Mathematics Test

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Mathematics Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

The set of solutions satisfying both x²+5x+6≥0 and x²+3x−4<0 is :

A
(−4,1)

B
(−4,−3] ∪ [−2,1)

C
(−4,−3) ∪ (−2,1)

D
[−4,−3] ∪ [−2,1]

Solution

\(x^{2}+5 x+6 \geq 0 \Rightarrow(x+2)(x+3) \geq 0 \Rightarrow x \in(-\infty,-3] \cup[-2, \infty)\)
and \(x^{2}+3 x-4<0 \Rightarrow(x+4)(x-1)<0 \Rightarrow x \in(-4,1)\)
Hence set of solution satisfying both is (-4,-3]\(\cup[-2,1).\)
Hence option B is the correct answer.
Question 2
1 / -0

Find the solution of the inequality \((x-3)<\sqrt{x^{2}+4 x-5}\) ?

A
(−∞,−5]∪[1,∞)

B
(−∞,−5]

C

\(\left(\frac{7}{5}, \infty\right)\)

D
(−∞,1]∪[5,∞)

Solution

\((x-3)<\sqrt{x^{2}+4 x-5}\)
\(\Rightarrow\)x-3²\(\Rightarrow-10 x<-14\)
\(\Rightarrow x>\frac{7}{5}\).......(i)
From the given inequality, it follows that
\(x^{2}+4 x-5 \geq 0\)
\(\Rightarrow(x+5)(x-1) \geq 0\)
\(\Rightarrow x \in(-\infty,-5] \cup[1, \infty)\).........(ii)
From ( 1 ) and (2) , it follows that
\(\Rightarrow x \in(-\infty,-5] \cup[1, \infty)\)
Hence option A is the correct answer.
Question 3
1 / -0

The figure of the function P(x)=ax²+bx+c is graphed. Find which of the following would be positive for the given scenario.

A
ab

B
b-a

C
-c

D
ac

Solution

The parabola is an open upward parabola.
Hence \(a>0\) Now the \(y_{\text {intercept }}\) is greater than \(0.\)
Hence, \(P(0)>0\) or \(c>0\)
Both the roots are positive. Therefore, sum of roots is positive. Hence, sum of roots \(>0\) or \(\frac{-b}{a}>0\)
But \(a>0\) already, hence \(b<0\)
Therefore, \(a>0, c>0\) but \(b<0\).

Hence option D is the correct answer.
Question 4
1 / -0

Let \(f(x)=x^{2}-x+1, x \geq\left(\frac{1}{2}\right)\) then the solution of the equation \(f(x)=f^{-1}(x)\) is :

A
x=1

B
x=2

C

\(x=\frac{1}{2}\)

D
None of these

Solution

\(y=x^{2}-x+1 \Rightarrow x^{2}-x+(1-y)=0\)
Here, \(a=1, b=-1\) and \(c=1-y\)
\(x=\frac{1 \pm \sqrt{1-4(1-y)}}{2}\left[\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right]\)
\(\because x>\frac{1}{2}, \quad \therefore x=\frac{1}{2}+\sqrt{y-\frac{3}{4}}\)
\(\Rightarrow f^{-1}(x)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}\)
Now, \(x^{2}-x+1=\frac{1}{2}+\sqrt{x-\frac{3}{4}}\)
since the graphs of the original and inverse functions can intersect only on the straight line \(y=x,\) therefore \(x=f(x) \quad \Rightarrow x=x^{2}-x+1\)
\(\Rightarrow x^{2}-2 x+1=0\)
\(\Rightarrow(x-1)^{2}=0\)
\(\Rightarrow x=1\)
Hence option A is the correct answer.
Question 5
1 / -0

If \(f(x)\) is a polynomial function such that \(f(x) f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)\) and \(f(3)=-80\) then \(f(x)\) is equal to:

A
x⁴+1

B
x⁴−1

C
1−x⁴

D
−1−x⁴

Solution

Consider \(f(x)=1-x^{n}\)
Therefore the above function complies with the rule that \(f(x)+f(1 / x)=f(x) \cdot f(1 / x)\)
\(=\frac{-(1-x)^{2 n}}{x^{n}}\)
Now substituting \(x=3\) we get \(f(x)\) as \(1-3^{n}\)
\(=-80\)
\(81=3^{n}\)
\(n=4\)
Therefore \(f(x)=1-x^{4}\)
Hence option C is the correct answer.
Question 6
1 / -0

If (3,1) is a solution of the equation 3x+2y=k find the value of k.

A
11

B
10

C
12

D
14

Solution

(3,1) is a solution of the equation 3x+2y=k
3x+2y=k
=>3(3)+2(1)=k
=>k=9+2=11
Hence option A is the correct answer.
Question 7
1 / -0

If f:R→R and g:R→R are defined by f(x)=x−[x] and g(x)=[x] for x∈R, where [x] is the greatest integer not exceeding x, then for every x∈R,f(g(x))=

A
x

B
0

C
f(x)

D
g(x)

Solution

We have, \(\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}]=\{\mathrm{x}\} \ldots\) (1)
where \(\{x\}\) is the fractional part of \(x\)
If \(\mathrm{g}(\mathrm{x})=[\mathrm{x}],\) where \([\mathrm{x}]\) is the greatest integer part of \(\mathrm{x},\) then
The value of \(\mathrm{g}(\mathrm{x})\) will always be an integer.
And \(\therefore \mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{f}([\mathrm{x}])=0\)
\(\therefore\) the fractional part of \(\mathrm{g}(\mathrm{x})\) i.e \([\mathrm{x}]\) is always \(0 .\)
Hence option B is the correct answer.
Question 8
1 / -0

The domain of \(f(x)=\sin ^{-1}\left(\frac{x-4}{3}\right)+\log (2-x)\) is :

A
[1,4]

B
[1,2)

C
[1,3)

D
[0,1)

Solution

We know that, \(-1 \leq \sin x \leq 1\)
For \(f\) to be defined,
\(-1 \leq \frac{x-4}{3} \leq 1\) and \(2-x>0\)
\(\Rightarrow-3 \leq x-4 \leq 3\) and \(x<2\)
\(\Rightarrow 1 \leq x \leq 7\) and \(x<2\)
From above required domain is [1,2).
Hence option B is the correct answer.
Question 9
1 / -0

If \(f(a)=\log \frac{2+a}{2-a}\) for \(0

A
f(x)

B
2f(x)

C
\(\frac{1}{2}\)f(x)

D
−f(x)

Solution

\(\operatorname{since} f(x)=\log \left(\frac{2+a}{2-a}\right),\) we have
\(\frac{1}{2} f\left(\frac{8 a}{4+a^{2}}\right)=\frac{1}{2} \log \left(\frac{2+\frac{8 a}{4+a^{2}}}{2-\frac{8 a}{4+a^{2}}}\right)=\frac{1}{2} \log \left[\frac{8+2 a^{2}+8 a}{8+2 a^{2}-8 a}\right]=\frac{1}{2} \log \left[\frac{4+a^{2}+4 a}{4+a^{2}-4 a}\right]=\)
Applying \(\log x^{a}=a \log x,\) we get
\(\frac{1}{2} \log \left(\frac{2+a}{2-a}\right)^{2}=\frac{2}{2} \log \left(\frac{2+a}{2-a}\right)\)
Hence, \(\frac{1}{2} f\left(\frac{8 a}{4+a^{2}}\right)=f(x)\)
Hence option A is the correct answer.
Question 10
1 / -0

If \(f: R \rightarrow R\) and \(g: R \rightarrow R\) are functions defined by \(f(x)=3 x-1 ; g(x)=\sqrt{x+6},\) then the value of \(\left(g \circ f^{-1}\right)(2009)\) is:

A
26

B
29

C
16

D
15

Solution

Given \(f(x)=3 x-1, g(x)=\sqrt{x+6}\)
Let \(f(x)=y\) \(\Rightarrow y=3 x-1 \Rightarrow y+1-3 x \Rightarrow x=\frac{y+1}{3}\)
Now, \(f^{-1}(y)=x=\frac{y+1}{3}\)
\(\Rightarrow f^{-1}(x)=\frac{x+1}{3}\) and \(g(x)=\sqrt{x+6}\)
Consider, \(\left(g \circ f^{-1}\right)(2009)=g\left[f^{-1}(2009)\right]\)
\(=g\left(\frac{2009+1}{3}\right)=g\left(\frac{2010}{3}\right)\)
\(=\sqrt{\frac{2010}{3}+6}=\sqrt{\frac{2028}{3}}=\sqrt{676}=26\)
Hence, option \(A\) is correct.

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 6

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Mathematics Test - 6

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SHARING IS CARING

Question 1

(−4,1)

(−4,−3] ∪ [−2,1)

(−4,−3) ∪ (−2,1)

[−4,−3] ∪ [−2,1]

Solution

Question 2

(−∞,−5]∪[1,∞)

(−∞,−5]

\(\left(\frac{7}{5}, \infty\right)\)

(−∞,1]∪[5,∞)

Solution

Question 3

ab

b-a

-c

ac

Solution

Question 4

x=1

x=2

\(x=\frac{1}{2}\)

None of these

Solution

Question 5

x4+1

x4−1

1−x4

−1−x4

Solution

Question 6

11

10

12

14

Solution

Question 7

x

0

f(x)

g(x)

Solution

Question 8

[1,4]

[1,2)

[1,3)

[0,1)

Solution

Question 9

f(x)

2f(x)

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

−f(x)

Solution

Question 10

26

29

16

15

Solution

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x⁴+1

x⁴−1

1−x⁴

−1−x⁴