Title Tangent & Normal (1).pdf ✅

MATHEMATICS Q.1 If the tangent at the point (at2 , at3 ) on the curve ay2 = x3 meets the curve again at P then P is - (A)         − 8 at . 4 at 2 3 (B)         − 8 at . 4 at 2 3 (C)         8 at . 4 at 2 3 (D) None of these [A] Q.2 If f(x) and g(x) are differentiable functions in [1,5] and  (x) = max {f(x), g(x)}, f(x) –g(x) = 0 has exactly one root  in [1,5] then - (A)  (x) continuous and differentiable at all points in [1, 5] (B)  (x) differentiable in [1, 5] (C)  (x) necessarily differentiable in [1,5] – {} (D)  (x) is not differentiable at x =  [C] Q.3 A curve is represented parametrically by the equations x = t + eat and y = –t + eat when t  R and a > 0. If the curve touches the axis of x at the point A, then the coordinates of the point A are (A) (1, 0) (B) (2e, 0) (C) (1/e, 0) (D) (e, 0) Sol.[B] let it be (t1 0) so 0 = – t1 + at1 e .........(i) Now dx dy = 0  dt dy = 0  – 1 + a at1 e = 0  at1 e = a 1 ....(ii) from (i) & (ii) t1 = a 1 Hence a 1 a. e = a 1  a 1 = e Hence x = a 1 + a 1 so x = a 2 = 2e pt A (2e, 0) e = 2 2 1 – b / a Q.4 Let f(x) be a differentiable function such that f(2) = 1, f (2) = 2. A tangent to g(x) = x f(x) is drawn at the point x = 2. Area of the triangle formed by the coordinate axes and a line perpendicular to the above tangent where it meets with y-axis is – (A) 80 sq. units (B) 180 sq. units (C) 160 sq. units (D) 120 sq. units [C] Q.5 A spotlight installed in the ground shines on a wall. A woman stands between the light and the wall casting a shadow on the wall. How are the rate at which she walks away from the light and rate at which her shadow grows related ? Spotlight Shadow Woman (A) One is a constant multiple of the other (B) It depends also on how close the woman is to the light (C) They are equal (D) They are independent [B] Q.6 P is a variable point on the curve y = f (x) and A is a fixed point in the plane not lying on the curve. If PA2 is minimum, then the angle between PA and the tangent at P is- (A) 4  (B) 3  (C) 2  (D) None of these [C] Sol. Obviously, AP is perpendicular on the tangent drawn to the curve. Q.7 The ends A and B of a rod of length 5 are sliding along the curve y = 2x2 . Let xA and xB be the x-coordinate of the ends. At the moment when A is at (0, 0) and B is at (1, 2) the derivative A B dx dx has the value equal to - (A) 1/3 (B) 1/5 (C) 1/8 (D) 1/9 [D] Sol. Given, y = 2x2 Now, (AB)2 = (XB – XA) + (2xB 2 – 2xA 2 ) = 5 or (xB – xA) 2 + 4 (xB 2 – XA 2 ) 2 = 5 On differentiating wrt XA and denoting A B dx dx =D
2(xB – xA) (D – 1) + 8(xB 2 – xA 2 ) (2xBD – 2xA) = 0 On putting xA = 0; xB = 1, then  2(1 – 0) (D – 1) + 8(1 – 0) (2D – 0) = 0  2D – 2 + 16D = 0  D = 9 1 . Q.8 A curve is represented parametrically by the equations x = t + eat and y = – t + eat when r  R and a > 0. If the curve touches the axis of x at the point A, then the coordinates of the point A are - (A) (1, 0) (B) (1/e, 0) (C) (e, 0) (D) (2e, 0) [D] Sol. x = t + eat; y = – t + eat dt dx = 1 + aeat ; dt dy = – 1 + aeat dx dy = at at 1 ae –1 ae + + At the point A, y = 0 and dx dy = 0 for some t = t1  1 at ae = 1 ...(i) Also, 0 = – t1 + 1 at e ;  1 at e = t1 ...(ii) On putting this value in Eq. (i), we get at1 = 1  t1 = a 1 ; Now, from Eq. (i), ae = 1  a = e 1 Hence, xA = t1 + 1 at e = e + e = 2e  A  (2e, 0). Q.9 The curve y = ax3 + bx2 + cx is inclined by 45o to x-axis at origin and it touches x-axis at (1,0). Then- (A) a = –2, b = 1, c = 1 (B) a = 1, b = 1, c = –2 (C) a = 1, b = –2, c = 1 (D) a = –1, b = 2, c = 1 [C] Sol. y' = 3ax2 + 2bx + c at (0, 0), y' = 1, c = 1 at (1, 0), y' = 0, 3a + 2b + 1 = 0  on solving a = 1, b = –2 Q.10 A continuous and differentiable function y = f(x) is such that its graph cuts line y = mx + c at n distinct points,. Then the minimum number of points at which f  (x) = 0 is/are (A) n – 1 (B) n – 3 (C) n – 2 (D) Cannot say Sol.[C] From LMVT there exist atleast (n – 1) point where f (x) = m.   atleast (n – 2) points where f (x) = 0 (using Rolle's theorem) Q.11 The points of contact of the vertical tangents to x = 2 – 3 sin , y = 3 + 2 cos  are (A) (2, 5), (2, 1) (B) (–1, 3), (5, 3) (C) (2, 5), (5, 3) (D) (–1, 3), (2, 1) Sol.[B] For vertical tangents d dx = 0 so, we have –3cos  = 0  = 2  or 2 3 . Corresponding to these values of , we have x = 2 – 3 sin 2  = –1, y = 3 + 2 cos 2  = 3; x = 2 – 3 sin 2 3 = 2 + 3 = 5, y = 3 + 2 cos 2 3 = 3 Thus the required points are (–1, 3), (5, 3). Q.12 The angle of intersection of curves y = [|sin x| + |cos x|] and x2 + y2 = 5 where [·] denotes the greatest integer function is (A) tan–1 (2) (B) tan–1       2 1 (C) tan–1 ( 2) (D) 2  Sol.[A] We know that 1  | sin x| + |cos x|  2 So that [|sin x| + |cos x|] will be constant function y = 1 No (–2, + 1) P x 2 + y 2 = 5 (2, 1) Q y = 1
Now intersection point P and Q are (–2, 1) and (2, 1) Slope of line y = 1 is zero and slope of tangent at P and Q are (–2) and (2) respectively Thus the angle of intersection is tan–1 (2) Q.13 The angle between the tangents at any point P and the line joining P to origin O, where P is a point on the curve ln(x2 + y2 ) = c tan–1 y/x, c is a constant, is (A) constant (B) varies at tan–1 (x) (C) varies as tan–1 (y) (D) None of these Sol.[A] P(x, y) be a point on the curve ln (x2 + y2 ) = c tan–1 y/x differentiating both side with respect to x 2 2 2 2 x y c(xy' y) (x y ) 2x 2yy' + − = + +  y ' = cx 2y 2x cy − + = m1 slope of OP = y/x = m2 So tan  = 1 2 1 2 1 m m m m + − = cx 2xy 2xy cy 1 x y cx 2y 2x cy 2 2 − + + − − + = 2/c  = tan–1 (2/c) which is independent of x and y Q.14 If the relation between subnormal SN and subtangent ST at any point S on the curve by2 = (x + a)3 is p(SN) = q(ST)2 , then p/q = (A) 27b a (B) 27b 8a (C) 27a 8b (D) 27 8b Sol.[D] by2 = (x + a)3 differentiating both sides 2by dx dy = 3(x + a)2 .1 dx dy = 2 3 by (x a) 2 + length of subnormal = SN = y dx dy = 2 3 b (x a) 2 + and length of subtangent = ST = y dy dx = 2 2 3(x a) 2by + q p = (SN) (ST) 2 (given) = 2 2 2 2 2 {3(x a) } .3(x a) (2by ) .2b + + = 27 8b 6 3 2 (x a) {(x a) } + + {using by2 = (x + a)3 = 27 8b q p = 27 8b Q.15 If the parabola y = f(x), having axis parallel to y-axis, touches the line y = x at (1, 1) then (A) 2f(0) + f(0) = 1 (B) 2f(0) + f(0) = 1 (C) 2f(0) – f(0) = 1 (D) 2f(0) – f(0) = 1 Sol.[B] Let f(x) = ax2 + bx + c  f (x) = 2ax + b Thus, f(0) = C and f(0) = b We have that information that y = f(x) is being touched by y = x at (1, 1)  ax2 + bx + c = x should have x = 1 at its repeated root.  ax2 + x(b –1) + c = 0 are 1, 1  a 1− b = 2, a c = 1  1 – b = 2a = 2c  1 – f (0) = 2f(0) or 2f (0) + f(0) = 1 Q.16 The eccentricity of the ellipse 4 x 2 + 3 y 2 = 1 is changed at the rate of 0.1 m/sec. The time at which it will touch the auxiliary circle is: (A) 2 seconds (B) 3 seconds (C) 6 seconds (D) 5 seconds Sol.[D] dt de = – 0.1 m/sec.  e = –0.1 t + e0 where e0 = 1/2  et = – 0.1 t + 2 1 , given et = 0  0.1t = 2 1 or t = 5 seconds.