Class 12th

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

$\equiv Req=2.5\Omega$

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

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

$\Rightarrow 300=60R$

$R=5\Omega$

Now for 3A, time = 10 sec

$H\text{'}=l^{2}Rt=3^2*5*10=450J$

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)

ceq = $\frac{24*8}{32}=6\mu F$

$\alpha =\frac{\Delta l_C}{\Delta l_E}=\frac{3.5}{4}=\frac{7}{8}$

$\beta =\frac{\alpha }{1-\alpha }=\frac{7/8}{1-7/8}=7$

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

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

? + 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

I = ∫ (cos 4x-1)/ (cot x-tan x) dx = ∫ (2 cos² 2x-1-1)/ (cos²x-sin²x)/ (sin x cos x) dx
= 2 ∫ (cos² 2x-1) sin x cos x) / (cos 2x) dx = ∫ (cos² 2x-1) sin 2x) / (cos 2x) dx
Put cos 2x = t
-2 sin 2x dx = dt
I = -1/2 ∫ (t²-1)/t dt = 1/2 ∫ (1/t - t) dt
= 1/2 (ln|t| - t²/2) + c
= 1/2 ln|cos 2x| - 1/4 cos² (2x) + c