6

Q1. The function f (x) = a^x, 0 < a < 1 is

• decreasing
• increasing
• strictly decreasing on R
• neither increasing or decreasing

Solution

f(x) =  Let x₁, x₂ ε R such that x₁ < x₂ . Then, x₁ < x₂      [ 0 < a < 1 & x₁ < x₂  ]  f (x₁) > f (x₂) Thus, x₁ < x₂ => f (x₁) > f (x₂) for all x₁, x₂ ε R Hence, f (x) is strictly decreasing function on R.  

Q2. Find two numbers whose sum is 24 and product is a large as possible.

• 10 and 14
• 20 and 4
• 12 and 12
• 8 and 16

Solution

We are given sum of numbers is 24 so we can assume the numbers to be  and . Product of these numbers is  which gives us the product function.             . To find extreme points take                         which gives To see whether this number maximise minimise.  We find second derivative  which is  and that is negative. So we conclude that the product is maximum if we choose number as 12, 12.

Q3. f(x) = x⁵ - 5x⁴ + 5x³ - 1. The local maxima of the function f(x) is at x =

• zero
• 1
• 3
• 5

Solution

Q4. Find the point on curve y = x² - 6x + 7 where tangent is parallel to x axis.

• (0, 0)
• (1, 2)
• (2, 4)
• (3, -2)

Solution

Q5. A point c in the domain of a function f is called a critical point of f if

• f is not differentiable at x = c
• f' (x) = 0 at x = c
• Either f' (c) = 0 or f is not differentiable
• f" (x) = 0, at x = c

Solution

Q6. Find the maximum and minimum values of f (x) = 2x³ – 24x + 107 in the interval [1, 3].

• 89, 75
• 59, 56
• 89, 69
• 89, -9

Solution

Q7. The equation(s) of normals(s) to the curve 3x² - y² = 8 which is parallel to the line x + 3y = 4 is.

• 3x - y + 8 = 0
• x + 3y + 8 = 0
• x + 3y + 18 = 0
• x + 3y + 3 = 0

Solution

Let the required normal be drawn at the point (x₁, y₁). The equation of the given curve is 3x² - y² = 8      (i) Differentiating both sides w.r.t. x, we get  = 0  =    = Since the normal at (x₁, y₁) is parallel to the line x + 3y = 4. Therefore, Slope of the normal at (x₁, y₁) = (Slope of the line x + 3y = 4)   =  =   y₁ = x_{1  .}......(ii)                      Since (x₁, y₁) lies on (i). Therefore,  = 8 ...... (iii) Solving (ii) and (iii), we get  = 8   = 4  x₁ = ± 2 Now, x₁ = 2  y₁ = 2  [Using (ii)] and, x₁ = -2  y₁ = -2          [Using (ii)] Thus, the coordinates of the point are (2, 2) and (-2, -2). The equation of the normal at (2, 2) is y -2 =         x + 3y - 8 = 0 The equation of the normal at (-2, -2) is y -(-2) =   x + 3y + 8 = 0

Q8. Find slope of normal to the curve y=5x²-10x + 7 at x=1

• not defined
• 1
• -1
• zero

Solution

Q9. The equation of tangent to the curve y = x³ + 2x + 6 which is perpendicular to the line x + 14y + 4 = 0 is :

• 14x - y + 5 = 0
• 14x -y + 1= 0
• 14x- y + 22 = 0
• 14x- y + 18 = 0

Solution

Let the coordinates of the point of contact be (x₁, y₁). As it lies on y = x³ + 2x + 6 \         y_­1 =           ....(i) The equation of the curve is y = x³ + 2x + 6 Differentiating both sides w.r.t.x, we get  = 3x² + 2 = Since the tangent at (x₁, y₁) is perpendicular to the line x + 14y + 4 = 0. Therefore, Slope of the tangent at (x₁, y₁) x Slope of the line = - 1   = - 1   = - 1   = 14  x₁ = ± 2 Now, x₁ = 2 & y₁ = 2³ + 2 x 2 + 6 = 18         [Using (i)] x₁ = - 2 & y₁ = (-2)³ + 2 (-2) + 6 = - 6           [Using (i)] So, the coordinates of the points of contact are (2, 18) and (- 2, - 6) The equation of the tangent at (2, 18) is y - 18 = 14 (x - 2)  14x - y - 10 = 0     The equation of the tangent at (- 2, - 6) is y - (- 6) = 14 (x - (-2)) 14x - y + 22 = 0

Q10. The equation of tangent line to y = 2x² + 7 which is parallel to the line 4x - y + 3 = 0 is.

• 4x - y + 15 = 0
• 4x - 2y + 5 = 0
• 4x - y - 5 = 0
• 4x - y + 5 = 0

Solution

Let the point of contact of the required tangent line be (x₁, y₁) The equation of the given curve is y = 2x² + 7 Differentiating both sides w.r.t. x, we get  = 4x   = 4x₁ Since the line 4x - y + 3 = 0 is parallel to the tangent at (x₁, y₁) \ Slope of the tangent at (x₁, y₁) = (Slope of the line 4x -y + 3 = 0)   =     4x₁ = 4 x₁ = 1 Now, (x₁, y₁) lies on y = 2x² + 7 \ y₁ = y₁ = 2 + 7 = 9         [As x₁ = 1] So, the coordinates of the point of contact are (1, 9) Hence, the required equation of the tangent line is    y - 9 = 4 (x - 1) 4x - y + 5 = 0

Q11. The sum of two positive numbers is 20. Find the numbers if their product is maximum

• 10,10
• 5,15
• 20,0
• 8,12

Solution

  Let the numbers be           Given           Let                     So numbers are 10, 10

Q12. Find point local maxima for the function f(x)=x³ +x² +x+1 

• x=1
• x=3
• x=0
• No point of local maxima

Solution

  To find local maxima or minima let us differentiate f(x) f'(x) = 3x² +2x +1 We see discriminant of this expression is negative so no real root exist for f'(x) so there are no extreme points and hence we will not get any point of either local maxima or minima.

Q13. The function f (x) = x³ – 8 on [1, 2]  is

• decreasing
• strictly decreasing
• increasing
• neither increasing or decreasing

Solution

We have  = x³ – 8    = 3x² Clearly,   > 0 for all x  [1, 2]. So f (x) is increasing on [1, 2]

Q14. Find the approximate change in total surface area  of a cube of side x  metre caused by increase in side by 1%.

• 12 m²
• 1.2x m²
• 12x m²
• 0.12x² m²

Solution

Let 'S' be the total surface area of the cube.  Here                                                     = approximate change in total surface area.

Q15. Find the point  on x²+ y² + 2x= 0 where normals parallel to x axis.

• (-2, 0)
• (2, 0)
• (-1, 1)
• (-1, -1)

Solution

Q16. The normal at any point q to the curve x = a (cos q +  q sin q), y = a (sin q  -  q cos q) is at distance from the origin that is equal to... .

• a
• 2a
• 3a
• 4a

Solution

Q17. Use differentials to approximate the value of cube root of 66

• 2.021
• 4.042
• 8.084
• 4.3

Solution

Q18. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x² + 22x + 35. Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.

• Rs 48
• Rs 92
• Rs 127
• 7

Solution

Q19. The equation of the normal to the curve x² = 4y which passes through the point (1, 2) is.

• 4x - y = 2
• 4x - 2y = 0
• x + y - 3 = 0
• 4x - 3y + 2= 0

Solution

Suppose the normal at P(x₁, y₁) on the parabola x² = 4y passes through the point (1, 2). We have, x² = 4y  2x =  =    = The equation of the normal at P(x₁, y₁) is y -y₁ =  y - y₁ =          (i) It passes through (1, 2) \ 2 -y₁ =  2 -y₁ =  x₁y₁ = 2 .....(ii) Also, P (x₁, y₁) lies on x² = 4y  \  = 4y₁       ...(iii) Eliminating y₁ between (ii) and (iii), we have  = 2  = 8  x₁ = 2 Putting x₁ = 2 in (ii), we get y₁ = 1 Putting the values of x₁ and y₁ in (i), we get y -1 = -1(x -2) = x + y -3 = 0 This is the required equation of the normal.

Q20. All normals to the curve x = a cos t + at sin t, y = a sin t – at cos t are at a distance a from the origin that is equal to….

• a
• 2a
• 3a
• 8a

Solution

We have, x = a cos t + at sin t and y = a sin t – at cos t Þ  = at cos t and  = at sin t Þ  =  =  = tan t The equation of the normal at any point t is given by y – (a sin t – at cos t) =  {x – (a cos t + at sin t)} Þ y – (a sin t – at cos t) = {x – (a cos t + at sin t)} Þ y sin t – (a sin² t – at sin t cos t) = – x cos t + acos² t + at sin t cos t Þ x cos t + y sin t = a …(i) \ Length of the perpendicular from the origin to (i) = Hence, all normals to the given curve are at a distance a from the origin.

Q21. If f (x) = a log |x| + bx² + x has extreme values at x = –1 and at x = 2, then  values of a and b are

• 2,-1/4
• 2,-1
• 2,-5/6
• 2,-1/2

Solution

               

Q22. For f function y = f(x) if we have f^'(c) = 0 and f^" > 0 then x = c is a point of

• local maxima
• local minima
• point of inflexion
• absolute minima

Solution

Local minima

Q23. The function f (x) = -3x + 12 on R.is

• decreasing
• strictly decreasing
• increasing
• neither increasing or decreasing

Solution

 Let x₁, x₂ R and  x₁ < x₂. Then, because x₁ < x₂  -3x₁ > - 3x₂ - 3x₁ + 12 > -3x₂ + 12  f (x₁) > f (x₂) Thus, x₁, < x₂  f (x₁) > f (x₂)  for all x₁, x₂  R. So, f (x) is strictly decreasing function on R.

Q24. The maximum and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 1 on the interval  [1,4]

• -58,257
• -49,258
• -63,257
•  -48,258

Solution

Q25. Without using derivatives find the maximum value of function g(x)=|x+2|-1

• 2
• No maximum value
• 1
• Zero

Solution

since |x+2| can be made as large as possible by taking  close to infinity so no maximum value exists for this function.

Q26. The coordinates of the points on the curve y = x² + 3x + 4, the tangents at which pass through the origin is ... .

• (2, 14)
• (-2, -2)
• (2, -2)
• (2, -14)

Solution

Let P(x₁, y₁) be a point on the given curve such that the tangent at P passes through the origin. Since P(x₁, y₁) lies on y = x² + 3x + 4 \ y₁ =    ..........(i) Now, y = x² + 3x + 4   = 2x + 3  = 2x₁ + 3 The equation of the tangent at P (x₁, y₁) is y - y₁ = (2x₁ + 3) (x - x₁) It passes through the origin i.e. (0, 0) \ 0 - y₁ = (2x₁ + 3) (0 - x₁) y₁ =        ............(ii) Subtracting (ii) from (i), we get  = 0 => x₁ = ± 2 From (ii), we obtain that IF x₁ = 2 , then y₁ = 4 + 6 + 4 = 14 If x₁ = -2 ,then y₁ = 4 - 6 + 4 = 2 Hence, the required points are (2, 14) and (-2, 2). If x₁ = 2 then y₁ = 4 + 6 + 4 = 14

Q27. If x + y = k is normal to the curve y² = 12x, then k is equal to

• 3
• 9
• -9
• -3

Solution

Q28. The function f(x) = e^x

• Has a local maximum  but no local minimum value
• Has a local minimum  but no local maximum value
• Has  both ,a local minimum  and a  local maximum value
• Has  neither a local minimum  nor a  local maximum value

Solution

Q29. Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24 x – 18x²

• 56
• 89
• 49
• 23

Solution

Q30. Find the equation of tangent to the curve y = (x-1)² which is parallel to the chord joining (1, 0) and (3, 4)

• 2x - y - 3 = 0
• x + y = 3
• x - y = 1
• 2x + y + 3 = 0