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a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

The number of points, where the function f : R R,

$f (x) = | x - 1 | c o s | x - 2 | s i n | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |,$ is NOT differentiable, is:

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Vishal Baghel

Contributor-Level 10

$f (x) = | x - 1 | c o s | x - 2 | s i n | x - 1 | + (x - 3) | x^{2} - 5 x + 4 |$

$f (x) = | x - 1 | c o s (x - 2) s i n | x - 1 | + (x - 3) | x - 1 | | x - 4 |$

x = 1, 4 (doubtful points)

Diff. at x = 1

$\underset{x \to 1}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1}{l i m} \frac{| x - 1 | (s i n | x - 1 | c o s (x - 2) + (x - 3) | x - 4 |)}{x - 1}$

$\begin{array}{l} R H D = \underset{x \to 4^{+}}{l i m} \frac{3 (s i n 3 c o s 2)}{0^{+}} = + \infty \\ L H D = \underset{x \to 4^{-}}{l i m} \frac{3 (s i n 3 c o s 2)}{0^{-}} = - \infty \end{array}} \Rightarrow$ Not diff. at x = 4

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a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

Let the focal chord of the parabola P : y² = 4x along the line L : y = mx + c, m > 0 meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola H : x² – y² = 4. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is:

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Vishal Baghel

Contributor-Level 10

L : y = mx + c, m > 0

y = m (x – 1)

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

$y = m x \pm \sqrt[]{4 m^{2} - 4}$

$\pm \sqrt[]{4 m^{2} - 4} = - m, m > 0$

$\Rightarrow \sqrt[]{4 m^{2} - 4} = m$

$M (\frac{5 + \sqrt[]{21}}{2}, \sqrt[]{3} + \sqrt[]{7}), N (\frac{5 - \sqrt[]{21}}{2}, \sqrt[]{3} - \sqrt[]{7})$

$=$ $2$ $(27)$ $=$ $2$

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a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ in the $x$ -direction. The magnitude of gravitational potential on the $x$ -axis at a distance $x$ , taking its value to be zero at infinity is:

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alok kumar singh

Contributor-Level 10

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eight

The area of the region ${(x, y) : | x - 1 | \leq y \leq \sqrt[]{5 - x^{2}}}$ is equal to:

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Vishal Baghel

Contributor-Level 10

$| x - 1 | \leq y \leq \sqrt[]{5 - x^{2}}$

$y = | x - 1 |, y^{2} + x^{2} = 5, y \geq 0 y = \sqrt[]{5 - x^{2}}$

${(x - 1)}^{2} + x^{2} = 5 \Rightarrow 2 x^{2} - 2 x - 4 = 0 \Rightarrow x^{2} - x - 2 = 0$

$(x - 2) (x + 1) = 0 \to x = - 1, 2$

$A = \int_{- 1}^{2} (\sqrt[]{5 - x^{2}} - | x - 1 |) d x$

$A r e a = \int_{- 1}^{2} (\sqrt[]{5 - x^{2}} d x - | x - 1 | d x)$

$= \int_{- 1}^{2} \sqrt[]{5 - x^{2}} d x + \int_{- 1}^{1} (x - 1) d x - \int_{1}^{2} (x - 1) d x$

$= \frac{5}{2} s i n^{- 1} (1) - \frac{1}{2} = \frac{5 π}{4} - \frac{1}{2}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Five

If f(X) = $\int_{1}^{α} \frac{l o g_{10} t}{1 + t} d t, α > 0, t h e n f (e^{3}) + f (e^{- 3}) i s e q u a l t o :$

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Vishal Baghel

Contributor-Level 10

$f (α) = \int_{1}^{α} \frac{l o g_{10} t}{1 + t} d t$

$f (\frac{1}{α}) = \int_{1}^{\frac{1}{α}} \frac{l o g_{10} t}{1 + t} d t = \int_{1}^{α} \frac{- l o g_{10} z}{1 + \frac{1}{z}} - \frac{1}{z^{2}} d z$

$= \int_{1}^{α} \frac{l o g_{10} z}{z (z + 1)} d z$

$= \int_{1}^{α} \frac{l o g_{10} t}{t} d t = \frac{1}{l n 10} \int_{1}^{α} \frac{l n t}{t} d t = \frac{1}{l n 10} {[\frac{{(l n t)}^{2}}{2}]}_{1}^{α} = \frac{{(l n α)}^{2}}{2 l n 10}$

$f (e^{3}) + f (e^{- 3}) = \frac{{(l n e^{3})}^{2}}{2 l n 10} = \frac{9}{2 l n 10}$

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New answer posted

a year ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Six

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200g. The bob is pulled aside until the string is at 60° with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be_________ ms^-1. (if g = 10m/s²)

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Payal Gupta

Contributor-Level 10

From conservation of Energy:

$m g l (1 - c o s 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow v^{2} = 2 g l (\frac{1}{2}) = g l$

$= 10 * \frac{25}{10} = 5 m / s$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Let $\hat{a} a n d \hat{b}$ be two unit vectors such that the angle between them is $\frac{π}{4} .$ If is the angle between the vectors $(\hat{a} + \hat{b}) a n d (\hat{a} + 2 \hat{b} + 2 (\hat{a} * \hat{b})),$ then the value of 164cos2 is equal to:

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Vishal Baghel

Contributor-Level 10

$(\hat{a} + \hat{b}) . (\hat{a} + 2 \hat{b} + 2 (\hat{a} * \hat{b})), \hat{a} . \hat{b} = \frac{1}{\sqrt[]{2}}$

= 1 + $\frac{1}{\sqrt[]{2}} + \sqrt[]{2}$ + 2 + 0 + 0

$= 3 + \frac{3}{\sqrt[]{2}}$

$| \hat{a} + 2 \hat{b} + 2 {(\hat{a} * \hat{b})}^{2} | = 1 + 4 + 4 (\frac{1}{2}) + 2 \sqrt[]{2} + 0 = 7 + 2 \sqrt[]{2}$

$\Rightarrow (2 + \sqrt[]{2}) (7 + 2 \sqrt[]{2}) c o s^{2} θ = 9 + \frac{9}{2} + 9 \sqrt[]{2} = \frac{27 + 18 \sqrt[]{2}}{2}$

$\Rightarrow c o s^{2} θ = \frac{27 + 18 \sqrt[]{2}}{2} * \frac{1}{18 + 11 \sqrt[]{2}} = \frac{(27 + 18 \sqrt[]{2}) (18 - 11 \sqrt[]{2})}{2 (324 - 242)}$

$= \frac{486 + 324 \sqrt[]{2} - 297 \sqrt[]{2} - 396}{2 * 82}$

164 cos² (θ)= 90 + $2$ $7$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k1, k2), then k1 + k2 is equal to:

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Vishal Baghel

Contributor-Level 10

${(a - 1)}^{2} + 4 = {(b - 1)}^{2} + 16$

$= {(a - 1)}^{2} + {(b - 1)}^{2}$

${(b - 1)}^{2} = 4 & {(a - 1)}^{2} = 16$

$\Rightarrow b = 1 \pm 2 a = 1 \pm 4$

= 3, 1= 5, 3

x + 2y = 3 …… (i)

3x – y = 8…… (ii)

x + 2y = 3

3x – y = 83 * 2

$\oplus \to 7 x = - 13 \Rightarrow x = \frac{- 13}{7}$

$y = \frac{- 39}{7} + 8 = \frac{17}{7}$

$k_{1} + k_{2} = x + y = \frac{4}{7}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

The statement $(p \land q) \Rightarrow (p \land r)$ is equivalent to:

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Vishal Baghel

Contributor-Level 10

$(p \land q) \to (p \land r)$

$\equiv \sim (p \land q) \lor (p \land r)$

$(A) (\sim q) \lor (p \land r) (B) (\sim p) \lor (p \land r)$

$(C) \sim (p \land r) \lor (p \land q) (D) \sim (p \land q) \lor r$

Option (D) is correct

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eight

The angle of elevation of the top of a tower from a point A due north of it and from a point B at a distance of 9 units due west of A is $c o s^{- 1} (\frac{3}{\sqrt[]{13}}) .$ If the distance of the point B from the tower is 15 units, then cot is equal to:

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Vishal Baghel

Contributor-Level 10

$100 c o t^{2} α = 1 5^{2} - 9^{2} = 144$

$\frac{O T}{15} = \frac{2}{3}$

$c o t α = \frac{6}{5} O T = 10$

$Δ A O T, O A = O T c o t α = 10 c o t α$

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