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Let f(x) = ∫e^x (x-1)(x-2)dx. Then f decreases in the interval

(1) (-∞, -2) (2) (- 2, - 1) (3) (1, 2) (4) (2, ∞) Solution: Given f(x) = ∫ex (x-1)(x-2)dx Differentiate both sides w.r.t.x... View Article

If for f(x) = 2x – x², Legranges’s theorem satisfies in [0, 1], then the value of c in [0,1] is

(1) c = 0 (2) c = ½ (3) c = ¼ (4) c = 1 Solution: According to Legranges’s theorem f’(c) = [f(b) - f(a)]/(b - a) Given f(x) = 2x... View Article

The function (e^2x – 1)/(e^2x + 1) is

(1) Increasing (2) Decreasing (3) Even (4) Strictly increasing Solution: Let f(x) = (e2x - 1)/(e2x + 1) = (e2x + 1 - 2)/(e2x +... View Article

If f(x) = x⁵ – 20x³ + 240x, then f(x) satisfies which of the following

(1) It is monotonically decreasing everywhere (2) It is monotonically decreasing only in (0, ∞) (3) It is monotonically increasing everywhere... View Article

The line 2x + √6y = 2 is a tangent to the curve x² – 2y² = 4. The point of contact is

(1) (4, -√6) (2) (7, -2√6) (3) (2, 3) (4) (√6, 1) Solution: Given equation of line 2x + √6y = 2 y = (2 - 2x)/√6 …(i) Equation of... View Article

The triangle formed by the tangent to the curve f(x) = x² + bx – b at the point (1, 1) and the co-ordinate axes, lies in the first quadrant. If its area is 2 then the value of b is

1) –1 2) 3 3) –3 4) 1 Solution: Given f(x) = x2 + bx - b Slope, m = dy/dx = 2x + b At (1,1) m = 2 + b The equation of the... View Article

The length of sub-tangent to the curve x²y² = a⁴ at the point (-a, a) is

1) 3a 2) 2a 3) a 4) 4a Solution: Given x2y2 = a4 y2 = a4/x2 Differentiate w.r.t.x 2y dy/dx = -2a4/x3 dy/dx = -2a4/2yx3 =... View Article

The abscissa of the points of curve y = x(x – 2) (x – 4) : where tangents are parallel to x-axis is obtained as

(1) x = 2 ± 2/√3 (2) x = 1 ± 1/√3 (3) x = 2 ± 1/√3 (4) x = ± 1 Solution: Given y = x(x - 2) (x - 4) = x(x2 - 6x + 8) = x3 - 6x2... View Article

For the curve yⁿ = a^n-1x, the subnormal at any point is constant. The value of n must be

1) 2 2) 3 3) 0 4) 1 Solution: Given yn = an-1x Differentiate w.r.t.x nyn-1dy/dx = an-1 dy/dx = an-1/nyn-1 Length of... View Article

On dropping a stone in stationary water circular ripples are observed. Rate of flow of ripples is 6 cm/sec. When radius of the circle is 10 cm, then fluid rate of increase in its area is

1) 120 sq. cm/sec 2) 12 sq. cm/sec 3) π sq. cm/sec 4) 120 π sq. cm/sec Solution: Given rate of flow of ripples is 6 cm/sec. dr/dt... View Article

The distance travelled by a particle moving in a straight line at time t is s = √(at² + 2bt + c). Acceleration of the particle is

(1) proportional to t (2) proportional to s (3) proportional to s-3 (4) none of these Solution: Given s = √(at2 + 2bt + c)... View Article

A stone, thrown vertically upward from the surface of the moon at a velocity of 24 m/sec reaches a height of s = 24 t – 0.8t² metre after t second. The acceleration due to gravity in m/sec² at the surface of the moon is

1) 0.8 2) -1.6 3) 2.4 4) -4.9 Solution: Given s = 24 t - 0.8t2 Velocity is the derivative of distance. ds/dt = 24 - 1.6 t... View Article

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Archives

Let f(x) = ∫e^x (x-1)(x-2)dx. Then f decreases in the interval

If for f(x) = 2x – x², Legranges’s theorem satisfies in [0, 1], then the value of c in [0,1] is

The function (e^2x – 1)/(e^2x + 1) is

If f(x) = x⁵ – 20x³ + 240x, then f(x) satisfies which of the following

The line 2x + √6y = 2 is a tangent to the curve x² – 2y² = 4. The point of contact is

The triangle formed by the tangent to the curve f(x) = x² + bx – b at the point (1, 1) and the co-ordinate axes, lies in the first quadrant. If its area is 2 then the value of b is

The length of sub-tangent to the curve x²y² = a⁴ at the point (-a, a) is

The abscissa of the points of curve y = x(x – 2) (x – 4) : where tangents are parallel to x-axis is obtained as

For the curve yⁿ = a^n-1x, the subnormal at any point is constant. The value of n must be

On dropping a stone in stationary water circular ripples are observed. Rate of flow of ripples is 6 cm/sec. When radius of the circle is 10 cm, then fluid rate of increase in its area is

The distance travelled by a particle moving in a straight line at time t is s = √(at² + 2bt + c). Acceleration of the particle is

A stone, thrown vertically upward from the surface of the moon at a velocity of 24 m/sec reaches a height of s = 24 t – 0.8t² metre after t second. The acceleration due to gravity in m/sec² at the surface of the moon is

If x = e^t sin t, y = e^t cos t, t is a parameter, then d²y/dx² at t = π is equal to

If x = a cos θ, y = b sin θ, then d³y/dx³ is equal to

Differential coefficient of sec^-1 1/(2x² – 1) w.r.t. √(1 – x²) at x = 1/2 is

If z = x√y + y/x^1/3, then ∂z/∂y =

If (x+y) sin u = x²y², then x ∂u/∂x + y ∂u/∂y is equal to

If [latex]u=e^{-x^{2}-y^{2}}[/latex], then

If u = tan^-1 (x+y), then x ∂u/∂x + y ∂u/∂y =

If u = xy² tan^-1(y/x), then xu_x+ yu_y=