Maths

Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that $∠ P R Q = 60 °,$ then the area of $Δ P Q R$ is equal to:

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x + 2y + z = 14

$⊥^{r} l i n e P Q : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{1} = t$

Q (1 + t, 2 + 2t, 3 + t)

x + 2y + z = 14 -> 1 + t + 4 + 4t + 3 + t = 14 Þ 6t = 6

t = 1

-> Q (2, 4, 4)

PQ = $\sqrt[]{1 + 4 + 1} = \sqrt[]{6}$

$t a n 60 ° = \frac{P Q}{Q R} \Rightarrow Q R = \frac{P Q}{\sqrt[]{3}} = \sqrt[]{2}$

ar (PQR) = $\frac{1}{2} \sqrt[]{6} \cdot \sqrt[]{2} = \sqrt[]{3}$

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Let A(a, -2), B(a, 6) and $C (\frac{α}{4}, - 2)$ be vertices of a $Δ A B C .$ If $(5, \frac{α}{4})$ is the circumcentre of $Δ A B C,$ then which of the following is NOT correct about $Δ A B C ?$

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Circumcentre (D) $\equiv (5, \frac{α}{4})$

${(5 - α)}^{2} + {(\frac{α}{4} + 2)}^{2} = {(5 - α)}^{2} + {(\frac{α}{4} - 6)}^{2} . . . . . . . . . . . . . . . (i)$

${(5 - \frac{α}{4})}^{2} + {(\frac{α}{4} + 2)}^{2} . . . . . . . . . . . . . . . . . . . . (i i)$

(i) -> $\frac{α}{4} + 2 = \pm (\frac{α}{4} - 6)$

(ii) -> 9 + 16 = 9 + 16

$\oplus \to x$

$(-) \to \frac{α}{2} = 4 \Rightarrow α = 8$

ar $(A B C) = 24$

2S = 24

R = 5, r = $\frac{Δ}{s} = \frac{24}{12} = 2$

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Maths Class 12th

The minimum value of the twice differentiable function f(x) = $\int_{0}^{x} e^{x - t} f' (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ is:

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$f (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t$

$f' (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t + e^{x} . \frac{f' (x)}{e^{x}} - [(2 x - 1) . e^{x} + (x^{2} - x + 1) . e^{x}]$

$\begin{array}{l} f (x) = (2 x + 1) . e^{x} - 2 e^{x} + c \\ f (x) = e^{x} (2 x - 1) c = 0 \end{array}$

$f (- \frac{1}{2}) = \frac{- 2}{\sqrt[]{e}}$

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Kindly consider the following questions

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Contributor-Level 10

$a_{n + 2} a_{n + 1} - a_{n + 1} a_{n} = 2$

Series will satisfy

$a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4}, . . . . . . a_{4} a_{5}$

$\begin{array}{l} 1.2 2.2 2.3 2.4 \\ \frac{a_{n} + \frac{1}{a_{n + 1}}}{a_{n + 2}} = \frac{a_{n + 2} - \frac{1}{a_{n + 1}}}{a_{n + 2}} \end{array}$

$= 1 - \frac{1}{2 (r + 1)} = \frac{2 r + 1}{2 (r + 1)}$

Now, proof = $\frac{30}{11} \frac{(2 r + 1)}{2 (r + 1)}$ $\frac{}{} \frac{}{}$

r = 1

$= \frac{(1.3.5 . . . . . . 61)}{2^{30} (2.3 . . . . . . 31)}$

$\frac{61}{2^{60} . 31.30} = α = - 60$

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Maths Functions

Let a, b and $γ$ be three positive real number. Let f(x) = ax⁵ + bx³ + $γ x, x \in R$ and g : R ® R be such that $g (f (x)) = x$ for all x $\in R .$ If $a_{1}, a_{2}, a_{3} . . . . ., a_{n}$ be in arithmetic progression with mean zero then the value of $f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i})))$ is equal to:

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Case I

= = 0 then

f (x) = yx

$g (x) = \frac{x}{y}$

$\frac{1}{x} \sum_{i = 1}^{n} f (a_{i}) \Rightarrow \frac{y}{x} (a_{1} + a_{2} + . . . . + a_{x}) = 0$

$\Rightarrow f (g (0)) \Rightarrow f (0)$

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Maths Sets

The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

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Contributor-Level 10

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

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Maths Class 12th

If the minimum value of f(x) = $\frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0,$ $\frac{^{}}{} \frac{}{^{}}$ is 14, then the value of is equal to:

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Contributor-Level 10

$\frac{5 x^{2}}{2} + \frac{α}{2 x^{5}} + \frac{α}{2 x^{5}} \geq 7 {(\frac{α^{2}}{2^{7}})}^{\frac{1}{7}}$

$\frac{7 . {(α)}^{2 / 7}}{2} = 14$

${(α^{2})}^{\frac{1}{7}} = 2^{2} \Rightarrow α = {(2^{2})}^{\frac{7}{2}} = 2^{7}$

=128

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a year ago

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + $3 (m_{1}^{2} + m_{2}^{2}) = 220 .$ If one vertex of the square is $(10 (c o s α - s i n α), 10 (s i n α + c o s α)),$ where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(c o s α - s i n α) x + (s i n α + c o s α) y = 10,$ then $72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13$ is equal to:

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Contributor-Level 10

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

...more

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

FIGURE

? = $72 (\frac{1}{16} + \frac{9}{16}) + 100 - 30 + 13 = \frac{720}{16} + 83 = 128$

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Maths Class 12th

The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

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