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Let f(x) = (sin(tan⁻¹ x) + sin(cot⁻¹ x) )² − 1, where |x| > 1.If [latex]frac{dy}{dx}=frac{1}{2}frac{d}{dx}(sin^{-1}f(x)))[/latex] and y(√3) = π/6, then y(-(√3) is equal to

a. π/3b. 2π/3c. - π/6d. 5 π/6Solution:f(x) = [sin(tan−1 x) + sin(cot−1 x) ]2 -1Put tan−1 x = ϕ, where ϕ ∈ (− π/2, - π/4) ᵁ (π/4, π/2)= [sin(tan−1... View Article

[latex]\lim_{x\rightarrow 0}\left ( \frac{3x^{2}+2}{7x^{2}+2} \right )^{1/x^{2}}[/latex] is equal to

a. eb. 1/e2c. 1/ed. e2Solution:Let,L= = 1/e2... View Article

The inversion function of f(x) = [latex]frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x epsilon (-1,1), is[/latex]

Solution: 84x = (1+y)/(1-y)... View Article

For > 0, let the curves ₁: ² = x and ₂ ∶ x² = intersect at origin and a point .

For a > 0, let the curves C1: y2 = ax and C2 ∶ x2 = ay intersect at origin and a point . Let the line x = b (0 < b < a) intersect the... View Article

The locus of a point which divides the line segment joining the point (0, −1) and a point on the parabola, x² = 4 , internally in the ratio 1: 2, is

a. 9x2 − 12y = 8b. 4x2 − 3y = 2c. x2 − 3y = 2d. 9x2 − 3y = 2Solution:Let point be (2 , 2) and be (ℎ, ). Now, eliminating from the above... View Article

Let : → be such that for all x ∈ , (2^1+x + 2^1−x), f(x) and (3^x + 3^−x) are in A.P., then the minimum value of f(x) is:

a. 0b. 4c. 3d. 2Solution:21−x + 21+x, (x), 3x + 3−x are in A.P. -------- (2)Adding (1) and (2), we getf(x) ≥ 1 + 2 = 3Thus, minimum value of... View Article

Which of the following is a tautology?

a. ( ∧ ( ⟶ )) ⟶ b. ∧ ( ∨ )c. ( → ( ∧ ( → ))d. ∨ ( ∧ )Solution:( ∧ ( ⟶ )) ⟶ = ( ∧ (∽ ∨ )) ⟶ = [( ∧∼ ) ∨ ( ∧ )] ⟶ =... View Article

If , and are the greatest values of ¹⁹ , ²⁰ , ²¹ respectively, then:

a. (a/11) = (b/22) = (c/42)b. (a/10) = (b/11) = (c/42)c. (a/11) = (b/22) = (c/21)d. (a/10) = (b/11) = (c/21)Solution: Therefore, max (19 Cp) =... View Article

Let y=(x) be a solution of the differential equation. [latex]sqrt{1-x^{2}}frac{dy}{dx}+sqrt{1-y^{2}}=0,left | x right |< 1[/latex].If y (1/2) = √3/2, then y(-1/√2) is equal to

a. -1/√2b. -√3/2c. 1/√2d. √3/2Solution: sin-1y + sin-1x = cIf x = ½, y = √3/2 then,sin-1(√3/2) + sin-1(1/2) = c(π/3) +(π/6) = cTherefore, c =... View Article

For which of the following ordered pairs ( , ), the system of linear equations

For which of the following ordered pairs (μ, δ), the system of linear equations x + 2y + 3z = 13x + 4y + 5z = μ4x + 4y + 4z = δis... View Article

A bakelite beaker has volume capacity of 500 cc at 30°C. When it is partially filled with V_m volume (at 30°C) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If ϒ_(beaker)= 6 × 10–6 °C^–1 and ϒ _(mercury)= 1.5 × 10^–4 °C^–1, where ϒ is the coefficient of volume expansion, then V_m (in cc) is close to

Solution:Answer. 20For the volume to remain constant at all temperatureExp. of the liquid = Exp. of solid∆V = vϒ∆TV1ϒ1= V2ϒ2500cc × 6 × 10-6 = Vm... View Article

When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to 0°, the surface tension of the liquid, in milliNewton m^–1, is [ρ_(liquid) =900 kgm^–3 , g = 10 ms^–2] (Give answer in closes integer)

Solution:Answer. 101r = 0.015 T = 101... View Article

A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizotally a distance of 0.2 m while launching the ball, the value of F(in N) is (g = 10 ms^–2)

Solution: F x 0.2 = mghF x 0.2 = 0.15 x 10 x 20F = (3/2)x [(10 x20)/2]F = 150 N... View Article

An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is N/100, where N is an integer, the value of N is _____.

Solution: , , , μ=1.58 μ = N/100 = 1.58 N = 158... View Article

A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm).

A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm).... View Article

In the circuit shown in the figure, the total charge is 750 μC and the voltage across capacitor C₂ is 20 V. Then the charge on capacitor C₂ is:

1) 650 μC2) 450 μC3) 590 μC4) 160 μCSolution:Answer. 3Potential difference across C2 & C3 are equalQ2= CV = 8 X 20 = 160 μCQ1= 750 – 160 = 590 μC... View Article