Mathematics Test

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Mathematics Test - 10

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Weekly Quiz Competition

Question 1
3 / -1

The area bounded by the parabola y² = 4x and its latusrectum is:

A
(8/3) sq. units

B
(3/8) sq. units

C
12 sq. units

D
(1/3) sq. units

Solution
Question 2
3 / -1

The equations of the circles, which touch both the axes and the line 4x+3y=12 and have centres in the first quadrant, are

A
x²+y²+x−y+1=0

B
x²+y²−2x−2y+1=0

C
x²+y²−6x−6y+36=0

D
None of these

Solution

Radius $(r)=$ perpendicular distance on line $4 x+3 y=12$ from centre $\Rightarrow r=r=\frac{|4 r+3 r-12|}{\sqrt{16+9}}$
$\Rightarrow|7 r-12|=5 r$
$\Rightarrow 7 r-12=\pm 5 r$
$\therefore 2 r=12 \Rightarrow r=6$
and $12 r=12$ $\Rightarrow r=1$
(i) When centre is (1,1) and radius is 1 then equation of circle is $(x-1)^{2}+(y-1)^{2}=1$
$\Rightarrow x^{2}+y^{2}-2 x-2 y+1=0$
Hence option B is the correct answer.
Question 3
3 / -1

The area bounded by the parabola y = 4x², X - axis between the ordinates x = 2, x = 4 is:

A
(224/3) sq. units

B
(125/3) sq. units

C
(122/3) sq. units

D
(121/3) sq. units

Solution
Question 4
3 / -1

The area enclosed by the curve |x| + |y| = 1 in sq. units is⋯

A
2

B
4

C
1

D
1/2

Solution

$| x | + | y | = 1$
Area bounded = 4 × (1/2) × 1 × 1 = 2
Hence option A is the correct answer.
Question 5
3 / -1

The function $f (x) = 2 | x | + | x + 2 | - | | x + 2 | - 2 | x ∣$ has a local minimum or a local maximum at x = ?

A
−2

B
2

C
2/3

D
None of these

Solution

$y=2|x|+|x+2|-|| x+2|-2| x||$
case- $1:-x<-2$
$y=-2 x-x-2-|-x-2+2 x|$
$y=-2 x-4 \ldots \ldots \ldots .(1)$
$\operatorname{case}-2:-2\(y=-x+2-|3 x+2|$
$\mathrm{So},$ for $x<\frac{-2}{3}$
$y=2 x+4$
And for $x \in\left[\frac{-2}{3}, 0\right]$
$y=-4 x \ldots \ldots$ (3)
Case $3:-x>0$
$y=3 x+2-|x-2|$
For $0\(y=4 x \ldots \ldots \ldots \ldots(4)$
And for $x>2$
$y=2 x+4 \ldots \ldots \ldots(5)$
After drawing this graph, we can conclude $x=-2,0$

Hence option A is the correct answer.
Question 6
3 / -1

The area bounded by y= x² + 2, X-axis, x = 1 and x = 2 is:

A
16/3

B
17/3

C
13/3

D
20/3

Solution

$y=x^{2}+2$
area of the shaded region. $=\int_{-}^{2} y d x$ $=\int_{1}^{2}\left(x^{2}+2\right) d x=\left.\frac{x^{3}}{3}\right|_{1} ^{2}+\left.2 x\right|_{1} ^{2}$
$=\frac{7}{3}+2(1)$
$=\frac{13}{3}$ squnits
Hence option C is the correct answer.
Question 7
3 / -1

The area of the region bounded by the curve y = x², x - axis and the ordinates x = 0, x = 2 is:

A
(8/3) sq. units

B
(3/8) sq. units

C
24 sq. units

D
25 sq. units

Solution
Question 8
3 / -1

A
$\tan^{2} x = \frac{2}{3}$

B
$\tan^{2} x = \frac{1}{3}$

C
$\frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{2}{125}$

D
None of these

Solution

Hence option A is the correct answer.
Question 9
3 / -1

If α and β are the roots of the equation x² + 6x + $λ = 0 3 α + 2 β = - 20$ then $λ$

A
-8

B
-16

C
16

D
8

Solution

$α + β = - 6 . . . . (i)$

$α β = λ . . . (i i)$

and given 3α + 2β = -20 ......(iii)

Solving (i) and (iii), we get β = 2, α = -8

Substituting these values in (ii), we get λ = -16

Hence option B is the correct answer.
Question 10
3 / -1

A
y²=xz

B
x>y>z

C
a=9,b=1

D
All of the above

Solution

$x, y$ and $z$ are $A M, G M$ and $H M$ between $a, b$ respectively.
$\Rightarrow x=\frac{a+b}{2} \ldots \ldots(1)$
$\Rightarrow y=\sqrt{a b} \ldots(2)$
$\Rightarrow z=\frac{2 a b}{a+b} \ldots \ldots \ldots(3)$
But, given $a=5 z$ and $x=y+2$
Substituting $a=5 z$ in (3) gives $a=9 b$
Substituting $a=9 b$ in (1) and (2)
$\Rightarrow x=5 b$ and $y=3 b$
$x=y+2 \Rightarrow b=1$
Also, we get $z=\frac{9}{5} b$
$\operatorname{since}, A M \geq G M \geq H M$ and $G M^{2}=(A M)(H M)$
Hence, option D is the correct.
Question 11
3 / -1

If the equation x²+y²−10x+21=0 has real roots x=α and y=β then

A
3≤x≤7

B
3≤y≤7

C
−2≤x≤2

D
None of these

Solution

Hence option A is the correct answer.
Question 12
3 / -1

A
2 sq. units

B
3 sq. units

C
4 sq. units

D
(1/2) sq. units

Solution
Question 13
3 / -1

If α and β are the roots of the equation x² - a(x + 1) - b = 0, then $(α + 1) (β + 1) =$

A
b

B
-b

C
1-b

D
b-1

Solution

\begin{equation}
\begin{array}{l}
\text { Given equation } x^{2}-a(x+1)-b=0 \\
\Rightarrow x^{2}-a x-a-b \Rightarrow a+\beta=a, a \beta=-(a+b) \\
\text { Now }(a+1)(\beta+1)=a \beta+a+\beta+1=-(a+b)+a+1=1-b
\end{array}
\end{equation}
Hence option C is the correct answer.
Question 14
3 / -1

If the sum of the roots of the equation $λ x^{2} + 2 k + 3 λ = 0$ be equal to their product, then λ =

A
4

B
-4

C
6

D
-(2/3)

Solution

Under condition $- \frac{2}{λ} = 3 \Rightarrow λ = - \frac{2}{3}$
Hence option D is the correct answer.
Question 15
3 / -1

The area of the region y = ax − bx² bounded by x-axis in sq. units is

A
a³/6b²

B
a³/6

C
a

D
b

Solution