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https://docs.google.com/document/d/1Byw7ogAgFCCtc9eN0bUa36lxWQe5yQzY/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true SYLLABUS Motion in a straight line, motion under gravity, Derivative as a rate measurer, geometrical interpretation of the derivative, tangent and normals, increasing and decreasing functions, maximum and minimum values of a function, applications of Rolle’s theorem and language’s mean value theorem. MOTION IN A STRAIGHT LINE VELOCITY The rate of change of displacement (or position) w.r.t time is called velocity. Thus ACCELERATION The rate of change of velocity w.r.t time is called acceleration. Thus at time t NOTE : The direction of acceleration is in the direction of velocity or opposite to it. When the direction of acceleration is opposite to the direction of velocity then it is called retardation. retardation means negative acceleration. Example 1% A particle moving in straight line covers s meter in t sec. If then calculate the following : ¼i½ Acceleration after 2 sec. ¼ii½ Time when velocity is 51 m/s. ¼iii½ Distance travelled in third sec. Solution % ¼i½ Acceleration after 2 sec. = 6 x 2 = 12 m/s. ¼ii½ sec. After 4 sec the velocity is 51 m/s ¼iii½ Distance after 3 sec = m Distance after 2 sec = Distance travelled in third sec. = 36 – 14 = 22 m Example 2% A particle moving in straight line covers s meter in t sec. If then calculate the velocity and acceleration when t = sec. Solution % s = 4t2 + t MOTION UNDER GRAVITY When a particle is thrown vertically upwards then it comes back due to the gravity. This motion of a particle is due to the acceleration g. When the particle is going upward, the value of g is negative and when it is coming back, the value of g is positive. At maximum height the velocity of a particle is zero. The value of g is 9.8 m/s2 or 980 cm/s2. Illustration 3% A ball is thrown upwards which comes back after 8 sec. on Earth. If the equation of motion is s = ut – 4.9 t2, where s is in meter and t in sec. then find the velocity at t = 0 and t = 2. Solution : Since it comes back after 8 sec.  distance covered in 8 sec. is 0.  s = ut – 4.9 t2 . . . (i) or or Differentiating, or  velocity at t = 0 m/s velocity at t = 2 m/s. Illustration 4% The equation of motion of a particle moving vertically is , where s is in meter and t in sec. If acceleration is –9.8 m@s2 and maximum height is 44 m then find the velocity after 1 sec. Solution : Given . . . (i) differentiating w.r.t. t acceleration  at maximum height v = 0  b – 9.8 t = 0 or  for maximum height  or velocity after 1 sec. Illustration 5% A man standing on 12 m high pillar, throw a ball upwards. The equation of motion is s = 16.9 t – 4.9 t2, where s is in meter and t in sec. Find the time taken by the ball in upward motion. Also find the time taken by the ball when it comes back to earth. Solution : s = 16.9 t – 4.9 t2 Let at time t in upward motion, the velocity of the ball is zero.So 0 = 16.9 – 9.8 t So the ball will take sec in upward motion. AS A RATE MEASURER Let y = f (x) be a single valued function of x. If increment in value of x is Δ x and that of in y is Δ y. Then the average rate of change of y w. r. t x in the interval (x, x + Δx) is the ratio (if it exists) is called the instantaneous rate of change of y w. r. t x at the point x. Therefore the derivative of a function at a point x represents the rate measurer of y w. r. t x at the point x. Illustration 6: On the curve x3=12y, find the interval at which the abscissa changes at a faster rate than the ordinate. Solution: Given x3=12y, on differentiating w. r. t x In the interval at which the abscissa changes at a faster rate than the ordinate we must have Illustration 7% A man of 2 m height moves away from a 3m high lamp post at the speed 2m/s. Find, at what rate (1) the length of his shadow increases. (2) the shadow is moving. Solution : (1) Let PQ is the lamp, AB is the man, shadow BS = y m and the distance of the man from the foot of the lamp post BQ = x m. speed (velocity) of the man m/s . . . (i) since ABS and PQS are similar . . . (ii) or or 3y = 2y + 2x or y = 2x and . . . (iii) The rate at which the shadow is increasing substituting the values from (i) and (iii) m/s ¼2½ If SQ = l then we have to find . From (ii), or or = 3x or now From (i) and (iv), m/s Illustration 8% The volume of a cube is increasing at a constant rate. Prove that the rate of change of surface is inversely proportional to the length of edge. Solution : Let the edge of the cube is x, volume V and surface is S. Now Given we know that now, GEOMETRICAL MEANING OF DERIVATIVE AT A POINT The derivative of the function y = f(x) at the point P(x, y) (when exists) is equal to the slope (or gradient) of the tangent line to the curve y = f(x) at P(x, y). Slope of tangent to the curve y = f(x) at the point (x, y) is m = tanθ = EQUATION OF TANGENT The equation of tangent to the curve y = f(x) at the point P(x1, y1) is given by Notes: (i) If = 0 then the tangent to curve y = f(x) at the point (x, y) is parallel to the x-axis. (ii) If → ∞, or = 0, then the tangent to the curve y = f(x) at the point (x, y) is parallel to the y-axis. (iii) If = tanθ > 0, then the tangent to the curve y = f(x) at the point (x, y) makes an acute angle with positive x-axis and vice versa. EQUATION OF NORMAL The normal to the curve at the point P(x1, y1) is a line perpendicular to the tangent at the point P(x1, y1) and passing through it. The angle between a tangent and a normal at a point is always 900. The equation of the normal to the curve y=f(x) at a given point P(x1, y1)is given by(x - x1) + (y - y1) = 0. EQUATION OF TANGENT AND NORMAL IF EQUATION OF THE CURVE IS GIVEN IN PARAMETRIC FORM If the equation of the curve is in the parametric form x = f(t) and y = g(t), then the equations of the tangent and the normal are ) and respectively Illustration 9: If at each point of the curve y = x3 – ax2 + x + 1 the tangents is inclined at an acute angle with the positive direction of the x-axis, then find the interval in which a lies. Key concept: since the tangent is always inclined at an acute angle with the x-axis, hence and the use the concept of quadratic equation that ax2 + bx + c > 0 for all x ∈R if a > 0 and D < 0 Solution: y = x3 – ax2 + x + 1 and the tangent is inclined at an acute angle with the positive direction of x-axis, ∴ 3x2 – 2ax + 1 >0, for all x ∈ R ⇒ (2a)2 – 4 (3)(1) <0⇒ 4(a2 – 3)< 0 ∴ Illustration 10: The tangent to the curve y = x – x3 at a point P meets the curve again at Q. Prove that one point of trisection of PQ lies on y-axis. Find the locus of the other point of trisection. Solution: For Hence the equation of the tangent at the point P(x1, y1) is; y – y1= (1 – 3x12)(x – x1) It meets the curve again at Q (x2, y2) Hence, (x2 – x23)– (x1 – x13) = (1 – 3x12)(x2 – x1) ⇒ (x2 – x1) [1- (x22 + x1x2 + x12)] = (x2 – x1)(1 – 3x12) ⇒ 1 – x22 – x1x2 – x12 = 0 or Since x1 ≠ x2, we have x2 = - 2x1 Q is (-2x1, - 2x1 + 8x13) If L1(α, β) is the point of trisection of PQ then . Hence L1lies on the y-axis. If L2(h, k) is the other point of trisection, then i.e. k = k = h - 5h3 ∴ locus of (h, k) is y = x – 5x3. ANGLE OF INTERSECTION OF TWO CURVES Let C1: y= f(x) and C2: y = g(x) be two curves. If two curves intersect at point P (x1, y1). Then the angle of intersection of two curves is defined as the angles between the tangents at their intersection. And is given by Note: (i) If the curves touch at P, then θ = 0 so that f ’ (x1) = g ‘ (x1) (ii) If the angle i.e. Two tangents are perpendicular to each other then the curves are said to cut orthogonally, then f ‘(x1). g ’(x1) =- 1. Illustration 11: If the curves x2-4y2+c=0 and y2 = 4x intersect orthogonally then find the range of c. Solution: Curves will intersect if x2 – 16 x + c = 0 has real roots. Thus c ≤ 64. For x2 – 4y2 + c = 0, For y2 = 4x, If curves intersect orthogonally then . But if y = 0, slope of both curves undefined. MONOTONOCITY Let y = f (x) be a given function with ‘D’ as it’s domain. Let then; Increasing Function: If a function f(x) is satisfying x1 > x2 ⇒ f(x1) > f (x2) for all x1, x2∈ D1, it means that the value of f (x) will keep on increasing with an increase in the value of x, then f is called increasing in D1. Decreasing Function: If a function f(x) is satisfying x1 > x2 ⇒ f(x1) < f (x2) for all x1, x2∈ D1, it means that the value of f (x) will keep on increasing with an increase in the value of x, then f is called decreasing in D1. Non-Decreasing Function: If a function f(x) is satisfying x1 > x2 ⇒ f(x1) ≥ f (x2) for all x1, x2∈ D1, it means that the value of f (x) will never decrease with an increase in the value of x, then f is called non-decreasing in D1. Non-Increasing Function: If a function f(x) is satisfying x1 > x2 ⇒ f(x1) ≤f (x2) for all x1, x2∈ D1, it means that the value of f (x) will never increase with an increase in the value of x, then f is called non-increasing in D1. Note: (i) If and points which make equal to zero (in between (a, b)) don’t form an interval, then f (x) would be increasing in [a, b] otherwise it will be non-decreasing function. (ii) Ifand points which make equal to zero (in between (a, b)) don’t form an interval, f (x) would be decreasing in [a, b], otherwise it will be non-increasing. Monotonic Function: A function which is either increasing or decreasing in its domain is called a monotonic function. Illustration 12: Prove that function f(x) = 2 cos x + cot x + 3x is decreasing in (0,π) Solution: f′(x) = – 2 sin x – cosec 2 x + 3 = - 2sin x + 1 > 0 in (0, π) hence f ′ > 0 function is increasing. Illustration 13: Find the intervals in which the function f (x) = 2x2 – ln |x| is (i) decreasing (ii) increasing Solution: f (x) = 2x2 – ln |x| f′ (x) = 4x – ⇒ f′ (x) = for increasing function f (x), f′ (x) > 0 ⇒ > 0 ⇒ x ∈ for decreasing function f (x), f′ (x) < 0 ⇒ < 0 ⇒ x ∈ . CONCEPT OF LOCAL MAXIMUM AND LOCAL MINIMUM LOCAL MAXIMUM A function f(x) is said to have a local maximum at x=a if the value of f(a) is greater than all the values of f(x) in a small neighbourhood of x=a. Mathematically, f (a) > f (a – h) and f (a) > f (a + h) where h > 0, then a is called the point of local maximum. Fig. Local Maxima LOCAL MINIMUM A function f(x) is said to have a local minimum at x = a, if the value of the function at x = a is less than the value of the function at the neighboring points of x = a. Mathematically, f (a) < f (a – h) and f (a) < f (a + h) where h > 0, then a is called the point of local minimum. A point of local maximum or a local minimum is also called a point of local extremum. Fig. Local Minima WORKING RULE TO DETERMINE THE POINTS OF LOCAL MAXIMA AND LOCAL MINIMA 1. First Derivative Test: If = 0 and changes it’s sign while passing through the point x = a , then (i) f(x) would have a local maximum at x = a if and . It means that should change it’s sign from positive to negative e.g. f (x) = –x2 has local maxima at x = 0. (ii) f(x) would have local minimum at x = a if and . It means that should change it’s sign from negative to positive. e.g. f (x) = x2 has local minima at x = 0. (iii) If f(x) doesn’t change it’s sign while passing through x = a, then f (x) would have neither a maximum nor minimum at x = a. e.g. f (x) = x3 doesn’t have any local maxima or minima at x = 0. 2. Second Derivative Test: Step I: Let f(x) be a differentiable function on a given interval and let f’’ be continuous at stationary point. Find f ‘ (x) and solve the equation f ‘ (x) = 0 given let x = a, b, … be solutions. Step II: Case (i) : If f ‘‘ (a) < 0 then f(a) is maximum. (Refer to above figure) Case (ii): If f ‘’(a) > 0 then f(a) is minimum. (Refer the above figure) Note: (i) If f’’(a) = 0 the second derivatives test fails in that case we have to go back to the first derivative test. (ii) If f’’(a) = 0 and a is not a point of local maximum nor local minimum then a is a point of inflection. 3. nth Derivative Test: It is nothing but the general version of the second derivative test. It says that if, f′ (a) = f″(a) = f″′ (a) =………. = fn (a) = 0 and fn+1 (a) ≠ 0 (all derivatives of the function up to order ‘n’ vanishes and (n + 1)th order derivative does not vanish at x = a), then f (x) would have a local maximum or minimum at x = a iff n is odd natural number and that x = a would be a point of local maxima if fn+1 (a) < 0 and would be a point of local minima if fn+1 (a) > 0. It is clear that the last two tests are basically the mathematical representation of the first derivative test. But that shouldn’t diminish the importance of these tests. Because at time it’s becomes very difficult to decide whether changes it’s sign or not while passing through point x = a, and the remaining tests may come handy in these type of situations. Illustration 14: Find the local maximum and local minimum for the following function : (i) (ii) Solution : (i) For local max. and local min. or or at negative at positive has maximum, has minimum, (ii) For local max. and local min.  at x = 1 f (x) has local max and at x = 6 there is local min, Local max value Local min value (iii) For local max. and local min. So at the function has local max and its value is So at the function has local min and its value is Illustration 15: If f(x) = where then find the points of local maxima and minima. Solution: Clearly, x = 1, 3 are the points of local minima and x = 2 is a local maxima. CONCEPT OF GLOBAL MAXIMA OR MINIMA Global maxima or minima in [a, b] Global maxima or minima of f(x) in [a, b] is basically the greatest or least value of f(x) in [a, b]. Global maxima or minima in [a, b] will always occur either at the critical points of f(x) within [a, b] or at the end points of the interval. Step to find out the global maxima or minima in [a, b] Step 1: Find out all the critical points of f(x) in (a, b). Let C1, C2,….Cn be the different critical points. Step 2: Find the value of the function at these critical points and also at the end points of the domain. Let the values of the function at critical points be f(C1), f(C2)………..f(Cn). Step 3: Find M1 =max{ f(a), f(C1), f(C2)………..f(Cn), f(b)} and M2= min{ f(a), f(C1), f(C2)………..f(Cn), f(b)}. Now M1is the maximum value of f(x) in [a, b] and M2 is the minimum value of f(x) in [a, b]. Global maxima or minima in (a, b): To find the global maxima and minima in (a, b) step 1 and 2 is same but after that we have to be little bit careful. After step 1 and 2 find M1 =max{ f(C1), f(C2)………..f(Cn)} and M2= min{f(C1), f(C2)………..f(Cn)}. Now if , f(x) would not have global maximum(or global minimum) in (a, b) but if and then M1 and M2 would respectively be the global maximum and global minimum of f(x) in (a,b) Illustration 16: Let f (x)= 2x3 – 9x2 + 12x + 6. Discuss the global maxima and minima of f (x) in [0, 2] and (1, 3). Solution: f (x) = 2x3 – 9x2 + 12 x + 6 (x) = 6x2 – 18x + 12 = 6 (x2 – 3x + 2) = 6 (x-1) (x-2) First of all let us discuss [0, 2]. Clearly the critical point of f (x) in [0, 2] is x = 1. f (0) = 6, f (1) = 11, f (2) = 10 Thus x = 0 is the point of global minimum of f(x) in [0, 2] and x = 1 is the point of global maximum. Now let us consider (1, 3). Clearly x = 2 is the only critical point in (1, 3). f (2) = 10. f (x) = 11 and f (x) = 15. Thus x = 2 is the point of global minimum in (1, 3) and the global maximum in (1, 3) does not exist. ROLLE’S THEOREM It is one of the most fundamental theorem of Differential calculus and has far reaching consequences. It states that if y = f (x) be a given function and satisfies, (i) f (x) is continuous in [a , b] (ii) f (x) is differentiable in (a , b ) (iii) f (a) = f (b) Then Note: (1) There can be more than one such c. (2) Think! The conditions of Rolle’s theorem are necessary or sufficient or both? The answer is conditions are only sufficient and necessary will be clear from the following examples: (a) Let Here condition (i) is violated However f ‘(x) = 0 if x = By defining f(0) = 0 and f(1) = 3 we can see that the result is true when (i) and (iii) are violated. (b) Let Clearly (ii) does not hold in (0, 2) and even so infact f ‘(x) = 0 for 1 < x < 2 (3) If f(x) is any polynomial then between any pair of roots of f (x) = 0 lies a root of f ’(x) = 0 Illustration 17: If ax2 + bx + c = 0, a, b, c ∈ R. Find the condition that this equation would have at least one root in (0, 1). Solution: Let f’(x) = ax2 + bx + c Integrating both sides, ⇒ ⇒ f(0) = d and Since, Rolle’s theorem is applicable ⇒ f(0) = f(1) ⇒ d = Hence required condition is 2a + 3b + 6c = 0 LAGRANGE’S MEAN VALUE THEOREM This theorem is in fact the general version of Rolle’s theorem. It says that if y = f(x) be a given function which is; (i) Continuous in [a , b] (ii) Differentiable in (a , b) Then . Let A ≡ (a , f (a)) and B ≡ (b , f (b)). Slope of Chord Illustration 18: If a, b, are two numbers with a < b, show that a real number c can be found between a and b such the 3c2 = b2 + ab + a2. Solution: Consider the function f(x) = x3 It is continuous and differentiable in (a, b). Hence by LMVT, there exist a point c such that a < c < b and f’(c) = . ASSIGNMENT LEVEL - I 1. The two curves (A) cut at right angles (B) touch each other (C) cut at an angle (D) cut at an angle SOLUTION: (A) Differentiating , we get Differentiating , we get Since, product of slopes is  The two curves cut at right angle. 2. The tangent to the curve x = a ( – sin ), y = a (1 + cos ) at the points  = (2n + 1), n Z are parallel to (A) x-axis (B) y-axis (C) y = x (D) x + y = 0. SOLUTION: (A)  The tangent is parallel to x-axis. 3. The critical points of the function f ' (x), where (A) –1 (B) 0 (C) 1 (D) 3 SOLUTION: (D) which shows that f '' (x) does not exist at x = 3.  Critical point of f ' (x) is 3. 4. If x lies in [0, 1], then minimum value of is (A) (B) 1 (C) 3 (D) None of these SOLUTION: (B)  f (x) is monotonically increasing on [0, 1]  Min. value of f (x) is f (0) =1 Note the method also. 5. If where m, then (A) x = a is a point of minimum (B) x = a is a point of maximum (C) x = a is not a point of maximum or minimum (D) None of these. SOLUTION: (C) f ' (x) = (x – a)2n (x – b)2m + 1  f ' (x) = 0  x = a, b f ' (x) does not change sign while passing through x = a. Hence 'a' is neither a point of maximum nor a point of minimum. 6. Tangents to the curve at x = –1 and x = 1 are (A) parallel (B) intersecting obliquely but not at an angle of 45° (C) perpendicular to each other (D) intersecting at an angle of 45°. SOLUTION: (A) and . Hence tangents are parallel. 7. The normal at the point (1, 1) on the curve is (A) x + y =0 (B) x + y + 1 = 0 (C) x – y + 1 = 0 (D) x – y = 0 SOLUTION: (D) slope of normal  slope of normal at Equation of normal is y – 1 = 1 (x – 1)  x – y = 0. 8. If the rate of increase of is twice the rate of decrease of it, then x is (A) 2 (B) 3 (C) 4 (D) 1 SOLUTION: (A)  by given condition x – 2 = –2 2 (x – 2)  x = 2. 9. For the curve , where the tangent is parallel to x-axis is (A) t = 0 (B) (C) (D) SOLUTION: (A) Tangent is perpendicular to x-axis if . 10. The point on the curve , the tangent at which makes an angle of 45° with x-axis will be given by (A) (B) (C) (2, 2) (D) SOLUTION: (D) (given) . 11. If tangent to the curve is perpendicular to x-axis then its point of contact is (A) (a, a) (B) (0, a) (C) (a, 0) (D) (0, 0) SOLUTION: (D) Point is (0, 0). 12. The number of values of k for which the equation has two different roots lying in the interval (0, 1) are (A) 3 (B) 2 (C) infinitely many (D) no value of k satisfies the requirement. SOLUTION: (D) Let . Let, if possible, a, b (0, 1) such that  no value of k satisfies the condition. 13. For the function , the value of x for which vanishes is (A) 3 (B) (C) (D) SOLUTION: (A) f (2) = 0 = f (4)  By Rolle's theorme, (2, 4) such that f ' (c) = 0  2c – 6 = 0  c = 3. 14. If the function f(x) = ax3 + bx2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and f′ = 0, then values of a and b are respectively (A) 1, - 6 (B) – 2, 1 (C) – 1, (D) – 1, 6. SOLUTION: (A) f(1) = f(3) a + b + 11 – 6 = 27a + 9b + 27 ⇒ 26a + 8b + 22 = 0 ⇒ 13a + 4b + 11 = 0 Also ⇒ 3a 15. The function assumes minimum value for x given by (A) 5 (B) 3 (C) (D) 2 SOLUTION: (B)  f ' (x) = 0  x = 3  f (x) is min. for x = 3. 16. xx has a stationary point at (A) x = e (B) (C) x = 1 (D) SOLUTION: (B) y = xx  log y = x log x 17. The absolute maximum of in [0, 2] is (A) 4 (B) 6 (C) 2 (D) 0 SOLUTION: (A)  absolute maximum is 4. 18. The sum of two numbers is 3, then max value of the product of the first and the square of the second is (A) 1 (B) 3 (C) 2 (D) 4 SOLUTION: (D) Let two numbers be x, 3 – x.   and But x = 3 is not possible  P is max. at x = 1 and maximum value is 1(3 – 1)2 = 4. 19. The maximum value of the function sin x (1 + cos x) is (A) 3 (B) (C) 4 (D) SOLUTION: (B)  y is maximum at and its value is . 20. If the parametric equation of a curve is given by , then the tangent to the curve at the point makes with the axis of x-axis the angle (A) 0 (B) (C) (D) SOLUTION: (D)  the tangent is perpendicular to x-axis. 21. The coordinates of the point on the curve the tangent at which passes through the origin is equal to (A) (2, 14), (–2, 2) (B) (2, 14), (–2, –2) (C) (2, 14), (2, 2) (D) None of these SOLUTION: (A)  equation of tangent is Y – y = (2x + 3) (X – x) It passes through (0, 0),  –y = – x(2x + 3)  x = 2, –2; y = 14, –2. 22. Let Then at x = 0, f has (A) a local maximum (B) no local maximum (C) a local minimum (D) no extremum SOLUTION: (D) f ' (x) does not exist as f '' (0–) = –1 and f ' (0+) = 1. 23. If , for every real x, then the minimum value of f (A) does not exist because f is unbounded (B) is not attained even through f is bounded (C) is equal to 1 (D) is equal to –1 SOLUTION: (D)  f ' (x) = 0  x = 0  x = 0 is a point of minima and minimum value is . LEVEL - II 24. The angle formed by the abscissa and the tangent to the parabola at the point is (A) tan–1 2 (B) tan–1 5 (C) tan–1 7 (D) None of these SOLUTION: (D) Slope of x-axis is 0.  slope of tangent to parabola at If  is the angle between x-axis and the tangent at P then tan  = 9  tan–1 9. 25. At (0, 0), the curve (A) touches x-axis (B) bisects the angle between the axes (C) makes an angle of 60° with ox (D) None of these SOLUTION: (B)  the curve bisects the angle between the axes. 26. If f and g are two increasing functions such that gof is defined, then (A) gof is an increasing function (B) gof is a decreasing function (C) gof is neither increasing nor decreasing (D) None of these SOLUTION: (A) Let such that ( f is increasing) ( g is increasing)  gof is increasing. 27. The function has no extrema if (n Z) (A) (B) (C) (D) None of these SOLUTION: (A)  f has no extrema if i.e. if sin ( – )  0 i.e. if sin ( – )  0, i.e. if  –   n, n Z. 28. Let f(x) satisfy the requirements of Lagrange’s mean value theorem in [0, 2]. If f(0) = 0 and f′(x) ≤ for all x in [0, 2], then (A) |f(x)| ≤ 2 (B) f(x) ≤ 1 (C) f(x) = 2x (D) f(x) = 3 for at least one x in [0, 2]. SOLUTION: (B) Given f’(x) ≤ ⇒ ⇒ h(x) = f(x) – is a deceasing that. Now h(0) = 0 and h(x) is decreasing ⇒ f(x) – ≤ 0 ⇒ f(x) ≤ ⇒ f(x) ≤ 1 since x ∈{0, 2} 29. A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the highest point of the ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of (A) 2 m/sec (B) 3 m/sec (C) 2.5 m/sec (D) 1.5 m/sec SOLUTION: (A) but when x = 4, y = 3  Height of the wall is decreasing at the rate of 2 m/sec. 30. A rod of length 13 metres has one end P on the x-axis and the other end Q on the y-axis. If P moves on the x-axis with the speed of 12 m/sec, then the speed of the other end Q when it is 12 m from the origin is (A) 3 m/sec (B) 5 m/sec (C) –5 m/sec (D) 4 m/sec SOLUTION: (C) (given) But, y = 12 (given)  speed of Q = –5 m/sec. 31. Equation of the tangent at the point P (t), where t is any parameter, to the parabola is (A) (B) (C) y = tx (D) SOLUTION: (A) Coordinates of the point P are (at2, 2at). Differentiating .  equation of tangent is . 32. The angle of intersection of the curves and is (A) (B) (C) (D) None of these SOLUTION: (C) Solving and , the point of intersection is = slope of tangent = = slope of tangent = . 33. Let f (x) and g (x) be differentiable for such that f (0) = 0, g (0) = 0, f (1) = 6. Let there exist a real number c in (0, 1) such that f ' (c) = 2g ' (c), then the value of g (1) must be (A) 1 (B) 3 (C) –2 (D) –1 SOLUTION: (B) Applying Rolle's theorem to F (x) = f (x) – 2g (x), F (0) = 0, F (1) – 2g (1) . 34. The function xx decreases on the interval (A) (0, e) (B) (0, 1) (C) (D) None of these SOLUTION: (C) is decreasing on . 35. The function has (A) only one maxima (B) only one minima (C) no maxima and minima (D) many maxima and minima SOLUTION: (B) and . Since  f has a local minima at x = 0. There is no other critical point and exists for all . 36. For all real x, the minimum value of is (A) 0 (B) (C) 1 (D) 3 SOLUTION: (B) where Now, y is min: when is max. is min. for x = 1  min. value of y is . 37. The least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is (A) (B) (C) (D) SOLUTION: (C) Find . 38. The triangle formed by the tangent to the curve at the point (1, 1) and the coordinate axes, lies in the first quadrant. If its area is 2, then the value of b is (A) –1 (B) 3 (C) –3 (D) 1 SOLUTION: (C)  equation of tangent at (1, 1) is  Length of x-intercept Length of y-intercept = – (b + 1)  Area of (given) . 39. If , then f (x) is (A) increasing on (B) decreasing R (C) increasing on R (D) decreasing on SOLUTION: (A) i.e. if i.e. . 40. If has its extremum values at x = – 1 and x = 2, then (A) a = 2, b = –1 (B) a = 2, (C) a = – 2, (D) None of these SOLUTION: (B) Now, at x = – 1, 2 Solving, . 41. The number of values of x where the function attains its maximum is (A) 0 (B) 1 (C) 2 (D) infinite SOLUTION: (B) has period and cos x has period 2. Hence, after attaining a maximum value at x = 0 (i.e. x = 0, y = 2), the function will not attain this value again. OBJECTIVE QUESTIONS OBJECTIVE LEVEL – 1 42. Total number of parallel tangents of f1(x) = x2 – x + 1 and f2(x) = x3 – x2 –2x + 1 is equal to (A) 2 (B) 3 (C) 4 (D) None of these SOLUTION: (D) f1’ (x1) = 2x1 – 1, f2’(x2) = 3x22 – 2x2 – 2 . Let tangents drawn to the curves y = f1(x) and y = f2(x) at (x1, f(x1)) and (x2, f(x2)) are parallel, then 2x1 – 1 = 3x22 – 2x2 – 2, which is possible for infinite number of order pair (x1, x2). 43. The function 2tan3x – 3tan2x + 12tanx + 3, x ∈ is (A) increasing (B) decreasing (C) increasing in (0, π/4) and decreasing in (π/4, π/2) (D) none of these SOLUTION: (A) Let f(x) = 2 tan3x – 3tan2x + 12 tanx + 3 f′(x) = (6tan2x – 6tanx + 12) sec2x = 6sec2x (tan2x – tanx + 2) > 0 Hence f(x) is always increasing. 44. Let f (x) = , then f has a (A) a local maxima at x = 0 (B) a local maxima at x = 2 (C) a local maxima at x = –2 (D) none of these SOLUTION: (A) f(x) = (4 – x2)2/3 f′(x) = at x = –2 local minima x = 0 local maxima x = 2 local minima 45. Let f (x) = x3 – 6x2 + 9x + 18, then f (x) is strictly decreasing in (A) (–∞, 1] (B) [3, ∞) (C) (–∞, 1] ∪ [3, ∞) (D) [1, 3] SOLUTION: (D) f(x) = x3 – 6x2 + 9x + 18 f′(x) = 3x2 – 12x + 9 = 3(x2 – 4x + 3) = 3(x – 1)(x – 3) ≤ 0 x ∈ [1, 3] 46. The absolute minimum value of x4 – x2 – 2x+ 5 (A) is equal to 5 (B) is equal to 3 (C) is equal to 7 (D) does not exist SOLUTION: (B) f(x) = x4 – x2 – 2x + 5 f′(x) = 4x3 – 2x – 2 = (x – 1) (4x2 + 4x + 2) Clearly at x = 1, we will set the minimum value which is 3. 47. Equation of the tangent to the curve y = e–|x| at the point where it cuts the line x=1 (A) is ey + x =2 (B) is x + y = e (C) is ex + y = 1 (D) does not exist SOLUTION: (A) y = e–|x| cut the line x = 1 at (1, 1/e) Tangent y = – = – (x – 1) ⇒ ey + x = 2 48. Rolle’s theorem holds for the function x3 + bx2 + cx, 1 ≤ x ≤ 2 at the point , the value of b and c are; (A) b = 8, c = - 5 (B) b = -5, c = 8 (C) b = 5, c = -8 (D) b = -5, c = -8. SOLUTION: (D) f′(4/3) = 0 ⇒ 16 + 8b + 3c = 0 Also f(1) = f(2) ⇒ 3b + c + 7 = 0 Hence b = – 5, c = –8 49. The number of value of k for which the equation x3 – 3x + k = 0 has two different roots lying in the interval (0, 1) are (A) 3 (B) 2 (C) infinitely many (D) no value of k satisfies the requirement. SOLUTION: (C) Let f(x) = x3 – 3x + k f′(x) = 3(x – 1) (x + 1) only are root may lie between –1 and 1 50. From mean value theorem: f(b) – f(a) = (b –a) f′ (x1); a < x1 < b if f(x) = , then x1 = (A) (B) (C) (D) . SOLUTION: (A) Since f(x) = f′(x1) = ⇒ ⇒ x1 = OBJECTIVE LEVEL – 2 51. The minimum value of ax + by, where xy = r2, is (r, ab >0) (A) 2r (B) 2ab (C) –2r (D) None of these SOLUTION: (A) Let f(x) = ax + f ’(x) = a – = 0 x = = 2r 52. If a, b, c, d are four positive real numbers such that abcd =1, then minimum value of (1+ a) (1 + b) (1 + c) (1 + d) is (A) 8 (B) 12 (C) 16 (D) 20 SOLUTION: (C) Applying AM ≥GM , Multiplying all focus (1 + a) (1 + b)(1 + c)(1 + d) ≥ 16 53. A lamp of negligible height is placed on the ground ‘l1’ metre away from a wall. A man ‘l2’ metre tall is walking at a speed of m/sec. from the lamp to the nearest point on the wall. When he is mid-way between the lamp and the wall, the rate of change in the length of this shadow on the wall is (A) m/sec. (B) m/sec. (C) m/sec. (D) m/sec. SOLUTION: (B) Clearly ⇒ hl1 – hx = l1l2 (Since h is decreasing put a – ve sign) ⇒ ⇒ at mid point x1 = l1/2, h = 2l2 ⇒ f ’(x) = 2ex + ae-x + 2a + 1≥ 0 clear a ≥ 0 54. Let f (x) = . If f (x) has a local minima at x = 1, then (A) a ≥ 5 (B) a > 5 (C) a > 0 (D) none of these SOLUTION: (A) f (x) = Local minimum value of f (x) at x = 1, will be 5 i.e. 1 –x + a ≥ 5 at x = 1 ⇒ a ≥ 5 55. Global minimum value of f (x) = x8 + x6 –x4 –2x3 –x2 –2x + 9 is (A) 0 (B) 1 (C) 5 (D) 9 SOLUTION: (C) We will get the minimum value of f(x) at x = 1 which is 5 OBJECTIVE LEVEL – 3 56. The greatest and least values of the function f(x) = ax + b √ x + c, when a > 0, b > 0, c > 0 in the interval [0,1] are (A) a + b + c and c (B) a/2 b√2+c, c (C) , c (D) None of these SOLUTION: (C) ∀x ∈ [0, 1] Hence f(x) will be minimum at 0 and maximum at x = 1 57. Maximum value of logx/x in interval [2 , ∞] is (A) log2/2 (B) 0 (C) 1/e (D) 1 SOLUTION: f′(x) = Clearly f(x) increasing for x < e and decreases for x > e ⇒ x = e is the point of local maxima ∴ maximum f(x) = . Hence (D) is the correct answer. 58. The curves C1: y1 = 1 – cos x, x ∈ (0, π) and C2 : y = |x| + a will touch each-other if (A) a = (B) a = (C) a = (D) a = SOLUTION: (A) Both the curve touch each other at sinx = or Hence point of contact is or For , we get a = For , we get a = 59. f1(x) = 2x, f2(x) = 3 sinx – x cosx then for x ∈ (A) f1(x) < f2(x) (B) f1(x) > f2(x) (C) f1(|x|) < f2(|x|) (D) none of these SOLUTION: (B) Let h(x) = f1(x) – f2(x) = 2x – 3sinx + xcosx h(0) = 0 h’(x) = 2 – 2cosx – x sinx h’’(x) = sinx – xcosx h’’’(x) = xsinx h’’’(x) > 0 ⇒ h’’(x)>0 ⇒h’(x) > 0 ⇒ h(x) > 0 60. A spherical balloon is pumped at the constant rate of 3 m3/min. The rate of increase of it’s surface area at certain instant is found to be 5 m2/min. At this instant, it’s radius is equal to (A) (B) (C) (D) SOLUTION: (C) We know and A = 4πr2 ⇒ ⇒ r = 6/5 m