If $t$ is a number greater than $1$ , then $t^{2}$ is how much greater than $t$ ?

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t (t - 1)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(t - 1) (t + 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $t (t - 1)$
Given, $t$ $>$ $1$
Let us consider that $^{'}$ $t^{2}$ $^{'}$ is greater than $^{'} t^{'}$ by $^{'} n^{'}$
Formally, $n$ $+$ $t$ $=$ $t^{2}$
$\Rightarrow n$ $=$ $t^{2}$ $-$ $t$
$\Rightarrow n$ $=$ $t$ $(t - 1)$
Therefore, $t^{2}$ is $^{'}$ $t$ $(t - 1)$ $^{'}$ greater than $t$ .

Suggest Corrections

Similar questions

T_{1}

T_{2}

x = {sin}^{- 1} t

y = log (1 - t^{2})

\frac{d^{2} y}{d x^{2}}

t = 1 / 2

A = [\begin{matrix} t & t + 1 t - 1 & t \end{matrix}]

A A^{T} = I_{2}

t^{2} + t + 1 = 0,

(t + \frac{1}{t}) + (t^{2} + \frac{1}{t^{2}}) + \dots + (t^{27} + \frac{1}{t^{^{27}}})

t^{2} + t + 1 = 0

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + {(t^{3} + \frac{1}{t^{3}})}^{2} . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}

Join BYJU'S Learning Program