Mathematics Test

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Mathematics Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The line x = c cuts the triangle with corners (0, 0), (1, 1) and (9, 1) into two regions. For the area of the two regions to be the same c must be equal to :

A
5/2

B
3

C
7/2

D
3 or 15

Solution

The area of the triagle formed by three given vertices is

Now the line x=c bisect the area of triangle in two equal parts so the area of the both the parts of the triangle must be four
Question 2
1 / -0

Which of the following functions is periodic?

A
f (x) = x - [x] where [x] denotes the greatest integer less than or equal to the real number x

B

\(f(x)=\sin \frac{1}{x}\) for \(x=0, f(0)=0\)

C
f(x) = x cos x

D
None of these

Solution

Clearly, \(f(x)=x-[x]=\{x\}\)
which has period 1
Let \(\sin \left(\frac{1}{\mathrm{x}}\right)\) be periodic with period \(\mathrm{T}\)
then \(\sin \left(\frac{1}{x+T}\right)=\sin \left(\frac{1}{x}\right)\)
\(\Rightarrow \frac{1}{\mathrm{x}+\mathrm{T}}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{1}{\mathrm{x}}\)
\(\Rightarrow x=\left(x^{2}+T x\right) n \pi+(-1)^{n}(x+T)\)
\(\Rightarrow T=\frac{x\left(1-(-1)^{n}\right)-x^{2} n \pi}{n \pi x+(-1)^{n}}\)
Now for a variable \(x\) and constant \(T\) the given relation cannot hold \(\forall\) allowable \(x,\) Hence \(\sin \frac{1}{x}\) is not periodic
Similarly for \(f(x)=x \cos x\) let \(T\) be the Period
hence, \((x+T) \cos (x+T)=x \cos x\)
\(\Rightarrow \mathrm{T} \cos (\mathrm{x}+\mathrm{T})=\mathrm{x}\{\cos \mathrm{x}-\cos (\mathrm{x}+\mathrm{T})\}\)
\(\Rightarrow \mathrm{T} \cos (\mathrm{x}+\mathrm{T})=2 x \sin \left(x+\frac{T}{2}\right) \sin \frac{T}{2}\)
\(\Rightarrow \frac{T}{\sin \frac{T}{2}}=\frac{2 x \sin \left(x+\frac{T}{2}\right)}{\cos (x+\mathrm{T})}\)
Note that LHS is a constant while RHS varies as \(\mathrm{x}\) varies for allowable values of \(\mathrm{x}\), hence no such \(\mathrm{T}\) is possible, so \(\mathrm{x} \cos \mathrm{x}\) is also non-periodic.
Hence, the correct option is (A)
Question 3
1 / -0

In triangle \(\Delta \mathrm{ABC}, \mathrm{CD}\) is the bisector of the angle \(\mathrm{C},\) If \(\cos C 2=1 / 3\) and \(\mathrm{CD}=6,\) then the value of \(1 / \mathrm{a}+1 / \mathrm{b}\). Where \(a, b\) and \(c\) are the lengths of the sides opposite to Angles \(a, b \& c\) respectively)

A
1/5

B
1/9

C
1/8

D
1/4

Solution
Question 4
1 / -0

The focal chord toy ² = 16xis tangent to\((x-6)^{2} y^{2}=2\), then the possible values of the slope of this chord, are

A
{-1, 1}

B
{-2, 2}

C
\(\{-2,1 / 2\}\)

D
\(\{2,-1 / 2\}\)

Solution

Here, the focal chord of \(y^{2}=16 x\) is tangent to circle \((x-6)^{2}+y^{2}=2\)
\(\Rightarrow\) focus of parabola as \((a, 0)\) i.e (4,0)
\(\Rightarrow\) centre and radius of circle is (6,0) and \(\sqrt{2}\) respectively
Thus the equation of focal chord will be
\((y-0)=m(x-4)\)
\(m x-y-4 m=0\)
As this line is tangent to the circle
\(\therefore \frac{|6 m-4 m|}{\sqrt{m^{2}+1}}=\sqrt{2}\)
\(\Rightarrow \frac{4 m^{2}}{m^{2}+1}=2\)
\(4 m^{2}=2 m^{2}+2\)
\(2 m^{2}=2 \Rightarrow m^{2}=1\)
\(m=\pm 1\)
therefore the slope of the focal chord as a tangent to circle \(=\pm 1\)
Hence, the correct option is (A)
Question 5
1 / -0

\(A=\left[\begin{array}{ccc}2 & +3 & +-4 \\ +1 & +0 & +6 \\ +-2 & +1 & +5\end{array}\right], B=\left[\begin{array}{ccc}5 & +1 & +2 \\ +6 & +-1 & +4 \\ +5 & +3 & +-4\end{array}\right]\), find \(2 A-3 B\)

A

\(\left[\begin{array}{ccc}-11 & +3 & +-14 \\ +-16 & +3 & +0 \\ +-19 & +-7 & +22\end{array}\right]\)

B

\(\left[\begin{array}{ccc}11 & +3 & +14 \\ +16 & +3 & +0 \\ +19 & +7 & +22\end{array}\right]\)

C

\(\left[\begin{array}{ccc}11 & +6 & +22 \\ +16 & +0 & +3 \\ +19 & +7 & +14\end{array}\right]\)

D
None of these

Solution

\(2 \mathrm{A}-3 \mathrm{B}=2 \mathrm{A}(-3) \mathrm{B}=2\left[\begin{array}{ccc}2 & +3 & +-4 \\ +1 & +0 & +6 \\ +-2 & +1 & +5\end{array}\right]+(-3)\left[\begin{array}{ccc}5 & +1 & +2 \\ +6 & +-1 & +4 \\ +5 & +3 & +-4\end{array}\right]\)
\(=\left[\begin{array}{cccc}4 & +6 & +-8 \\ +2 & +0 & +12 \\ +-4 & +2 & +10\end{array}\right]+\left[\begin{array}{cccc}-15 & +-3 & +-6 \\ +-18 & +3 & +-12 \\ 4-15 & +6-3 & +-8-6 \\ +2-18 & +0+3 & +12-12 \\ +-4-15 & +2-9 & +10+12\end{array}\right]=\left[\begin{array}{ccc}-11 & +3 & +-14 \\ +-16 & +3 & +0 \\ +-19 & +-7 & +22\end{array}\right]\)
Hence, the correct option is (A)
Question 6
1 / -0

If \(f(x)\) is a polynomial satisfying \(f(x) f(1 / x)=f(x) f(1 / x)\) and \(f(2)>1\) then \(\lim _{x \rightarrow 1} f(x)\) is :

A
2

B
1

C
- 1

D
None of these

Solution

\(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(1 / \mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}(1 / \mathrm{x}), \mathrm{f}(2)>1\)
let \(\mathrm{f}(\mathrm{x})=\mathrm{a}_{0} \mathrm{x}^{\mathrm{n}}+\mathrm{a}_{1} \mathrm{x}^{\mathrm{n}-1}+\ldots+\mathrm{a}_{\mathrm{n}-1} \mathrm{x}+\mathrm{a}_{\mathrm{n}}, \mathrm{a}_{0} \neq 0\)
Then by comparing the coefficients of like powers, we get
\(a_{n}=1, a_{0}^{2}=1, a_{1}=a_{2}=_{-m}=a_{n-1}=0 \therefore \quad f(x)=-x^{n}+1\)
or \(f(x)=x^{n} 1\)
Then \(\quad \mathrm{f}(2)>1 f(x)=x^{n}+1\)
\(\therefore \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1}\left(x^{n} 1\right)=2\)
Hence, the correct option is (A)
Question 7
1 / -0

If (1 + x + x² )²⁵ = a₀ + a₁ x + a₂ x² + .............a₅₀ x⁵⁰ then a₀ + a₂ + a₄ + .........+ a₅₀ is --

A
odd and of the form 3n

B
odd and of the form (3n - 1)

C
odd and of the form (3n + 1)

D
even

Solution

Putting \(x=1 \&-1 \&\) adding
\begin{equation*}
\begin{array}{l}
\qquad \begin{array}{l}
a_{0}+a_{2}+\ldots \ldots+a_{50}=\frac{3^{25}+1}{2}=\frac{(1+2)^{25}+1}{2} \\
=\frac{25 C _{0}+{ }^{25} C _{1} \cdot 2+{ }^{25} C _{2} \cdot 2^{2}+\ldots+{ }^{25} C _{25} \cdot 2^{25}+1}{2}=\frac{2\left[1+{ }^{25} C _{1}+{ }^{25} C _{2} \cdot 2+\ldots+{ }^{25} C _{25} \cdot 2^{24}\right]}{2}
\end{array} \\
=2\left[13+25 C _{2}+\ldots \ldots+25 C _{25} \cdot 2^{23}\right] \Rightarrow \text { even }
\end{array}
\end{equation*}
Concepts:
Main Concept. Summation of Binomial Coefficients(1) The sum of binomial coefficients in the expansion of \((1 x)^{n}\) is \(2^{n}\) \(2^{n}=C_{0} C_{1} C_{2} \ldots . . C_{n}\)
(2) Sum of binomial coefficients with alternate signs:
\(C_{0}-C_{1} C_{2}-C_{3-}=0\)
(3) Sum of the coefficients of the odd terms in the expansion of \((1 x)^{n}\) is equal to sum of the coefficients of even terms and each is equal to \(2^{n-1}\) Hence, the correct option is (D)
Question 8
1 / -0

If 5a + 4b + 20c = t, then the value of t for which the line ax + by + c - 1 = 0 always passes through a fixed point is

A
0

B
20

C
30

D
None of these

Solution

\(a x+b y+c=1\)
\(\Rightarrow a x+b y+\frac{t-5 a-4 b}{20}=1\)
\(\Rightarrow 20 a x+20 b y+t-5 a-4 b=20\)
\(\Rightarrow a(20 x-5)+b(20 y-4)+(t-20)=0\)
If \(t=20,\) then line passes through \(\left(\frac{1}{4}, \frac{1}{5}\right)\)
Hence, the correct option is (B)
Question 9
1 / -0

The area of the curvilinear triangle bounded by the \(y\) -axis \(\&\) the curve \(y=\tan x \& y=2 / 3 \cos x\) is

A

\(\frac{1}{3} \log \frac{\sqrt{3}}{2}\)

B

\(\frac{1}{3}-\log \left(\frac{\sqrt{3}}{2}\right)\)

C

\(-\frac{1}{3} \log \left(\frac{\sqrt{3}}{2}\right)\)

D
None of these

Solution
Question 10
1 / -0

The equation of the pair of straight lines parallel to the \(y\) - axis and which are tangents to the circle \(x^{2}+y^{2}-6 x-4 y\) \(12=0\) is:

A
x² - 4x - 21 = 0

B
x² - 5x + 6 = 0

C
x² - 6x - 16 = 0

D
None of these

Solution

Given circle is \(x^{2} y^{2}-6 x-4 y-12=0\)

\(\Rightarrow(x-3)^{2}+(y-2)^{2}=25=5^{2}\)
Centre \(=(3,2)\) radius \(=5\)
If a line parallel to \(y\) -axis, \(x=\lambda\) touches the given circle, then
\(\Rightarrow \frac{3-\lambda}{\sqrt{1}}=\pm 5 \quad\) (radius)
\(\Rightarrow 3-\lambda=\pm 5\)
\(\lambda=3 \pm 5=-2,8\)
Pair of lines \(\quad(x+2)(x-8)=0\)
\(\Rightarrow \quad x^{2}-6 x-16=0\)
Hence, the correct option is (C)

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Mathematics Test - 30

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Mathematics Test - 30

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SHARING IS CARING

Question 1

5/2

3

7/2

3 or 15

Solution

Question 2

f (x) = x - [x] where [x] denotes the greatest integer less than or equal to the real number x

\(f(x)=\sin \frac{1}{x}\) for \(x=0, f(0)=0\)

f(x) = x cos x

None of these

Solution

Question 3

1/5

1/9

1/8

1/4

Solution

Question 4

{-1, 1}

{-2, 2}

\(\{-2,1 / 2\}\)

\(\{2,-1 / 2\}\)

Solution

Question 5

\(\left[\begin{array}{ccc}-11 & +3 & +-14 \\ +-16 & +3 & +0 \\ +-19 & +-7 & +22\end{array}\right]\)

\(\left[\begin{array}{ccc}11 & +3 & +14 \\ +16 & +3 & +0 \\ +19 & +7 & +22\end{array}\right]\)

\(\left[\begin{array}{ccc}11 & +6 & +22 \\ +16 & +0 & +3 \\ +19 & +7 & +14\end{array}\right]\)

None of these

Solution

Question 6

2

1

- 1

None of these

Solution

Question 7

odd and of the form 3n

odd and of the form (3n - 1)

odd and of the form (3n + 1)

even

Solution

Question 8

0

20

30

None of these

Solution

Question 9

\(\frac{1}{3} \log \frac{\sqrt{3}}{2}\)

\(\frac{1}{3}-\log \left(\frac{\sqrt{3}}{2}\right)\)

\(-\frac{1}{3} \log \left(\frac{\sqrt{3}}{2}\right)\)

None of these

Solution

Question 10

x2 - 4x - 21 = 0

x2 - 5x + 6 = 0

x2 - 6x - 16 = 0

None of these

Solution

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x² - 4x - 21 = 0

x² - 5x + 6 = 0

x² - 6x - 16 = 0