

\(b^{\frac{1}{n+1}}\)

\(-b^{\frac{1}{n+1}}\)

Solution

Question 4

√3

1/√3

1

None of these

Solution

Question 5

In 2

In 3

In 4

e3

Solution

Question 6

a/2

a

2a

4a

Solution

Question 7

1/√10

-1/√10

1/10

-1/10

Solution

Question 8

Symmetric only

Equivalence relation

Reflexive only

None of these

Solution

Question 9

-2

√(-4)

2

None of these

Solution

Question 10

\(\frac{2 f(x) g(x)}{1+(f(x))^{2}}\)

\(\frac{2 f(x) g(x)}{1+(g(x))^{2}}\)

\(\frac{2 f(x)}{1+(f(x))^{2}}\)

\(\frac{2 f(x)}{1+(f(x))^{2}}\)

Solution

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e³

$\sqrt{a}$