Commerce (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics

Find a particular solution of the following differential equation:- x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x² + 1) dx, x ≠ 0.

Find the equation of a curve passing through the point (−2, 3), given that the slope of the tangent to the curve at any point (x, y) is `(2x)/y^2.`

Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]

Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.

If A is a square matrix of order 3 with |A| = 4 , then the write the value of |-2A| . 

Find the intervals in which function f given by f(x)  = 4x³ - 6x² - 72x + 30 is (a) strictly increasing, (b) strictly decresing .

Find the area of the region. 

{(x,y) : 0 ≤ y ≤ x² , 0 ≤ y ≤ x + 2 ,-1 ≤ x ≤ 3} .

If sin y = xsin(a + y) prove that `(dy)/(dx) = sin^2(a + y)/sin a`

If A = `[[0 , 2],[3, -4]]` and kA = `[[0 , 3"a"],[2"b", 24]]` then find the value of k,a and b.

If tan^-1 x - cot^-1 x = tan^-1 `(1/sqrt(3)),`x> 0 then find the value of x and hence find the value of sec^-1 `(2/x)`.

`"If y" = (sec^-1 "x")^2 , "x" > 0  "show that"  "x"^2 ("x"^2 - 1) (d^2"y")/(d"x"^2) + (2"x"^3 - "x") (d"y")/(d"x") - 2 = 0`

Find: ∫ sin x · log cos x dx

Solve: tan^-1 4 x + tan^-1 6x `= π/(4)`.

Prove that the function f : N → N, defined by f(x) = x² + x + 1 is one-one but not onto. Find the inverse of f: N → S, where S is range of f.

Solve the differential equation: `(d"y")/(d"x") - (2"x")/(1+"x"^2) "y" = "x"^2 + 2`

Solve the differential equation:  ` ("x" + 1) (d"y")/(d"x") = 2e^-"y" - 1; y(0) = 0.`

If `"A" = [(1,1,1),(1,0,2),(3,1,1)]`, find A^-1. Hence, solve the system of equations x + y + z = 6, x + 2z = 7, 3x + y + z = 12.

Solve for x : `tan^-1 ((2-"x")/(2+"x")) = (1)/(2)tan^-1  ("x")/(2), "x">0.`

Solve the differential equation : `("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2 d"y" = 0  "given that"  "y" = 0  "when"  "x" = 1`.