Class 11th

Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to:

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Contributor-Level 9

x = 2t $A (2 t, \frac{t^{2}}{3})$

$y = \frac{t^{2}}{3}$ S (0, 3)

$3 y = {(\frac{x}{2})}^{2}$ $B (0, λ)$

$3 k = \frac{t^{2}}{3} + 3 + λ = \frac{2 t^{2}}{3} + 3 - \frac{12 t^{2}}{9 - t^{2}}$

$\underset{t \to 1}{l i m} 3 k = \frac{2}{3} + 3 - \frac{12}{8} = 3 - \frac{5}{6} = \frac{13}{6}$

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New answer posted

3 months ago

If y = m₁x + c₁ and y = m₂x + c₂, m₁ $\neq$ m₂ are two common tangents of circle x² + y² = 2 and parabola y² = x, Then the value of $8 | m_{1} m_{2} |$ is equal to:

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Class 11th Maths Binomial Theorem

Contributor-Level 9

$y^{2} = 4 (\frac{1}{4}) x$

$x - t y + \frac{1}{4} t^{2} = 0$

$\frac{1}{t} = m, m_{2} \Rightarrow (8 m_{1} m_{2}) = 3 \sqrt[]{2} - 4$

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New answer posted

3 months ago

If $\frac{1}{2 . 3^{10}} + \frac{1}{2^{2} . 3^{9}} + . . . . + \frac{1}{2^{10} . 3} = \frac{K}{2^{10} . 3^{10}},$ then the remainder when K is divided by 6 is:

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Contributor-Level 9

$t_{n} = \frac{1}{2^{n} . 3^{11 - n}} = {(\frac{1}{3})}^{11} . {(\frac{3}{2})}^{n}$

$S_{10} = {(\frac{1}{3})}^{11} . \frac{(\frac{3}{2}) . ({(\frac{3}{2})}^{10} - 1)}{\frac{3}{2} - 1}$

$= \frac{1}{3^{10}} . \frac{3^{10} - 2^{10}}{2^{10}}$

$= 1 +^{10} C_{1} . 2^{1} +^{10} C_{2} . 2^{2} + . . . . +^{10} C_{9} . 2^{9}$

k = 21 + 6m ® R = 3

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New answer posted

3 months ago

Consider the following two propositions:

$P 1 : \sim (p \to \sim q)$

$P 2 : (p \land \sim q) \land ((\sim p) \lor q)$

If the proposition p ® $((\sim p) \lor q)$ is evaluated as FALSE, then :

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Contributor-Level 9

$p_{1} : \sim (p \leftrightarrow \sim q)$ $p_{2} : (p \land \sim q) \land ((\sim p) \lor q)$

$= \sim ((\sim p) \lor (\sim q))$

$= \sim (\sim (p \land q))$

$p \land q$

$(\sim p) \lor ((\sim p) \lor q)$ is false (given)

p q p₁ p₂ $(\sim p) \lor q$

T F T

...more

$p_{1} : \sim (p \leftrightarrow \sim q)$ $p_{2} : (p \land \sim q) \land ((\sim p) \lor q)$

$= \sim ((\sim p) \lor (\sim q))$

$= \sim (\sim (p \land q))$

$p \land q$

$(\sim p) \lor ((\sim p) \lor q)$ is false (given)

p q p₁ p₂ $(\sim p) \lor q$

T F T

T F F

F T F T

F T

less

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New answer posted

3 months ago

The value of

$-^{15} C_{1} + 2 .^{15} C_{2} - 3 .^{15} C_{3} + . . . . - 15 .^{15} C_{15} +^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + . . . . +^{14} C_{11} i s :$

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Contributor-Level 10

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ = $\sum r .^{15} C_{r} x^{r - 1}$

put = x = -1

$\Rightarrow 0 = -^{15} C_{1} + 2 .^{15} C_{2} - . . . . . . . = A$

$B =^{14} C_{13} = 2^{13}$

$∴ A + B = 2^{13} - 14$

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New answer posted

3 months ago

If $e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + . . . \infty) l o g_{e} 2}$ satisfies the equation t² – 9t + 8 = 0, then the value of $\frac{2 s i n x}{s i n x + \sqrt[]{3} c o s x} (0 < x < \frac{π}{2}) i s$

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Contributor-Level 10

$e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + . . . . . \infty)} l o g_{e} 2$

$= e^{\frac{c o s^{2} x}{1 - c o s^{2} x} l o g_{e} 2} = e^{c o t^{2} x l o g_{e} 2} = 2^{c o t^{2} x}$

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t = 2^{c o t^{2} x} = 8 = 2^{3}$

$\Rightarrow c o t^{2} x = 3 = c o t^{2} \frac{π}{6}$

$\frac{2 s i n x}{s i n x + \sqrt[]{3} c o s x} = \frac{2 * \frac{1}{2}}{\frac{1}{2} + \sqrt[]{3} * \frac{\sqrt[]{3}}{2}} = \frac{1}{2}$

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New answer posted

3 months ago

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36,$ which is outside the parabola y² = 9x is:

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Contributor-Level 10

x² + y² = 36 ……(i)

y² = 9x .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A = \int_{0}^{3} (\sqrt[]{36 - x^{2}} - 3 \sqrt[]{x}) d x$

$= {[\frac{x \sqrt[]{36 - x^{2}}}{2} + 18 s i n^{- 1} \frac{x}{6}]}_{0}^{3} - 3 . {\frac{x^{3 / 2}}{3 / 2}]}_{0}^{3}$

$= 3 π - \frac{3 \sqrt[]{3}}{2}$

$∴$ Required area = $= \frac{1}{2} π {(6)}^{2} + 2 (3 π - \frac{3 \sqrt[]{3}}{2}) = (24 π - 3 \sqrt[]{3}) s q . u n i t$

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New answer posted

3 months ago

Two vertical poles are 150m apart and the height of one is three times that of other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

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Contributor-Level 10

BC = CN = x = 75

$c o t θ = \frac{3 h}{75}$

$t a n θ = \frac{h}{75}$

$\frac{3 h^{2}}{75 * 75} = 1 \Rightarrow h = \frac{75}{\sqrt[]{3}} = 25 \sqrt[]{3}$

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New answer posted

3 months ago

Let p and q be two positive numbers such that p + q = 2 and p⁴ + q⁴ = 272. Then p and q are roots of the equation:

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Contributor-Level 10

p + q = 2

p⁴ + q⁴ = 272

${(p^{2} + q^{2})}^{2} - 2 p^{2} q^{2} = 272$

${[{(p + q)}^{2} - 2 p q]}^{2} - 2 p^{2} q^{2} = 272$

Let pq = t -> (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

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New answer posted

3 months ago

The statement among the following that is a tautology is:

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