Mathematics Test

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Mathematics Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

\(\lim _{x \rightarrow 0} \frac{3^{x}-1}{\sqrt{x+1}-1}\) is equal to

A
log_e9

B
log_e3

C
0

D
1

Solution

\(\lim _{x \rightarrow 0} \frac{3^{x}-1}{\sqrt{x+1}-1}\)
\(=\lim _{x \rightarrow 0} \frac{3^{x}-1}{\sqrt{x+1}-1} \times \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\)
\(\lim _{x \rightarrow 0} \frac{3^{x}-1}{x} \lim _{x \rightarrow 0} \sqrt{x+1}+1\)
\(=\log 3 \times 2=2 \log 3=\log 9\)
Question 2
1 / -0

If cos⁻¹x−sin⁻¹x = 0 , then x is equal to

A
\(\pm \frac{1}{\sqrt{2}}\)

B
1

C
√2

D
1/√2

Solution

Given, \(\sin ^{-1}(x)-\cos ^{-1}(x)=0\)
\(\sin ^{-1} x+\cos ^{-1} x=\pi / 2\)
\(\frac{\pi}{2}=2 \cos ^{-1}(x)\)
\(\cos ^{-1}(x)=\frac{\pi}{4}\)
\(x=\cos \left(\frac{\pi}{4}\right)\)
\(x=\frac{1}{\sqrt{2}}\)
Question 3
1 / -0

A company has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?

A
165

B
185

C
205

D
225

Solution

The number of ways of selecting a group of eight is 5 men and 3 women \(=\) \({ }^{5} C_{5} \times{ }^{6} C_{3}=20\)
4 men and 4 women \(=\) \({ }^{5} C_{4} \times{ }^{6} C_{4}=75\)
3 men and 5 women \(=\) \({ }^{5} C_{3} \times{ }^{6} C_{5}=60\)
2 men and 6 women \(=\) \({ }^{5} C_{2} \times{ }^{6} C_{6}=10\)
Thus the total possible cases is \(20+75+60+10=165\)
Option \(\mathrm{A}\) is correct answer.
Question 4
1 / -0

The value of \(\lim _{h \rightarrow 0} \frac{\ln (1+2 h)-2 \ln (1+h)}{h^{2}}\) is

A
1

B
-1

C
0

D
none of these

Solution

Substituting gives us indeterminate form. So, applying L'Hospitals Rule, \(L=\lim _{x \rightarrow 0} \frac{\frac{2}{1+2 h}-\frac{2}{1+h}}{2 h}\)
Substituting gives us indeterminate form so, applying L'Hospitals Rule again, \(=\lim _{x \rightarrow 0} \frac{\frac{-4}{(1+2 h)^{2}}+\frac{2}{(1+h)^{2}}}{2}\)
\(=\frac{-4+2}{2}\) (direct substitution) \(=-1\)
Question 5
1 / -0

The set of points where x²|x| is differentiable thrice is

A
R

B
R − 0, ± 1

C
R − 0

D
None of the above

Solution

Let \(f(x)=x^{2}|x|\) which can be expressed as \(f(x)=\left\{\begin{aligned}-x^{3}, & x<0 \\ 0, & x=0 \therefore f^{\prime}(x)=\left\{\begin{aligned}-3 x^{2}, & x<0 \\ 0, & x=0 \\ 3 x^{2}, & x \geq 0 \end{aligned}\right.\\ x^{3}, \quad x \geq 0 \end{aligned}\right.\)
So, \(f^{\prime}(x)\) exists for all real \(x\). \(f^{\prime \prime}(x)=\left\{\begin{aligned}-6 x, & x<0 \\ 0, & x=0 \\ 6 x, & x \geq 0 \end{aligned}\right.\)
So, \(f^{\prime}(x)\) exists for all real \(x . f "(x)=\left\{\begin{array}{ll}-6, & x \leq 0 \\ 0, & x=0 \text { However, } f^{\prime \prime}(0) \text { does not } \\ 6, & x \geq 0\end{array}\right.\)
exists since \(f^{\prime \prime \prime}\left(0^{-}\right)=-6\) and \(f^{\prime \prime \prime}\left(0^{+}\right)=6\) which are not equal. Thus, the set of points where \(f(x)\) is thrice differentiable is \(R-0\) Hence, option 'C' is correct.
Question 6
1 / -0

\(f(x)=\left\{\begin{array}{cl}x^{2} & \text { for } \quad x \geq 1 \\ a x+b & \text { for } \quad x<1\end{array}\right.\)
If \(f\) is a differentiable function, then

A
a =−1,b = 2

B
a = 2, b = −1

C
\(a=-\frac{1}{2}, b=3\)

D
None of these

Solution

For \(\mathrm{f}\) to be differentiable it has to be continuous \(\Rightarrow \mathrm{LHS}=\mathrm{RHS}=f(1)\)
\(\Rightarrow a+b=1 \Rightarrow(1)\)
Also for \(f\) to be differentiable at \(x=1\), we have
\(\mathrm{LHD}=\mathrm{RHD}\)
\(\Rightarrow(2 x)_{x=1}=a \Rightarrow a=2\)
Now using ( 1\() b=-1\)
Question 7
1 / -0

Solve 5 (2x + 1) ≤ 35, where x is a natural number.

A
x ≤ 3

B
x = {1, 3}

C
x = 3

D
x = 3,4,5,6....... ∞

Solution

The given inequation is \(5(2 x+1) \leq 35\)
\(\Rightarrow 2 x+1 \leq 7\)
\(\Rightarrow 2 x \leq 6\)
\(\Rightarrow x \leq 3\)
\(\because \quad x \in N\)
\(\therefore \quad x=\{1,2,3\}\)
Question 8
1 / -0

The value of \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\) is equal to

A
7π/6

B
5π/6

C
π/3

D
π/6

Solution

The value of \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\)
\(=\cos ^{-1}\left[\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right]\)
\(=\cos ^{-1}\left(\cos \frac{5 \pi}{6}\right)\)
\(=\frac{5 \pi}{6}\)
Question 9
1 / -0

A bag contains 5 red balls and 8 blue balls. It also contains 4 green and 7 black balls. If a ball is drawn at random, then find the probability that it is not green.

A
5/6

B
1/4

C
1/6

D
7/4

Solution

Toatal number of favourable
cases that a ball is not green
\(=20\)
Total number of cases \(=24\) Probability \(=\frac{20}{24}=\frac{5}{6}\)
Question 10
1 / -0

Three numbers are chosen at random without replacement from {1,2,3,...,8}. The probability that their minimum is 3, given that their maximum is 6, is:

A
3/8

B
1/5

C
1/4

D
2/5

Solution

let event \(A=\) minimun is 3
let event \(B =\) maximum is 6
total ways \(=\cdot{ }^{8} C_{3}\) \(P(A \cap B)=\frac{2}{.^{8} C_{3}}=\frac{2}{\frac{8 !}{3 ! 5 !}}\)
\(=\frac{2 \times 3 !}{8 \times 7 \times 6}=\frac{1}{28}\)
\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
\(=\frac{\frac{1}{28}}{\frac{s^{5} C_{2}}{s C_{3}}}\)
\(=\frac{\frac{2}{s^{8} C_{3}}}{\frac{s C_{2}}{s C_{3}}}\)
\(=\frac{2}{5^{5} C_{2}}\)
\(=\frac{2 \times 2}{5 \times 4}=\frac{1}{5}\)
option 2 is correct

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 12

Result

Mathematics Test - 12

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SHARING IS CARING

Question 1

loge9

loge3

0

1

Solution

Question 2

\(\pm \frac{1}{\sqrt{2}}\)

1

√2

1/√2

Solution

Question 3

165

185

205

225

Solution

Question 4

1

-1

0

none of these

Solution

Question 5

R

R − 0, ± 1

R − 0

None of the above

Solution

Question 6

a =−1,b = 2

a = 2, b = −1

\(a=-\frac{1}{2}, b=3\)

None of these

Solution

Question 7

x ≤ 3

x = {1, 3}

x = 3

x = 3,4,5,6....... ∞

Solution

Question 8

7π/6

5π/6

π/3

π/6

Solution

Question 9

5/6

1/4

1/6

7/4

Solution

Question 10

3/8

1/5

1/4

2/5

Solution

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