Tutor profile: Dilip T.
Q.1. In a right triangle ABC with angle A equal to 90 degree , find angle B and C so that sin(B) = cos(B) Q.2. In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).
Q.1. Ans: Let b be the length of the side opposite angle B and c the length of the side opposite angle C and h the length of the hypotenuse. sin(B) = b/h and cos(B) = c/h sin(B) = cos(B) means b/h = c/h which gives c = b The two sides are equal in length means that the triangle is isosceles and angles B and C are equal in size of 45 degree. Q.2. Ans: Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse. tan(A) = opposite side / adjacent side = a/b = 3/4 We can say that: a = 3k and b = 4k , where k is a coefficient of proportionality. Let us find h. Pythagora's theorem: h2 = (3k)2 + (5k)2 Solve for h: h = 5k sin(A) = a / h = 3k / 5k = 3/5 and cos(A) = 4k / 5k = 4/5.
Q.1. If f(x) and g(x) are differentiable functions such that f '(x) = 3 x and g '(x) = 2 x 2 then the limit lim [(f(x) + g(x)) - (f(1) + g(1))] / (x - 1) as x approaches 1. is equal to ? Q.2. Find derivative w.r.t. x of y = √x/(x + 3); x > 0 Q.3. Find derivative w.r.t. x of y = x3 - 4x2 + 1/x; x ≠ 0
Q.1. Ans: The given limit is the definition of the derivative of f(x) + g(x) at x = 1. The derivative of the sum is equal to the sum of the derivatives. Hence the given limit is equal to f '(1) + g '(1) = 5. Q.2. Ans: (3 - x)/2√x(x + 3)2 Q.3. Ans: 6x - 8 + 2/x3.
Q.1. If x <2, simplify |x - 2| - 4|-6| Q.2. Find the equation of the line that passes through the points (-1 , -1) and (-1 , 2). Q.3. Solve the equation |-2x + 2| -3 = -3
Q.1. Ans: If x < 2 then x - 2 < 2 and if x - 2 < 2 the |x - 2| = -(x - 2). Substitute |x - 2| by -(x - 2) and |-6| by 6 . |x - 2| - 4|-6| = -(x - 2) -4(6) = -x -22 Q.2. Ans: m = (y2 - y1) / (x2 - x1) = [2 -( -1)] / [(-1 -( -1)] = 3 / 0 The slope is indefined which means the line is perpendicular to the x axis and its equation has the form x = constant. Since both points have equal x coordinates -1, the equation is given by: x = -1 Q.3. Ans: The equation to solve is given by. |-2x + 2| -3 = -3 Add 3 to both sides of the equation and simplify. |-2x + 2| = 0 |-2x + 2| is equal to 0 if -2x + 2 = 0. Solve for x to obtain x = 1
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