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Q1. A window is in the form of a rectangle surmounted by a semi circular opening. The total perimeter of the window is 10 m. Find the dimensions o the window to admit maximum light through the whole opening. Do you think that by getting maximum light, we can save electricity? Do you agree that we should save electricity?

Solution

Q2. Find the points at which the f given f (x) = ( x –2)⁴   (x + 1)³  has local maxima

Solution

We have,  f (x) = (x – 2)⁴ (x + 1)³   = 4 (x – 2)³ (x + 1)³ + 3 (x – 2)⁴ (x + 1)²  = (x – 2)³ (x + 1)² (7x – 2)   = (x – 2)² (x + 1)²(x – 2) (7x – 2) Now,  = 0  x = Since (x – 2)² (x + 1)²is always positive. So, sign of   depends upon the sign of (x – 2) (7x – 2). Clearly,  changes its sign from positive to negative as increases through So, x =  is a point of local maximum.

Q3. Show that the normal at any point θ to the curve x = a cos θ + aθ sin θ and y = a sin θ - aθ cos θ is at constant distance from origin.

Solution

We first find slope of tangent for this differentiate x and y with respect to θ. and So So slope of normal would be negative reciprocal of  i.e. slope of normal; would be -cot θ So equation of normal becomes. on simplification this reduces to x cos θ + y sin θ = 0 So length of perpendicular from origin to this line is So normal at point θ is at constant distance from origin for whatever θ you choose.

Q4.  If Then show that  is strictly increasing.

Solution

Differentiate to get On solving this we get Clearly  because x is strictly between 0 and 1, 0 < x < 1 Therefore function is strictly increasing

Q5. The slope of the curve 2y² = ax² + b at (1, – 1) is – 1. Find a, b

Solution

The equation of the curve is 2y² = ax² + b    …(i) Differentiating w.r.t. x, we get   It is given that the slope of the tangent at (1, –1) is –1. Therefore, Since the point (1, –1) lies on (i). Therefore, 2 (–1)² = a (1)² + b Þ a + b = 2 Putting a = 2 in a + b = 2, we obtain b = 0. Hence a = 2 and b = 0.

Q6. A balloon which is spherical is being inflated by pumping gas in at the rate of 8 cm³/sec. Find the rate at which the surface area increases when the radius is 4cm? On the balloon Make India Polio Free is printed. Do you think that we should work seriously towards Polio eradication?

Solution

When radius is 4 cm, rate of change of surface area is 4cm²/sec                                                        OR   'Polio eradication' should be our ultimate goal. Because of polio, lives of many children are ruined.

Q7. Find the equation of tangent to the curve y = x³ + 3x² - 7x - 20 which is perpendicular to x + 2y = 6

Solution

Here are will first find point at which tangent is to be found out. Slope of line  since tangent is perpendicular to this line so slope of tangent should be negative reciprocal of  so slope of tangent is 2. Now differentiating the equation of curve Let (x₁, y₁) be the point on curve at which we want to find tangent Therefore slope of tangent at  3x₁²  + 6x₁  - 7  Now the slope should be 2 Therefore 3x₁² + 6x₁ - 7 =2 3x₁² + 6x₁ - 9 = 0 3(x₁ + 2x₁ - 9) = 0 So 3(x₁ + 3)(x₁ - 1) = 0  So x₁ = -3, x₁ = 1 Taking equation of curve y = x³ + 3x² - 7x - 20 When x₁= -3, y₁ = (-3)³ + 3(-3)² - 3 - 20 When x₁ = 1, y₁ = -23 So pts. are (-3, 1) and (1, -23) and slope is 2 so equations are y - 1 = 2 (x + 3) i.e. 2x - y + 7 = 0              and y + 23 = 2(x - 1) i.e. 2x - y - 25 = 0

Q8. Find all the points of local maxima and minima and the corresponding maximum and minimum values of the function f (x) =

Solution

We have, f (x) =    = – 3x³ – 24x² – 45x = –3x (x² + 8x + 15) For local maximum or local minimum, we must have  = 0 – 3x (x² + 8x + 15) = 0  – 3x (x + 3) (x + 5) = 0 x = 0, –3, –5 Thus, x = 0, x = –3 and x = –5 are the possible points of local maxima or minima. For x=0, f'(x) changes from positive to negative. So x=0 is the point of local maxima.  For x=-3, f'(x) changes its sign from negative to positive. So x=-3 is a point of local minima. For x=-5, f'(x) changes sign from positive to negative. So x=-5 is a point of local maxima.   Corresponding values of f(x), we get x=0, f(x) = 105 x=-3, f(x) = 57.75 x=-5, f(x) = 73.75

Q9. Find the approximate value of f(4.02) where the function is given by f(x)=x³ - 3x² + x + 2   

Solution

 

Q10. Show that the height of cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle x is one third that of the cone and the greatest volume of cylinder is 

Solution

Q11. A square piece of tin of side 18 cm is to made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum. On four sides of this box, write four life skills.

Solution

Let x cm be the length of each side of the square which should be cut off from corners of the square tin sheet.           Let V be the volume of the open box                               So Volume is maximum when x=3           So volume is maximum when the side of the square to be cut off is 3 cm. Four life skills: Responsible, polite, love for peace, spirit of brotherhood

Q12. Find absolute maxima or absolute minima for the following function in the given interval

Solution

Let us take the function As before to locate the points of local maxima or minima we differentiate this function with respect to . Taking Which implies Which Let us now evaluate values of given function at these extreme points as well as boundary points. Comparing these results we see absolute maximum value is  at  and absolute minimum value is  at Note thatchanges sign from negative to positive as  passes through  so it is also a point of local minima so here points of absolute maximum or minimum values are also the points of local maxima or minima.

Q13. Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Solution

As before let us first find the point of intersection of curves. y².y = k means y = k^1/3  So x = k^2/3 therefore point of intersection of curves is (k^2/3 , k^1/3 )  differentiating x = y² with respect to x we get So slope of tangent to this curve at (k^2/3 , k^1/3 ) is    Similarly differentiating xy = k with respect to x So slope of tangent to curve  xy = k at (k^2/3 , k^1/3 ) is                                          If these curves intersect at right angles then m₁m₂ = -1      So      Which gives 1 = 2k^2/3. Taking cubes we get the condition 8k² = 1     

Q14. Find the angle of intersection of the curves x² = y and y² = x

Solution

We first need to find their points of intersection. So solving these equations of curves we get x⁴ = x Which gives x = 0, x = 1 Therefore points of intersection of the curves are (0, 0), (1, 1) Differentiating  with respect to x we get   Similarly differentiating  we get  which gives To find angle of intersection at (0, 0) So m₁ = slope of tangent to curve x² = y at (0,0)             = 2x ]_(0,0) = 0  So tangent is parallel to  axis     m₂= slope of tangent  to curve y² = x at (0,0)            = not defined So tangent is parallel to y axis   Now one tangent is parallel to x axis and other is parallel to y axis. So angle between tangents is right angle. So we can say curves intersect orthoganally at (0, 0)   Similarly to find angle between curves at (1, 1) m₁ = slope of tangent to curve x² = y at (1,1)       = 2 m₂ = Slope of tangent to curve x = y ² at (1,1)         If θ is the angle of intersection of curves then So θ is given by tan θ = 3/4

Q15. Show that f (x) = tan^-1 (sin x + cos x) is an increasing function on the interval

Solution

Q16. Find the equation of tangent to the curve y = cos (x + y), 0 < x < ∏ which is perpendicular to 2x - y = 7

Solution

Differentiating y = cos(x + y) with respect to x we get So  Let (x₁, y₁) be any point on the curve so slope of tangent at (x₁, y₁) is given by   Also slope of given line is 2 and tangent required is perpendicular to this line so slope of tangent would be Therefore So Therefore on solving we get sin (x₁ + y₁) = 1  So  where n is an integer For this value of x + y, But  = 0 for all integral values of n So y is always zero.   And since 0 < x <  we must have So   So coordinates of pts on the curve where we want to find tangent is  So equation of tangent becomes or 2x + 4y =  is the required tangent.

Q17. Find the intervals in which the function f (x) =  is increasing or decreasing

Solution

We have, f (x) =  Clearly, f (x) is defined for all x satisfying x + 1 > 0 i.e. x > – 1 So, domain (f) = (–1, ¥) Now, f (x) =  Þ f ¢ (x) = Þ f ¢ (x) = Þ f ¢ (x) = For f (x) is be increasing, we must have f¢ (x) > 0 Þ  Þ x - 1 > 0 Þ x > 1 Þ x Î (1, ¥) So, f (x) is increasing on (1, ¥) Hence, f (x) is increasing in its domain

Q18. Evaluate

Solution

Take x=4 and Δx=0.01 Note here that we have taken as 4 as its square root exists.     Which means   Now, Therefore,        Hence,

Q19. Find the equation of the normal to the curve y = (1 + x)^y + sin^–1 (sin² x) at x = 0.

Solution

We have, y = (1 + x)^y + sin^–1 (sin² x)       …(i) Putting x = 0, we get y = (1 + 0)^y + sin^–1 (sin² 0) Þ y = 1 Thus, we have to write the equation of the normal to (i) at P (0, 1). Differentiating (i) w.r.t. x, we get    Putting x = 0 and y = 1, we obtain     Hence, the equation of the normal at P (0, 1) is y – 1 = – 1 (x – 0) or x + y= 1

Q20. Look at the conical funnel whose vertical angle is 90^o. Water is dripping out from it at uniformly at the rate of 8 cm³/sec through a tiny hole at the vertex.  Find the rate at which the slant height of the water is decreasing when the slant height of the water is 4cm. Is leaking of water leads to wastage of water? Do you think we can afford to waste our natural resources?

Solution

Let V be the volume of the Water at time 't'.   Then, where 'r' and 'h' are the radius and height respectively.  From the figure it is clear  Therefore the volume is                    Differentiating with respect to t we get  Now we are given Therefore,  When the slant height is 4cm Slant height is decreasing at the rate of Yes, leaking of water leads to wastage of water. We must do everything possible to save our natural resources.

Q21. Find the intervals in which the function f (x) is increasing, f (x) = 2x³ - 9x² + 12x + 15

Solution

We have, f (x) = 2x³ - 9x² + 12x + 15  & f (x) = 6x² - 18x + 12 = 6(x² - 3x + 2) (i)   For f (x) to be increasing, we must have                 f'(x) > 0 => 6 (x² - 3x + 2) > 0 => x² - 3x + 2 > 0     [ 6 > 0 \ 6(x² - 3x + 2) > 0 & x² - 3x + 2 > 0) => (x - 1) (x - 2) > 0 =>  x < 1 or x > 2 (¥, 1)υ(2, ¥) So, f(x) is increasing on (¥, 1)υ(2, ¥)

Q22. Show that the following function is neither strictly increasing nor strictly decreasing in  

Solution

Differentiating the function we get Equate to zero to get . So divides the real line into disjoint intervals In the interval ,  cot x is greater than 1 So cot x - 1 is negative Therefore f'(x) is negative and the function is strictly decreasing in . Similarly, cot x is less than 1 in So cot x -1 is positive. Therefore f'(x) is positive and the function is strictly increasing in . So overall in the interval , the function is neither increasing nor decreasing.

Q23. Separate  into subintervals in which f (x)  = sin 3x is increasing or decreasing

Solution

We have, f (x) = sin 3x Þ f ¢ (x) = 3 cos 3x Now,  or Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases Case I : When  In this case, we have  Þ cos 3x > 0 Þ 3 cos 3x > 0 Þ f ¢ (x) > 0 Thus, f ¢ (x) > 0 for  i.e., So, f (x) is increasing on Case II : When In this case, we have  Þ cos 3x < 0 Þ 3 cos 3x < 0 Þ f ¢ (x) < 0 Thus, f ¢ (x) < 0 for  i.e., So, f (x) is decreasing on Hence, f (x) is increasing on  and decreasing on