If (nr−tn+14)If (nr−tn+14){\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right) be a perfect square, then the values of t are = ?" /> If (nr−tn+14)If (nr−tn+14){\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right) be a perfect square, then the values of t are = ?" /> If (nr−tn+14)If (nr−tn+14){\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right) be a perfect square, then the values of t are = ?" />

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$If (n^{r} - t n + \frac{1}{4})$ ${\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right)$ be a perfect square, then the values of t are = ?

1± 2
21, 2
32, 3
4± 1

Answer:- 1

Explanation:-

Solution:

\begin{aligned} If (n^{r} - t n + \frac{1}{4}) be a perfect square \\ r = 2 t = \pm 1 \\ (If t = 1) n^{2} - n + \frac{1}{4} \\ = n^{2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \\ = {(n - \frac{1}{2})}^{2} \\ (If t = - 1) n^{2} + n + \frac{1}{4} \\ = n^{2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \\ = {(n + \frac{1}{2})}^{2} \end{aligned}

$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr}$

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be a perfect square, then the values of t are = ?", "text": "

If (n^{r} - t n + \frac{1}{4})

${\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right)$ be a perfect square, then the values of t are = ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} If (n^{r} - t n + \frac{1}{4}) be a perfect square \\ r = 2 t = \pm 1 \\ (If t = 1) n^{2} - n + \frac{1}{4} \\ = n^{2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \\ = {(n - \frac{1}{2})}^{2} \\ (If t = - 1) n^{2} + n + \frac{1}{4} \\ = n^{2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \\ = {(n + \frac{1}{2})}^{2} \end{aligned}

$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} If (n^{r} - t n + \frac{1}{4}) be a perfect square \\ r = 2 t = \pm 1 \\ (If t = 1) n^{2} - n + \frac{1}{4} \\ = n^{2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \\ = {(n - \frac{1}{2})}^{2} \\ (If t = - 1) n^{2} + n + \frac{1}{4} \\ = n^{2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \\ = {(n + \frac{1}{2})}^{2} \end{aligned}

$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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