New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Quadratic Equation and Inequalities question

2023 · 13 Apr · Shift 2 · Q42

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Quadratic Equation and Inequalities
2023 · 13 Apr · Shift 2 · Q42

Quadratic Equation and Inequalities question

2023 · 13 Apr · Shift 2 · Q42

JEE MainMathematicsQuadratic Equation and InequalitiesNumerical+4 / −1

Let

[\alpha]

denote the greatest integer

\leq \alpha

. Then

[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]

is equal to

\underline{\hspace{2cm}}

Numerical answer

View written solutionFree

Correct answer: 825

We need to compute $S=[\sqrt1]+[\sqrt2]+[\sqrt3]+\cdots+[\sqrt{120}].$ Here $[x]$ denotes the greatest integer function.
Observe that $[\sqrt n]=k$ whenever $k\le \sqrt n < k+1.$ Squaring, $k^2\le n < (k+1)^2.$ So for each integer $k$ , the value $k$ occurs for all integers $n=k^2,k^2+1,\dots,(k+1)^2-1,$ whose count is $((k+1)^2-1)-k^2+1=(k+1)^2-k^2=2k+1.$
Now list the ranges up to $120$ .

$[\sqrt n]=1$ for $n=1,2,3$ : $3$ terms
$[\sqrt n]=2$ for $n=4$ to $8$ : $5$ terms
$[\sqrt n]=3$ for $n=9$ to $15$ : $7$ terms
$[\sqrt n]=4$ for $n=16$ to $24$ : $9$ terms
$[\sqrt n]=5$ for $n=25$ to $35$ : $11$ terms
$[\sqrt n]=6$ for $n=36$ to $48$ : $13$ terms
$[\sqrt n]=7$ for $n=49$ to $63$ : $15$ terms
$[\sqrt n]=8$ for $n=64$ to $80$ : $17$ terms
$[\sqrt n]=9$ for $n=81$ to $99$ : $19$ terms
$[\sqrt n]=10$ for $n=100$ to $120$ : $21$ terms

Since $11^2=121>120$ , the maximum value is $10$ .

Therefore, $S=1\cdot3+2\cdot5+3\cdot7+4\cdot9+5\cdot11+6\cdot13+7\cdot15+8\cdot17+9\cdot19+10\cdot21.$
Compute this sum: \begin{align*} S&=3+10+21+36+55+78+105+136+171+210\ &=13+21+36+55+78+105+136+171+210\ &=34+36+55+78+105+136+171+210\ &=70+55+78+105+136+171+210\ &=125+78+105+136+171+210\ &=203+105+136+171+210\ &=308+136+171+210\ &=444+171+210\ &=615+210\ &=825. \end{align*}
Hence, $[\sqrt1]+[\sqrt2]+\cdots+[\sqrt{120}]=825.$
Comparison with stored answer: Stored correct answer = $825$ . This matches our derived answer.

More from Quadratic Equation and Inequalities

The number of real roots of the equation $x ∣ x ∣ - 5∣ x + 2∣ + 6 = 0$ , is :2023 · MCQ
The equation $x^{2} - 4 x + [x] + 3 = x [x]$ , where $[x]$ denotes the greatest integer function, has :2023 · MCQ
Let $λ \in R$ and let the equation E be $∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ = 0$ . Then the largest element in the set S = { $x + λ : x$ is an integer solution of E} is 2023 · Numerical
The number of real solutions of the equation $3 (x^{2} + \frac{1}{x ^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ , is2023 · MCQ
Let $α \in R$ and let $x^{2} + 6 0^{\frac{1}{4}} x + a = 0$ be the roots of the equation $x^{2} + 6 0^{\frac{1}{4}} x + a = 0$ . If $α^{4} + β^{4} = - 30$ , then the product of all possible values of $a$ is .2023 · Numerical
Let $λ e 0$ be a real number. Let $14 x^{2} - 31 x + 3 λ = 0$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $35 x^{2} - 53 x + 4 λ = 0$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0$ . Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ …2023 · MCQ
Let $α_{1}, α_{2}, ...., α_{7}$ be the roots of the equation $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$ and $∣ α_{1} ∣ \geq ∣ α_{2} ∣ \geq ... \geq ∣ α_{7} ∣$ . Then $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6}$ is…2023 · Numerical
If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and $x^{2} - a x - 5 = 0$ have a common real root is $\frac{3}{2 β}$ then $β$ is equal to .2023 · Numerical