Mathematics Test

Result

Mathematics Test - 1

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Question 1
1 / -0

A

B

C

D

Solution
Question 2
1 / -0

The equations of common tangents to hyperbola x²/a² – y²/b² = 1 and y²/a² – x²/b² = 1, are

A
y = ± x ± Ö(b² – a²)

B
y = ± x ± Ö(a² – b²)

C
y = ± x ± (a² – b²)

D
None of these

Solution
Question 3
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If a,b,c be in A.P. and b,c,d be in H.P., then

A
ab = cd

B
a+b=b+c

C
bc = ad

D
abcd = 1

Solution
Question 4
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If two roots of the equation $x^{3} - 3 x + 2 = 0$ are same, then the roots will be

A
2, 2, 3

B
1, 1, -2

C
- 2, 3, 3

D
-2, -2, 1

Solution
Question 5
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One root of the equation cos x - x + 1/2 = 0 lies in the interval

A
[0, π/2]

B
[-π/2, 0]

C
[π/2, π]

D
[π, 3π/2]

Solution
Question 6
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A
0

B
1

C
1/60

D
1/120

Solution
Question 7
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If the roots of the equation q $x^{2}$ + p $x$ + q = 0 where p, q are real, are complex, then the roots of the equation $x^{2}$ - 4q $x$ + $p^{2}$ = 0 are

A
Real and unequal

B
Real and equal

C
Imaginary

D
nothing can be said in particular

Solution
Question 8
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If f(x) = x (√x - √(x+1) then:

A
f is continuous but not differentiable at x = 0

B
f is differentiable at x = 0

C
f is differentiable but not continuous at x = 0

D
f is not differentiable at x = 0

Solution
Question 9
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In a $\Delta \mathrm{ABC}$, if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$, and the side $\mathrm{a}=2$, then area of the triangle is :

A

$1$

B
$2$

C
$\frac{\sqrt{3}}{2}$

D

$\sqrt{3}$

Solution

$\because \frac{\cos A}{a}=\frac{\cos B}{b}$
$\Rightarrow \frac{b^{2}+c^{2}-a^{2}}{2 a b c}=\frac{a^{2}+c^{2}-b^{2}}{2 a b c}$
$ \Rightarrow a^{2}=b^{2}$
$ \Rightarrow a=b$
Similarly, $\frac{\cos B}{b}=\frac{\cos C}{c}$
$ \Rightarrow \mathrm{b}=\mathrm{c}$
So,
$a=b=c=2$
Triangle is equilateral.
So, Area $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}=\sqrt{3}$
Question 10
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If a₁, a₂, a₃ are in H.P., then

A
a₁a₂ + a₂a₃ + a₃a₁ = 0

B
a₁a₂ + a₂a₃ = a₃a₁

C
a₁a₂ + a₂a₃ = 2a₁a₃

D
a₁a₂ + a₂a₃ + 2a₁a₃ = 0

Solution

If a₁, a₂, a₃ are in H.P.
Let 1/a₁ , 1/a₂ , 1/a₃ will be in A.P.
2/a₂ = 1/a₁ + 1/a₃
2a₁. a₃ = a₁.a₂ + a₂a₃