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

Let f : R -> R be a continuous function such that f(3x) – f(x) =. If f(8) = 7, then f(14) is equal to:

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Vishal Baghel

Contributor-Level 10

f (3x)- f (x) = x

Replace $x \to \frac{x}{3} \Rightarrow f (x) - f (\frac{x}{3}) = \frac{x}{3}$

Again replace $x \to \frac{x}{3} \Rightarrow f (\frac{x}{3}) - f (\frac{x}{3^{2}}) - f (\frac{x}{3^{2}}) = \frac{x}{3^{2}}$

$\Rightarrow f (3 x) - f (0) = \frac{3 x}{2} p u t t i n g x = \frac{8}{3} \Rightarrow f (8) - f (0) = 4 ∴ f (0) = 3$

Also putting x = $\frac{14}{3}$ in f (3x) – 3 = $\frac{3 x}{2} \Rightarrow$ F (14) – 3 = 7 f (14) = 10

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

The value of $l o g_{e} 2 \frac{d}{d x} (l o g_{c o s x} c o s e c x) a t x = \frac{π}{4} i s$

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alok kumar singh

Contributor-Level 10

$L e t f (x) = l o g_{c o s x} c o s e c x = \frac{l o g c o s e c x}{l o g c o s x}$

$f' (x) = \frac{l o g c o s x . s i n x (- c o s e c x c o t x - (l o g c o s e c x) \frac{1}{c o s x} . (s i n x))}{{(l o g c o s x)}^{2}}$

$A t x = \frac{π}{4} f' (\frac{π}{4}) = \frac{- l o g (\frac{1}{\sqrt[]{2}}) + l o g \sqrt[]{2}}{{(l o g \frac{1}{\sqrt[]{2}})}^{2}} = \frac{2}{l o g \sqrt[]{2}} ∴ a t x = \frac{π}{4}, l o g_{e} (2 f' (x)) = 4$

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Class 12th Magnetism and Matter

At a certain place the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at the certain place, is:

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alok kumar singh

Contributor-Level 10

$B c o s δ = B_{H}$

B cos 30° = 0.5

$B = \frac{0 . * 2}{\sqrt[]{3}} = \frac{1}{\sqrt[]{3}}$

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

Let = $\underset{x \to 0}{l i m} \frac{α x - (e^{3 x} - 1)}{α x (e^{3 x} - 1)}$ for some $a \in R .$ Then the value of α + β is:

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alok kumar singh

Contributor-Level 10

$β = \frac{α x - (e^{3 x} - 1)}{α x (e^{3 x} - 1)}, α \in R \underset{x \to 0}{l i m} \frac{\frac{α}{3} - (\frac{e^{3 x} - 1}{3 x})}{α x (\frac{e^{3 x} - 1}{3 x})}$

$= \underset{x \to 0}{l i m} \frac{1 - \frac{(1 + 3 x + \frac{9 x^{2}}{2} + . . . . . . . . . . - 1)}{3 x}}{1 + 3 x + \frac{9 x^{2}}{2} + . . . . . . . . - 1} = - \frac{1}{2} ∴ α + β = \frac{5}{2}$

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Class 12th Gravitation

The escape velocity of a body on a planet 'A' is 12 kms^-1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is:

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alok kumar singh

Contributor-Level 10

$M_{A} = ρ_{A} * \frac{4}{3} π R_{A}^{3}$ $ρ_{B} = 4 ρ_{A}$

$M_{B} = ρ_{B} * \frac{4}{3} π R_{B}^{3}$ $R_{B} = \frac{R_{A}}{2}$

$\frac{M_{A}}{M_{B}} = 2, \frac{R_{A}}{R_{B}} = 2$ $\frac{V_{E A}}{V_{E B}} = \sqrt[]{\frac{2 G_{1} M_{A}}{R_{A}} * \frac{R_{B}}{2 G_{1} M_{B}}}$

$v_{E A} = v_{E B} = 12 k m s e c^{- 1}$

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

If the maximum value of a, for which the function f_a(x) = tan^-1 2x – 3ax + 7 is non-decreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\bar{a}$ , then $f_{a} (\frac{π}{8})$ is equal to

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alok kumar singh

Contributor-Level 10

$f_{a} (x) = t a n^{- 1} 2 x - 3 a x + 7 \Rightarrow f_{a}' (x) = \frac{2}{1 + 4 x^{2}} - 3 a \Rightarrow f_{a}' (x) \geq 0 \Rightarrow 3 a \leq \frac{2}{1 + 4 x^{2}}$

$a_{m a x} = \frac{2}{3} (\frac{1}{1 + 4 * \frac{π^{2}}{36}}) = \frac{6}{9 + π^{2}} = \bar{a} ∴ f_{a} (\frac{π}{8}) = t a n^{- 1} \frac{2 π}{8} - 3 \frac{π}{8} \frac{6}{9 + π^{2}} + 7 = 8 - \frac{9 π}{4 (9 + π^{2})}$

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Class 12th Motion in a Plane

Two vectors $\vec{A} a n d \vec{B}$ have equal magnitudes. If magnitude of $\vec{A} + \vec{B}$ is equal to two times the magnitude of $\vec{A} - \vec{B},$ then the angle between $\vec{A} a n d \vec{B}$ will be:

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alok kumar singh

Contributor-Level 10

$| \vec{A} + \vec{B} | = 2 | \vec{A} - \vec{B} |$

Given A = B

Squaring equation (1) both side.

${| \vec{A} + \vec{B} |}^{2} = {(2 | \vec{A} - \vec{B} |)}^{2}$

$A^{2} + B^{2} + 2 \vec{A} . \vec{B} = 4 (A^{2} + B^{2} - 2 \vec{A} . \vec{B})$

2A² + 2A² cosq = 4 (2A² – 2A² cosq)

2A² + 2A² cosq = 8A² – 8A² cosq

10A² cosq = 6A²

cosq = $\frac{3}{5}$

q = cos^-1 (3/5)

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Class 12th Straight Lines

Let P and Q be any points on the curves (x – 1)² + (y + 1)² = 1 and y = x², respectively. The distance between P and Q is minimum for some vale of the abscissa of P in the interval

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alok kumar singh

Contributor-Level 10

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f $(\frac{1}{4}) f (\frac{1}{3}) < 0 \Rightarrow t \in (\frac{1}{4}, \frac{1}{3})$

Let P(1 – sin q, -1 + cos q) $∴$ slope of normal = slope of CP $- \frac{1}{2 t} = \frac{c o s θ}{- s i n θ} \Rightarrow 2 t$ Þ = tan q according to question $x = 1 - s i n θ = 1 - \frac{2 t}{\sqrt[]{1 + 4 t^{2}}} = g (t) ∴ g (t) = 1 - \frac{2 t}{\sqrt[]{1 + 4 t^{2}}}$ ,

Þ g'(t) < 0 g(t) is decreasing function in $t \in (\frac{1}{4}, \frac{1}{3}) \Rightarrow g (t) \in (0.440, 0.485) \in (\frac{1}{4}, \frac{1}{2})$

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Class 12th Thermal Properties of Matter

In van dar Wall equation $[P + \frac{a}{V^{2}}] [V - b] = R T; P$ is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants $\frac{a}{b}$ is dimensionally equal to: