Calculus

f'(t)=(-(1+t)-(1-t))/((1+t)²)=-(2)/((1+t)²) f'(1)=-(2)/((2)²)=-\frac12

f(x)=-(x²-6x+8)=-(x-3)²+1 Maximum value is 1 at x=3. Range is (-∞,1].

Given y=e⁻ˣ x. Characteristic equation: (D+1)²+1=0 D²+2D+2=0 Thus the differential equation is: y''+2y'+2y=0 Differentiating twice more: y⁽⁴⁾+2y⁽³⁾+2y''=0 y⁽⁴⁾=-2y⁽³⁾-2y'' Comparing with y₄+ky=0, we get: k=-2

The function is undefined when: |x|=0 x=0 |x|=0 |x|=1 x=±1 Thus points of discontinuity are: -1,0,1

Using integration by parts: u=⁻¹x, dv=xdx x⁻¹xdx=\frac12(x²+1)⁻¹x-\frac12x+C

Using: dx√(a²-x²)=⁻¹(x)/(a) Here a=\sqrt2: ⁻¹(1)/(\sqrt2)-⁻¹0=(π)/(4)

₀ⁿ [x]dx = =₀ⁿ⁻¹ k = (n(n-1))/(2) Given: (n(n-1))/(2)=66 n²-n-132=0 n=12

For G.P.: a=kp, b=kq, c=kr Rows are proportional ⇒ determinant = 0.

Using identity: 2⁻¹x=⁻¹(2x)/(1-x²) 2⁻¹\frac15=⁻¹(5)/(12), 2⁻¹\frac18=⁻¹(16)/(63) Now: ⁻¹(5)/(12)+⁻¹\frac12+⁻¹(16)/(63) =⁻¹(1)=(π)/(2)

Step 1: Use the trigonometric identity to simplify the integrand:
Step 2: Factor out the constant and set up the definite integral:
Step 3: Compute the integral. The antiderivative of is :
Step 4: Evaluate the definite integral from 0 to :
Step 5: Substitute the limits: - At , , so : - At , , so : - Difference: