MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Quantitative Aptitude

Progressions

Which of the ...

Question

Which of the following pairs is not the additive inverse of each other?
A. $\frac{1}{7}, - \frac{1}{7}$
B. $- 1 \frac{1}{3}, 1 \frac{1}{3}$
C. $0, \frac{1}{7}$
D. $- \frac{1}{5}, \frac{1}{5}$

Open in App

Solution

-

Suggest Corrections

thumbs-up

0

Similar questions

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

If $f = \frac{x}{1 + x^{2}} + \frac{1}{3} {(\frac{x}{1 + x^{2}})}^{3} + \frac{1}{5} {(\frac{x}{1 + x^{2}})}^{5} + . . .$ to $\infty$ and $g = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} - \frac{2}{9} x^{9} + . . .$ , then $f = d \times g$ . Find 4d.

ω

is complex cube root of unity and a, b, c are three real numbers such that

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

can be written as product of two determinants.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Fear of the Dark

QUANTITATIVE APTITUDE

Watch in App

explore_more_icon

Explore more

Other Quantitative Aptitude

Join BYJU'S Learning Program

CrossIcon