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Class 12th

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Class 12th Maths

Let g(x) = x + 2| - 3|. If a denotes the number of relative minima, b denotes the number of relative maxima and c denotes the product of the zeros. Then the value of (a + 2b - c) is __________

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Vishal Baghel

Contributor-Level 10

c = -5, a = 2, b = 1

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Class 12th Maths

In a geometric progression the ratio of the sum of the first 5 terms to the sum of their reciprocals is 49 and sum of the first and the third term is 35. The fifth term of the G.P. is p, the value of 4p is __________

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Vishal Baghel

Contributor-Level 10

Let five terms in G.P. be a/r², a/r, a, ar, ar²
Then, a (r? ² + r? ¹ + 1 + r + r²) / (1/a) (r? ² + r? ¹ + 1 + r? ¹ + r? ²) = 49
⇒ a² = 49 ⇒ a = ±7
Also, a/r² + a = 35
Therefore, a = -7 is not possible
Now, fifth term = ar² = a (7/28) = p ⇒ 4p = 7

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Class 12th Physics

The total current supplied to the circuit is shown in figure by the 5 V battery is…………….A.

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Raj Pandey

Contributor-Level 9

$\equiv R e q = 2.5 Ω$

$l = \frac{v}{R_{e q}} = \frac{5}{2.5} = 2 A$

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Class 12th Physics

A resistor develops 300 J of thermal energy in 15s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10s is…………….J.

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Raj Pandey

Contributor-Level 9

$H = I^{2} R t = 2^{2} * R * 15$

$\Rightarrow 300 = 60 R$

$R = 5 Ω$

Now for 3A, time = 10 sec

$H' = l^{2} R t = 3^{2} * 5 * 10 = 450 J$

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Class 12th Inverse Trigonometric Functions

The value of cot^-1(2² + 1/2) + cot^-1(2³ + 1/2²) + ……∞ is equal to tan^-1 λ, then the value of λ is equal to __________

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Vishal Baghel

Contributor-Level 10

T? = cot? ¹ (2²? ¹ + 1/2? ) = cot? ¹ (1 + 2²? ¹)/2? ) = tan? ¹ (2? / (1 + 2²? ¹)
T? = tan? ¹ (2? ¹ - 2? ) / (1 + 2? ¹ . 2? ) = tan? ¹ (2? ¹) – tan? ¹ (2? )
S∞ = π/2 - tan? ¹ (2) = cot? ¹ (2) = tan? ¹ (1/2)

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a year ago

Class 12th Physics Electrostatic Potential and Capacitance

The equivalent capacitance between points A and B in below shown figure will be…………. $μ F .$

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Raj Pandey

Contributor-Level 9

ceq = $\frac{24 * 8}{32} = 6 μ F$

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Class 12th Physics

If an emitter current is changed by 4mA, the collector current changes by 3.5mA. The value of b will be:

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alok kumar singh

Contributor-Level 10

$α = \frac{Δ l_{C}}{Δ l_{E}} = \frac{3.5}{4} = \frac{7}{8}$

$β = \frac{α}{1 - α} = \frac{7 / 8}{1 - 7 / 8} = 7$

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Class 12th Limits and Derivatives

f(1) = 2, g(1) = 2, f'(x), g'(x) exist, then the value of lim(x→1) [f(1)g(x)-f(1)-g(1)f(x)+g(1)] / [f(1)g(x)-f(x)g(1)] is __________

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Vishal Baghel

Contributor-Level 10

lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1

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Class 12th Physics Ray Optics and Optical Instruments

The focal length f is related to the radius of curvature r of the spherical convex mirror by:

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alok kumar singh

Contributor-Level 10

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + $\frac{1}{2} r .$

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a year ago

Class 12th Vector Algebra

Vectors a, b, c are such that every pair is non-collinear and the vector a + b is collinear with c and vector (b + c) is collinear with a. The resultant of a, b and c is

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Vishal Baghel

Contributor-Level 10

? + b? = λ? c? (i)
b? + c? = λ? (ii)
Form (i) – (ii),
? – c? = λ? c? – λ?
(1 + λ? )? = (1 + λ? )c?
? ? and c? are non – collinear
⇒ 1 + λ? = 0, 1 + λ? = 0
λ? = λ? = -1
⇒? + b? + c? = 0

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