Class 12th

Let f(x) = 2x² – x – 1 and $S = {n \in Z : | f (n) | < 800} .$ Then, the value of $\sum_{n \in S}^{} f (n)$ is equal to

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Contributor-Level 10

$| f (x) | \leq 800 \Rightarrow 2 n^{2} - n - 1 \leq 800$

$\Rightarrow 2 n^{2} - n - 801 \leq 0$

$\sum_{x \in S}^{} f (x) = \sum_{}^{} (2 x^{2} - x - 1)$

$= 2 (1 9^{2} + 1 8^{2} + . . . . . . . . 1^{2} + 0^{2} + 1^{2} + . . . . . + 2 0^{2})$

= 10620

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New answer posted

a year ago

Let M and N be the number of points on the curve y⁵ – 9xy + 2x = 0, where the tangents to the curve are parallel to x-axis and y-axis, respectively. Then the value of M + N equals

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Contributor-Level 10

y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{d y}{d x}$ 9y + 2 = 0

$\frac{d y}{d x} = \frac{9 y - 2}{5 y^{4} - 9 x}$

For horizontal tangent $\frac{d y}{d x} = 0 \Rightarrow y = \frac{2}{9}$ which does not satisfy the equation so no horizontal

For vertical tangent $5 y^{4} - 9 x = 0$

m = 0, N = 2

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New answer posted

a year ago

Let y = y(x) be the solution curve of the differential equation $s i n (2 x^{2}) l o g_{e} (t a n x^{2}) d y + (4 x y - 4 \sqrt[]{2} x s i n (x^{2} - \frac{π}{4})) d x = 0,$ $0 < x < \sqrt[]{\frac{π}{2},}$ which passes through the point $(\sqrt[]{\frac{π}{6}, 1})$ . Then $| y (\sqrt[]{\frac{π}{3}}) |$ is equal to

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Contributor-Level 10

$s i n (2 x^{2}) . 10 9_{e} (t a n x^{2}) d y + 4 x y d x = 4 \sqrt[]{2} . x . (s i n x^{2} c o s \frac{π}{4} - c o s x^{2} s i n \frac{π}{4}) d x$

$\Rightarrow l n (t a n x^{2}) d y + \frac{4 x}{s i n (2 x^{2})} y d x = \frac{4 x (s i n x^{2} - c o s x^{2})}{s i n (2 x^{2})} d x$

Integrate

$\Rightarrow y . l n (t a n x^{2}) = 2 . l n (\frac{s i n x^{2} + c o s x^{2} - 1}{s i n x^{2} + c o s x^{2} + 1}) + C$

$x = \sqrt[]{\frac{π}{6}}, y = 1$ calculate C.

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New answer posted

a year ago

Let the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z - 3}{- 4}$ intersect the plane containing the lines $\frac{x - 4}{1} = \frac{y + 1}{- 2} = \frac{z}{1}$ and 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3, a $\in R$ at the point $P (α, β γ) .$ Then the value of + + $γ$ equals

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Class 12th Inverse Trigonometric Functions

Contributor-Level 10

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is $x (4 a + 2 λ) + y (- 1 - 5 λ) + z (5 - λ) = 7 a + 3 λ$

This plane contains 4, -1, 0

9a + 1 + 10 = 0…… (i)

Plane contains the line $\frac{x - 4}{1} = \frac{y + 1}{- 2} = \frac{z}{1}$

$4 a + 11 λ + 7 = 0$ ……. (ii)

From (i) & (ii) a = 1, $λ$ =1

Equation of plane $π \equiv x + 2 y + 3 z - 2 = 0$

$\Rightarrow$ $7$ $P$ $+$ $3$ $-$ $2$ $P$ $+$ $4$ $-$ $1$ $2$ $P$ $+$ $9$ $-$ $2$ $=$ $0$ $\Rightarrow$ $P$ $=2$

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New answer posted

a year ago

For $k \in R,$ let the solutions of the equation $c o s (s i n^{- 1} (x c o t (t a n^{- 1} (c o s (s i n^{- 1} x)))))$ =, 0 < $| x | < \frac{1}{\sqrt[]{2}}$ be and , where the inverse trigonometric functions take only principal values. If the solutions of the equation x2 – bx – 5 = 0 are $\frac{1}{α^{2}} + \frac{1}{β^{2}} a n d \frac{α}{β} t h e n \frac{b}{k^{2}}$ is equal to

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Contributor-Level 10

use $s i n^{- 1} x = c o s^{- 1} \sqrt[]{1 - x^{2}}$

$t a n^{- 1} \sqrt[]{1 - x^{2}} = c o t^{- 1} \frac{1}{\sqrt[]{1 - x^{2}}}$

$s i n^{- 1} \frac{x}{\sqrt[]{1 - x^{2}}} = c o s^{- 1} \frac{\sqrt[]{1 - 2 x^{2}}}{\sqrt[]{1 - x^{2}}}$

Sum of roots b = 1 + 2 $\frac{(k^{2} - 1)}{k^{2} - 2}$

Product of roots 5 = 2 $(\frac{k^{2} - 1}{k^{2} - 2})$

b = 4, k² = $\frac{1}{3}$

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New answer posted

a year ago

In the following reaction

The % yield for reaction I is 60% and that of reaction II is 50%. The overall yield of the complete reaction is___________% [Nearest integer]

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Contributor-Level 10

Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 * 0.5 n = 0.3 n

So, overall % age yield of complete reaction = $\frac{0.3 n}{n} * 100 = 30 %$

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New answer posted

a year ago

Optical activity of an enantiomeric mixture is + 12.6° and the specific rotation of (+) isomer is +30°. The optical purity is_____________%

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Contributor-Level 10

$% O p t i c a l p u r i t y = \frac{O b s e r v e d r o t a t i o n o f m i x t u r e * 100}{r o t a t i o n o f p u r e e n a u t i o m e r}$

$= \frac{+ 12.6}{+ 30} * 100 = 42 %$

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New answer posted

a year ago

In the titration of KMnO₄ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is____________.

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Class 12th Chemistry Coordination Compounds

Contributor-Level 10

$2 K M n O_{4} + 5 H_{2} {\overset{+ 3}{C}}_{2} O_{4} + 3 H_{2} S O_{4} (d i l) \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C {\overset{+ 4}{O}}_{2} + 8 H_{2} O$

Oxidation state of carbon changes from +3 to +4

$∴$ change in O. S is 1

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New answer posted

a year ago

The conductivity of a solution of complex with formula CoCl₃(NH₃)₄ corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is__________________.

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Contributor-Level 10

Complex is $[C o {(N H_{3})}_{4} C l_{2}] C l$

Primary valency = No. of ionisable species = 1

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New answer posted

a year ago

At 310 K, the solubility of CaF₂ in water is 2.34 * 10^-3 g/ 100 mL. The solubility product of CaF₂ is ____________* 10^-8 (mol/L)³. (Give molar mass : CaF₂ = 78 gmol^-1)

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