Mathematics Test

Result

Mathematics Test - 33

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $α, β$ are eccentric angles of the extremities of a focal chord of an ellipse, then eccentricity of the ellipse is :

A
$(\cos \alpha+\cos \beta) / \cos (\alpha+\beta)$

B
$(\sin \alpha-\sin \beta) / \sin (\alpha-\beta)$

C
$sec α + sec β$

D
$(\sin \alpha+\sin \beta) / \sin (\alpha+\beta)$

Solution

Let $P(a \cos \alpha, b \sin \alpha)$ and $Q(a \cos \beta, b \sin \beta)$ are the ends of focal chord of the
ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$
then equatiion of $\mathrm{PQ}$ is
$$
\frac{\mathrm{x}}{\mathrm{a}} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{\mathrm{y}}{\mathrm{b}} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)
$$
its pass through $(\pm a e, 0)$
$$
\pm e \cos \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)
$$
Taking +ve sign, then
$$
\begin{array}{l}
\mathrm{e}=\frac{\cos \left(\frac{a-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} \\
=\frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)} \\
=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}
\end{array}
$$
Hence, the correct option is (D)
Question 2
1 / -0

The value of definite integral $\int_{0}^{1} \frac{\sin ^{-1} \sqrt{\mathrm{x}}}{\mathrm{x}^{2}-\mathrm{x} 1} \mathrm{d}$ x is equal to

A
$\pi^{2} / 6 \sqrt{3}$

B
-π²/6√3

C
-π²/2√3

D
π²/2√3

Solution

put $\mathrm{x}=\sin ^{2} \theta \Rightarrow \mathrm{dx}=\sin 2 \theta \mathrm{d} \theta$
$\mathrm{I}=\int_{0}^{\pi / 2} \frac{(\theta \sin 2 \theta)}{\sin ^{4} \theta-\sin ^{2} \theta+1} \mathrm{d} \theta$
$\mathrm{I}=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-\theta\right) \sin 2 \theta}{\cos ^{4} \theta-\cos ^{2} \theta+1} \mathrm{d} \theta$
$=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-\theta\right) \sin 2 \theta}{\left(1-\sin ^{2} \theta\right)^{2}-\left(1-\sin ^{2} \theta\right)+1} \mathrm{d} \theta$
$=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-\theta\right) \sin 2 \theta}{\sin ^{4} \theta-\sin ^{2} \theta+1} \mathrm{d} \theta$
adding (1) and (2)
$2 \mathrm{I}=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\sin 2 \theta}{\sin ^{4} \theta-\sin ^{2} \theta+1} \mathrm{d} \theta$
put $\sin ^{2} \theta=t$
$2 \mathrm{I}=\frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{dt}}{t^{2}-t+1}=\frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\left(t-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$2 \mathrm{I}=\left.\frac{\pi}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\left(t-\frac{1}{2}\right) \cdot 2}{\sqrt{3}}\right)\right|_{0} ^{1}$
$2 \mathrm{I}=\frac{\pi}{\sqrt{3}}\left(\frac{2 \pi}{6}\right) \Rightarrow \mathrm{I}=\frac{\pi^{2}}{6 \sqrt{3}}$
Hence, the correct option is (A)
Question 3
1 / -0

tan⁻¹(5/6)+12tan⁻¹(11/60)=

A
π

B
π/2

C
π/4

D
3π/4

Solution

$\tan ^{-1} \frac{5}{6}+\frac{1}{2} \tan ^{-1} \frac{11}{60}$
$\frac{1}{2} \tan ^{-1}\left(\frac{\frac{5}{6} \times 2}{1-\left(\frac{5}{6}\right)^{2}}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{11}{60}\right)\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$
$=\frac{1}{2} \tan ^{-1}\left(\frac{60}{11}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{11}{60}\right)$
$=\frac{1}{2}\left[\cos ^{-1}\left(\frac{11}{60}\right)+\tan ^{-1}\left(\frac{11}{60}\right)\right]\left[\tan ^{-1} x=\cos ^{-1} \frac{1}{x}\right]$
$=\frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{4}$ Because $\left[\cos ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}\right]$
Hence, the correct option is (C)
Question 4
1 / -0

If $\frac{1}{\sqrt{2}}

A
2cos⁻¹(x−π/4)

B
2cos⁻¹x

C
π/4

D
0

Solution

Formula \(\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) x, y>0$
$\therefore \cos ^{-1} x+\cos ^{-1}\left(\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right)=\cos ^{-1}\left[\frac{x\left(x+\sqrt{1-x^{2}}\right)}{\sqrt{2}}-\sqrt{1-x^{2}}\left(\sqrt{1-\frac{1+2 x \sqrt{1-x^{2}}}{2}}\right)\right]$
$=\cos ^{-1}\left(\frac{x\left(x+\sqrt{1-x^{2}}\right)}{\sqrt{2}}-\sqrt{1-x^{2}} \sqrt{\frac{1-2 x \sqrt{1-x^{2}}}{2}}\right)$
$=\cos ^{-1}\left(\frac{x\left(x+\sqrt{1-x^{2}}\right)}{\sqrt{2}}-\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\left(x-\sqrt{1-x^{2}}\right)\right)$
$=\frac{\pi}{4}$
$=\cos ^{-1}\left(\frac{x^{2}+x \sqrt{1-x^{2}}}{\sqrt{2}}-\frac{\left(x-\sqrt{1-x^{2}} \sqrt{1-x^{2}}\right)}{\sqrt{2}}\right)$
$\left.=\frac{x^{2}+x \sqrt{1-x^{2}}}{\sqrt{2}}-\left(\frac{x \sqrt{1-x^{2}}+x^{2}-1}{\sqrt{2}}\right)\right)$
$=\frac{1}{\sqrt{2}}$
Hence, the correct option is (C)
Question 5
1 / -0

If $\tan ^{-1}\left(x+\frac{2}{x}\right)-\tan ^{-1}\left(x-\frac{2}{x}\right)=\tan ^{-1}\left(\frac{4}{x}\right)$ then $x=$

A
±√2

B
±2

C
±1

D
±√3

Solution

$\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) x, y,>0$
$\therefore \tan ^{-1}\left(x+\frac{2}{x}\right)-\tan ^{-1}\left(x-\frac{2}{x}\right)=\tan ^{-1}\left(\frac{\frac{4}{x}}{1+\left(x^{2}-\frac{4}{x^{2}}\right)}\right)=\tan ^{-1}\left(\frac{4}{x}\right)$
[Given]
$\Rightarrow \frac{\frac{4}{x}}{1+x^{2}-\frac{4}{x^{2}}}=\frac{4}{x}$
$\Rightarrow 1=1+x^{2}-\frac{4}{x^{2}}$
$x^{4}=4$
$x^{2}=2$
$x=\pm \sqrt{2}$
Hence, the correct option is (A)
Question 6
1 / -0

If sin⁻¹(x)+sin⁻¹(2x)=π/3, then x=

A
3/28

B
√3/28

C
√(3/28)

D
3/√28

Solution

Given $\sin ^{-1} x+\sin ^{-1}(2 x)=\frac{\pi}{3}$
$\sin ^{-1} 2 x=\frac{\pi}{3}-\sin ^{-1} x$
take sin on both sides, $2 x=\sin \frac{\pi}{3} \cos \sin ^{-1} x-\cos \frac{\pi}{3} \sin \sin ^{-1} x$
$2 x=\sin \frac{\pi}{3} \cos \sin ^{-1} x-\cos \frac{\pi}{3} \sin \sin ^{-1} x$
$2 x=\frac{\sqrt{3}}{2} \sqrt{1-x^{2}}-\frac{1}{2} x$
$2 x+\frac{x}{2}-\frac{\sqrt{2\left(1-x^{2}\right)}}{2}$
$\frac{5 x}{2}=\frac{\sqrt{3\left(1-x^{2}\right)}}{2}$
$25 x^{2}=3-3 x^{2}$
$28 x^{2}=3$
$x^{2}=\frac{3}{28}$
$x=\sqrt{\frac{3}{28}}$
Hence, the correct option is (C)
Question 7
1 / -0

If sin⁻¹x+4cos⁻¹x=π , then x=

A
1/2

B
1/√2

C
√3/2

D
1

Solution

$\sin ^{-1} x+4 \cos ^{-1} x=\pi$
$\frac{\pi}{2}+3 \cos ^{-1} x=\pi$
$3 \cos ^{-1} x=\frac{\pi}{2}$
$\cos ^{-1} x=\frac{\pi}{6}$
$\Rightarrow x=\frac{\sqrt{3}}{2}$
Hence, the correct option is (C)
Question 8
1 / -0

$\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=$

A
3π/20

B
7π/10

C
7π/20

D
17π/20

Solution

$\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \left(\frac{9 \pi}{10}\right)-\sin \frac{9 \pi}{10}\right)\right]$
$=\cos ^{-1}\left(\cos \frac{\pi}{4}+\frac{9 \pi}{10}\right)[\because \cos A \cos B-\sin A \sin B=\cos (A+B)]$
$=\cos ^{-1}\left(\cos \frac{46 \pi}{40}\right)$
$=\cos ^{-1}\left(\cos \frac{23 \pi}{20}\right)=\cos ^{-1}\left(\cos \left(\pi+\frac{3 \pi}{20}\right)\right)$
$=\cos ^{-1}\left(-\cos \frac{3 \pi}{20}\right)\left[\cos ^{-1}(-x)=\pi-\cos ^{-1} x\right]$
$=\pi-\cos ^{-1} \cos \left(\frac{3 \pi}{20}\right)$
$=\pi-\cos ^{-1} \cos 18^{\circ}$
as $0<\frac{3 \pi}{20}<\frac{\pi}{2}$
$\therefore \cos ^{-1}(\cos x)=x x \in\left[0, \frac{\pi}{2}\right]$
$\therefore \cos ^{-1}\left(\frac{1}{\sqrt{2}}\left(\cos \left(\frac{9 \pi}{10}\right)-\sin \frac{9 \pi}{10}\right)\right)=\pi-\frac{3 \pi}{20}=\frac{17 \pi}{20}$
Hence, the correct option is (D)
Question 9
1 / -0

If $\Delta A B C$ is right angled at $C$, then $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=$

A
π/4

B
π/3

C
π/6

D
π/2

Solution

$k=\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{a+c}\right)$
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$k=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{a+c}}{1-\frac{a b}{(a+c)(b+c)}}\right]$
$=\tan ^{-1}\left(\frac{a^{2}+a c+b^{2}+b c}{a b+a c+b c+c^{2}-a b}\right)$
By cosine rule for triangle $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 b c}$
$=\tan ^{-1}\left(\frac{2 b \cos c+c^{2}+a c+b c}{a c+b c+c^{2}}\right)$
Given $C=90^{\circ}, \cos C=0$
$=\tan ^{-1}\left(\frac{c^{2}+a c+b c}{a c+b c+c^{2}}\right)$
$=\tan ^{-1}(1)$
$=\frac{\pi}{4}$
Hence, the correct option is (A)
Question 10
1 / -0

sin−1(sin5)+cos−1(cos7)−tan−1(tan5)=

A
7-2

B
π

C
3π-1

D
2π-1

Solution

As $\sin ^{-1}(\sin \theta)=\theta$ for $\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\cos ^{-1}(\cos \theta)=\theta$ for $\theta \in[0, \pi]$
and $\tan ^{-1}(\tan \theta)=\theta$ for $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\therefore \sin ^{-1}(\sin 5)+\cos ^{-1}(\cos 7)-\tan ^{-1}(\tan 5)$
$=\sin ^{-1}(\sin (5-2 \pi))+\cos ^{-1}(\cos (7-2 \pi))-\tan ^{-1}(\tan (5-2 \pi))$
$=5-2 \pi+7-2 \pi-5+2 \pi$
$=7-2 \pi$
Hence, the correct option is (A)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 33

Result

Mathematics Test - 33

Score

-

Rank

-

SHARING IS CARING

Question 1

\((\cos \alpha+\cos \beta) / \cos (\alpha+\beta)\)

\((\sin \alpha-\sin \beta) / \sin (\alpha-\beta)\)

sec α + sec β

\((\sin \alpha+\sin \beta) / \sin (\alpha+\beta)\)

Solution

Question 2

\(\pi^{2} / 6 \sqrt{3}\)

-π2/6√3

-π2/2√3

π2/2√3

Solution

Question 3

π

π/2

π/4

3π/4

Solution

Question 4

2cos−1(x−π/4)

2cos−1x

π/4

0

Solution

Question 5

±√2

±2

±1

±√3

Solution

Question 6

3/28

√3/28

√(3/28)

3/√28

Solution

Question 7

1/2

1/√2

√3/2

1

Solution

Question 8

3π/20

7π/10

7π/20

17π/20

Solution

Question 9

π/4

π/3

π/6

π/2

Solution

Question 10

7-2

π

3π-1

2π-1

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$sec α + sec β$

-π²/6√3

-π²/2√3

π²/2√3

2cos⁻¹(x−π/4)

2cos⁻¹x