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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Find the equations of the tangent and normal to the parabola `y^2=4a x`at the point `(a t^2,2a t)`. |
| Answer» Correct Answer - `x - ty + at^(2) = 0, tx + y = at^(3) + 2at` | |
| 102. |
If `f(x)=x^(2)+(16)/(x^(2))`, thenA. f has minima at `x= pm 2`B. f has maxima at `x=pm 2`C. f has maximum value 8D. f has minimum value -8 |
| Answer» Correct Answer - A | |
| 103. |
If `y = 2^(x) " then " (dy)/(dx) =`?A. `x(2^(x -1))`B. `(2^(x))/((log 2))`C. `2^(x) (log 2)`D. none of these |
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Answer» Correct Answer - C `(dy)/(dx) = 2^(x) (log 2)` |
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| 104. |
If `f(x)=x^(3)-3x,` thenA. f has minima at x=1B. f has maxima at x=1C. f has maximum value -2D. f has minimum value 2 |
| Answer» Correct Answer - A | |
| 105. |
Find the point on the curve `y=x^3-11 x+5`at which thetangent is `y" "=" "x" "" "11`. |
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Answer» Equation of curve ` y = x^(3) - 11x + 5` ….(1) ` rArr (dy)/(dx) =3x^(2) - 11` Slope of tangent at point `(x, y) = 3x^(2) - 11` Slope of tangent y = x - 11 is = 1 `:. 3x^(2) - 11 = 1` ` rArr x^(2) = 4` `rArr x = 2 or x=-2` put x = - 2 in equation (1) `y= 2^(3) - 11 xx 2 + 5 =-9` but the equation (1) ` y = (-2)^(3) - 11 xx (-2) + 5 = 19` but the equation y = x - 11 is not satisfied by the point (-2, 19). `:. ` Required point = (2, -9) |
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| 106. |
For the curve `y=4x^3-2x^5,`find all the points at which the tangents pass through the origin. |
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Answer» The equation of the given curve is `y = 4x^(3) - 2x^(5)`..(i) On differentiating (i) w.r.t. x, we get `(dy)/(dx) = 12 x^(2) - 10x^(4)` Let the required point be `P (x, y)` The equation of tangent to be given curve at P(x, y) is given by `Y - y = (12x^(2) - 10x^(4)) (X - x)`...(ii) If it passes through (0, 0) then, we have `y = (12x^(2) - 10x^(4)) x` `rArr 4x^(3) - 2x^(5) = 12x^(3) - 10x^(5)` [using (i)] `8x^(3) - 8x^(5) = 0 rArr 8x^(3) (1-x^(2)) = 0` `rArr x = 0 or x = 1 or x = -1` Putting x = 0 in (i), we get y = 0 Putting x = 1 in (i), we get y = 2 Putting `x = -1` in (i), we get `y = -2` So, the required points are (0, 0) , (1, 2) and `(-1, -2)` |
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| 107. |
Find the points on the curve `4x^2+9y^2=1`, where the tangents are perpendicular to the line `2y+x=0`. |
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Answer» The equation of the given line is `y = - (1)/(2)x` `:.` slope of this line `= - (1)/(2)`..(i) Let the required point be `(x_(1), y_(1))` Now, `4x^(2) + 9y^(2) = 1 rArr 8x + 18y .(dy)/(dx) = 0` `rArr (dy)/(dx) = ((-4x)/(9y)) rArr ((dy)/(dx))_((x_(1)"," y_(1))) = (-4x_(1))/(9 y_(1))` `:.` slope of the tangent at `(x_(1), y_(1)) = (-4x_(1))/(9y_(1))`...(ii) From (i) and (ii), we have `(-4x_(1))/(9y_(1)) xx (-(1)/(2)) = - 1 or y_(1) = - (2)/(9) x_(1)`..(iii) Since `(x_(1), y_(1))` lies on the curve `4x^(2) +9y^(2) = 1`, we have `4x_(1)^(2) + 9y_(1)^(2) = 1 rArr 4x_(1)^(2) + 9 xx (4)/(81) x_(1)^(2) = 1` [using (iii)] `:. x_(1)^(2) = (9)/(40) or x_(1) = +- (3)/(2sqrt10)` So, `y_(1) = +- (2)/(9) xx (3)/(2 sqrt10) = +- (1)/(3 sqrt10)` Hence, the required point are `((3)/(2 sqrt10), (1)/(3 sqrt10)) and ((-3)/(2 sqrt10) , (-1)/(3 sqrt10))` |
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| 108. |
Tangents are drawn from the origin to the curve `y = sin x`. Prove that their points of contact lie on the curve `x^(2) y^(2) = (x^(2) - y^(2))` |
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Answer» Let the point of contact be `(x_(1), y_(1))` Now, `y = sin x rArr (dy)/(dx) = cos x rArr ((dy)/(dx))_((x_(1)"," y_(1))) = cos x_(1)` `:.` the equation of the tangent at `(x_(1), y_(1))` is `(y - y_(1))/(x - x_(1)) = cos x_(1)`. Since the tangent passes through the origin, we have `(-y_(1))/(-x_(1)) = cos x_(1)`, i.e., `(y_(1))/(x_(1)) = cos x_(1)`..(i) But, `(x_(1) , y_(1))` lies on the curve `y = sin x` `:. y_(1) = sin x_(1)`...(ii) Squaring (i) and (ii) and adding, we get `(y_(1)^(2))/(x_(1)^(2)) + y_(1)^(2) = 1 rArr y_(1)^(2) + x_(1)^(2) y_(1)^(2) = x_(1)^(2)` `rArr x_(1)^(2) y_(1)^(2) = (x_(1)^(2) - y_(1)^(2))` This shows that `(x_(1), y_(1))` lies on the curve `x^(2) y^(2) = (x^(2) - y^(2))` |
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| 109. |
If `y = log_(10) x " then "(dy)/(dx)`= ?A. `(1)/(x)`B. `(1)/(x) (log 10)`C. `(1)/(x(log10))`D. none of these |
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Answer» Correct Answer - C `y = (log x)/(log 10)` |
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| 110. |
If `f(x)=x^(3)-3x,` thenA. f has minima at x=-1B. f has maxima at x=1C. f has maximum value -2D. f has minimum value -2 |
| Answer» Correct Answer - D | |
| 111. |
Find the equation of tangent of the curve `yx^(2)+x^(2)-5x+6=0` at that point at which curve crosses the X-axis. |
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Answer» At X-axis, y=0 ` :. Yx^(2)+x^(2)-5x+6=0` `rArr 0+x^(2)-5x+6=0` `rArr (x-2)(x-3)=0` `rArr x=2 or x=3` `:. " Points are "(2,0) or(3,0)` Again `yx^(2)+x^(2)-5x+6-0` `rArr x^(2)*(dy)/(dx)+2xy + 2x -5 = 0` `rArr (dy)/(dx) = (5-2x-2xy)/(x^(2))` Slope of tangent at point (2,0) `m= (5-4-0)/4 = 1/4` and equation of tangent ` y-0= 1/4 (x-2)` ` rArr 4y=x-2 rArr x-4y=2` Slope of tangent at point (3, 0). `m=(5-6-0)/3^(2) = - 1/9` and equation of tangent ` y-0 =-1/9 (x-3)` ` rArr 9y =- x+3 rArr x+9y=3`. |
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| 112. |
Find the equation of the normal to the curve `y = 2 sin^(2) 3x "at " x = (pi)/(6)` |
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Answer» When `x = (pi)/(6)`, we have `y = 2 sin^(2) ((3pi)/(6)) = 2` So, the point of contact is `((pi)/(6), 2)` Now, `y = 2 sin^(2) 3x rArr (dy)/(dx) = (4 sin 3x) xx 3 xx (cos 3x)` `rArr (dy)/(dx) = 12 sin 3 x cos 3 x = 6 sin 6x` `:. ((dy)/(dx)) " at" (x = (pi)/(6)) = 6 sin (6 xx (pi)/(6)) = 6 sin pi = 0` So, the tangent is parallel to the x-axis. Thus, the normal is parallel to the y-axis and passes through the point `((pi)/(6), 2)` So, the equation of the normal is `x = (pi)/(6)` |
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| 113. |
Find the equations of the tangent and the normal to the curve `y (x -2) (x -3) - x + 7 = 0` at the point where it cuts the x-axis |
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Answer» The curve cuts the x-axis, where `y = 0` Putting y = 0 in the given equation, we get `x = 7` `:.` the point of contact is (7, 0) Now, `y (x -2) (x -3) - x + 7 = 0 rArr y(x^(2) - 5x + 6) - x + 7 = 0` `rArr (x^(2) - 5x + 6).(dy)/(dx) + y (2x -5) - 1 = 0 rArr (dy)/(dx) = (1 + 5y - 2xy)/(x^(2) - 5x + 6)` `:. ((dy)/(dx))_((7","0))= ((1 + 5 xx 0 - 2 xx 7 xx 0)/(49 - 35 + 6)) = (1)/(20)` So, the equation of the tangent is `(y - 0)/(x -7) = (1)/(20) or x - 20y - 7 = 0` Also, the equation of the normal is `(y - 0)/(x -7) = - 20 or 20x + y - 140 = 0` |
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| 114. |
find all the lines that pass through the point ` (1, 1)` and are tangent to the curve represent parametrically as `x = 2t-t^2` and `y = t + t^2`A. `(2sqrt(43))/(9)`B. 2C. 3D. `(2sqrt(53))/(9)` |
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Answer» Correct Answer - D `x=2t-t^(2)" (i)"` and `y=t+t^(2)" (ii)"` `(dy)/(dx)=(dy//dt)/(dx//dt)=(1+2t)/(2-2t)` Slope of tangent using the point `(1,1)` and `(2t-t^(2),t+t^(2))" (iii)"` Equating (ii) and (iii), `(t+t^(2)-1)/(2t-t^(2)-1)=(1+2t)/(2-2t)` `rArr" "3t^(2)-4t+1=0` `rArr" "t=1,(1)/(3)` For `t=1`, point is `P(1,2)` and for `t=(1)/(3)`, point is Q`((5)/(9),(4)/(9))` `therefore" "PQ=(2sqrt(53))/(9)` |
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| 115. |
If `f(x)=x^(3)-3x,` thenA. f has minima at x=-1B. f has maxima at x=1C. f has maixmum value 2D. f has minimum value 2 |
| Answer» Correct Answer - C | |
| 116. |
If `y = e^(1//x) " then " (dy)/(dx) =` ?A. `(1)/(x).e^((1//x-1))`B. `(-e^(1//x))/(x^(2))`C. `e^(1//x) log x`D. none of these |
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Answer» Correct Answer - B `(dy)/(dx) = (e^(1//x))/(-x^(2))` |
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| 117. |
Find the equation of the tangent to the curve `y = (sec^(4) x - tan^(4) x) " at " x = (pi)/(3)` |
| Answer» Correct Answer - `13y - 48 sqrt3x + 16 sqrt3pi - 21 = 0` | |
| 118. |
Find the equation of the tangent to the curve `sqrtx+sqrt y = a` at the point `(a^2/4,a^2/4)` |
| Answer» Correct Answer - `2(x + y) = a^(2)` | |
| 119. |
If `y=sqrt(x+sqrt(x+sqrtx+............oo))`, then `(dy)/(dx)`A. `(1)/((2y -1))`B. `(1)/((y^(2) -1))`C. `(2y)/((y^(2) -1))`D. none of these |
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Answer» Correct Answer - A `y = sqrt(x +y) rArr y^(2) = x + y rArr 2y (dy)/(dx) = 1 + (dy)/(dx)` |
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| 120. |
Find the maximum and minimum values of `f(x) = (-x + 2sin x) " on " [0, 2pi]` |
| Answer» Correct Answer - max. value is `(-(pi)/(3) + sqrt3) " at " x = (pi)/(3)` and min. value is `((5pi)/(3) + sqrt3) " at " x = (5pi)/(3)` | |
| 121. |
if `y=x^x ` then `(dy)/(dx)`A. `x^(x) log x`B. `x^(x) (1 + log x)`C. `x(1 + log x)`D. none of these |
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Answer» Correct Answer - B `log y = x log x`. Now differentiative w.r.t. x |
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| 122. |
If `f(x)=x^(3)-3x,` thenA. f has maxima at x=-1B. f has minima at x=-1C. f has maximum value 4D. f has minimum value 2 |
| Answer» Correct Answer - A | |
| 123. |
A wire of length 36cm is cut into the two pieces, one of the pieces is turned in the form of a square and other in form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum |
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Answer» Correct Answer - `(144 sqrt3)/(4 sqrt3 + 9) cm, (324)/(4 sqrt3 + 9) cm` Let the perimeter of the square be x cm Then the perimeter of the triangle is `(36 -x)cm` `:.` side of the square `= (x)/(4)` cm And, side of the triangle `= (1)/(3) (36 -x) cm` `:. A = (x^(2))/(16) + (sqrt3)/(4) (12 - (x)/(3))^(2) = (x^(2))/(16) + (sqrt3)/(4) (144 + (x^(2))/(9) - 8x)` `rArr A = ((sqrt3)/(36) + (1)/(16)) x^(2) - 2 sqrt3x + 36 sqrt3` `rArr (dA)/(dx) = ((4 sqrt3 + 9))/(144) xx 2x - 2sqrt3 and (d^(2)A)/(dx^(2)) = (4 sqrt3 + 9)/(72) gt 0` `:. (dA)/(dx) = 0 hArr x = (144 sqrt3)/((4 sqrt3 + 9)) cm` |
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| 124. |
Find points at which the tangent to the curve `y=x^3-3x^2-9x+7`is parallel to the x-axis. |
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Answer» Equation of curve ` y = x^(3) - 3x^(2) - 9x +7`…(1) `rArr (dy)/(dx)= 3x^(2) - 6x-9` The tangent of the curve will be parallel to x-axis at those point where slope = 0. `:. 3x^(2) - 6x - 9 =0` `rArr x^(2) - 2x - 3 = 0` ` rArr(x-3)(x+1) = 0` ` rArr x= 3 or x =- 1` put x = 3 in equation (1) ` y = 27 - 27 - 27 +7 =-20` Put x =- 1 in equation (1) ` y =- 1- 3+9+7=12` `:.` Required point = (3, -20) and ( -1, 12) |
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| 125. |
If `y = x^(sin x) " then " (dy)/(dx) =` ?A. `(sin x).x^((sin x-1))`B. `(sin x cos x).x^((sin x-1))`C. `x^(sin x) {(sin x +x log x. cos x)/(x)}`D. none of these |
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Answer» Correct Answer - C `log y = sin (log x)` |
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| 126. |
If `y = cot^(-1) ((1 -x)/(1 +x)) " then " (dy)/(dx) =` ?A. `(-1)/((1 +x^(2)))`B. `(1)/((1 + x^(2)))`C. `(1)/((1 + x^(2))^(.^(3)//_(2)))`D. none of these |
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Answer» Correct Answer - B Put `x = tan theta`, They, `y = cot^(-1) .tan ((pi)/(4) - theta) = cot^(-1) [cot {(pi)/(2) - ((pi)/(4) - theta)}] = ((pi)/(4) + theta)` `:. y = (pi)/(4) + tan^(-1) x` |
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| 127. |
If `f(x)=x^(3)-3x,` thenA. f has extreme values x=1, -1B. f has minimum value 30C. f has minima at x=-1D. f has maximum value -2 |
| Answer» Correct Answer - A | |
| 128. |
Find the equation of the tangent and the normal to the curve `y = x^(2) + 4x + 1` at the point where x = 3 |
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Answer» When `x = 3`, we have `y = (3^(2) + 4 xx 3 +1) = 22` So, the point of contact is (3, 22) Now `y = x^(2) + 4x + 1 rArr (dy)/(dx) = 2x + 4 rArr ((dy)/(dx))_((3, 22)) = (2 xx 3 + 4) = 10` Equation of the tangent is `(y - 22)/(x -3) = (-1)/(10) rArr x + 10y - 278 = 0` And, equation of the normal is `(y - 22)/(x -3) = (-1)/(10) rArr x + 10y - 223 = 0` |
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| 129. |
The equation of normal to the curve `y=3x^(2)-x+1` at (1,3) isA. x+5y+16=0B. x+5y-16=C. x-5y+16=0D. x-5y-16=0 |
| Answer» Correct Answer - B | |
| 130. |
If `(x +y) = sin (x + y) " then " (dy)?(dx)=` ?A. `-1`B. 1C. `(1 - cos (x + y))/(cos^(2) (x + y))`D. none of these |
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Answer» Correct Answer - A `(x + y) = sin (x + y) rArr 1 + (dy)/(dx) = cos (x + y) .[1 + (dy)/(dx)]` |
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| 131. |
The equation of normal to the curve `y=x^(2)+4x` at the point whose ordinate is -3, isA. x-2y-5=0, x+2y+9=0B. x+2y+5=0, x-2y-9=0C. x-2y-7=0, x+2y+3=0D. x+2y+7=0, x-2y-3=0 |
| Answer» Correct Answer - D | |
| 132. |
The equation of normal to the curve `y=3x^(2)+4x-5" at "(1,2)` isA. x+10y-21=0B. x-10y+21=0C. x-10y-21=0D. x+10y+21=0 |
| Answer» Correct Answer - A | |
| 133. |
If `y = cos^(-1) (4x^(3) -3x) " then " (dy)/(dx)=` ?A. `(3)/(sqrt(1 -x^(2)))`B. `(-3)/(sqrt(1 -x^(2)))`C. `(4)/(sqrt(1 -x^(2)))`D. `(-4)/((3x^(2) -1))` |
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Answer» Correct Answer - B Putting `x = cos theta`, we get `y = cos^(-1) (cos 3 theta) = 3 theta = 3 cos^(-1) x` |
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| 134. |
If `y = sqrt((sec x -1)/(sec x + 1)) " then " (dy)/(dx) =` ?A. `sec^(2)x`B. `(1)/(2) "sec"^(2) (x)/(2)`C. `(-1)/(2) "cosec"^(2) (x)/(2)`D. none of these |
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Answer» Correct Answer - B `y = ((1 - cos x)/(1 + cos x))^((1)/(2)) = {(2 sin^(2) (.^(x)//_(2)))/(2 cos^(2) (.^(x)//_(2)))}^((1)/(2)) = "tan " (x)/(2)` |
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| 135. |
If `e^(x +y) = xy " then " (dy)/(dx)=` ?A. `(x (1 -y))/(y (x -1))`B. `(y(1 -x))/(x(y -1))`C. `((x -xy))/(xy -y)`D. none of these |
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Answer» Correct Answer - B `(x + y) = log x + log y rArr 1 + (dy)/(dx) = (1)/(x) + (1)/(y).(dy)/(dx)` |
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| 136. |
If `x^(y) = y^(x) " then " (dy)/(dx)=` ?A. `((y - x log y))/((x - y log x))`B. `(y(y -x log y))/(x(x - y logx))`C. `(y(y + x log y))/(x(x + y logx))`D. none of these |
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Answer» Correct Answer - B `x^(y) = y^(x) rArr y log x = x log y`. Now, differentiate both side w.r.t.x. |
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| 137. |
`y = sin^(-1) {(sqrt(1 +x) + sqrt(1 -x))/(2)} " then " (dy)/(dx) =` ?A. `(-1)/(2 sqrt(1 -x^(2)))`B. `(1)/(2 sqrt(1 -x^(2)))`C. `(1)/(2(1 +x^(2)))`D. none of these |
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Answer» Correct Answer - A Put `x = cos theta`. Then, `y = (pi)/(4) + (1)/(2) cos^(-1) x` |
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| 138. |
If `y = cos^(-1) ((x^(2) -1)/(x^(2) +1)) " then " (dy)/(dx) =`?A. `(2)/((1 +x^(2)))`B. `(-2)/((1 + x^(2)))`C. `(2x)/((1 + x^(2)))`D. none of these |
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Answer» Correct Answer - B Putting `x = cot theta`, we get `y = cos^(-1) ((1 - tan^(2) theta)/(1 + tan^(2) theta)) = cos^(-1) (cos 2 theta) = 2 theta = 2 cot^(-1)x` |
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| 139. |
If `y = sin^(-1) (3x -4x^(3)) " then " (dy)/(dx) =` ?A. `(3)/(sqrt(1 -x^(2)))`B. `(-4)/(sqrt(1 -x^(2)))`C. `(3)/(sqrt(1 + x^(3)))`D. none of these |
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Answer» Correct Answer - A Putting `x = sin theta`, we get `y = sin^(-1) (sin 3 theta) = 3 theta = 3 sin^(-1) x` |
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| 140. |
If `x = a cos^(2) theta, y = b sin^(2) theta " then "(dy)/(dx)=` ?A. `(-a)/(b)`B. `(a)/(b) cot theta`C. `(-b)/(a)`D. none of these |
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Answer» Correct Answer - C `(dx)/(d theta) = - 2 a cos theta sin theta, (dy)/(d theta) = 2b sin theta cos theta rArr (dy)/(dx) = ((.^(dy)//_(d theta)))/((.^(dx)//_(d theta))) = (-b)/(a)` |
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| 141. |
If `y = cos^(-1) x^(3) " then " (dy)/(dx)=` ?A. `(-1)/(sqrt(1 - x^(6)))`B. `(-3x^(2))/(sqrt(1 -x^(6)))`C. `(-3)/(x^(2) sqrt(1 -x^(6)))`D. none of these |
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Answer» Correct Answer - B `y = cos^(-1) x^(3) rArr (dy)/(dx) = (-3x^(2))/(sqrt(1 -x^(6)))` |
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| 142. |
If `y = sec^(-1) ((1)/(2x^(2) -1)) " then " (dy)/(dx)`= ?A. `(-2)/((1 + x^(2)))`B. `(-2)/((1 -x^(2)))`C. `(-2)/(sqrt(1 -x^(2)))`D. none of these |
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Answer» Correct Answer - C Put `x = cot theta`. Then, `y = sec^(-1) ((1)/(2 cos^(2) theta -1)) = sec^(-1) (sec 2 theta) = 2 theta = 2 cos^(-1) x` |
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| 143. |
If `y = sec^(-1) ((x^(2) + 1)/(x^(2) -1)) " then " (dy)/(dx) =` ?A. `(-2)/((1 + x^(2)))`B. `(2)/((1 + X^(2)))`C. `(-1)/((1 - X^(2)))`D. none of these |
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Answer» Correct Answer - A Put `x = cot theta`. Then, `y = sec^(-1) ((1 + tan^(2) theta)/(1 - tan^(2) theta)) = sec^(-1) ((1)/(cos 2 theta)) = sec^(-1) (sec 2 theta) = 2 theta = 2 cot^(-1) x` |
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| 144. |
If `y = cos^(-1) x^(3) " then " (dy)/(dx)=` ?A. `(-1)/((1 + x))`B. `(2)/(sqrt((1 + x)))`C. `(-1)/(2 sqrtx (1 + x))`D. none of these |
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Answer» Correct Answer - C `y = cot^(-1) sqrtx rArr (dy)/(dx)= (-1)/((1 + x)) .(1)/(2 sqrtx) = (-1)/(2 sqrtx (1 + x))` |
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| 145. |
If `y = sqrt((1 + x)/(1 -x)) " then " (dy)/(dx) =` ?A. `(2)/((1 -x)^(2))`B. `(x)/((1 -x)^(.^(3)//_(2)))`C. `(1)/((1 -x)^(.^(3)//_(2)).(1 + x)^(½))`D. none of these |
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Answer» Correct Answer - C `log y = (1)/(2) {log (1 + x) - log (1 -x)} rArr (1)/(y).(dy)/(dx) = (1)/(2) {(1)/((1 +x)) + (1)/((1 -x))}` |
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| 146. |
If `x = a sec theta, y = b tan theta " then " (dy)/(dx) =` ?A. `(b)/(a) sec theta`B. `(b)/(a) "cosec" theta`C. `(b)/(a) cot theta`D. none of these |
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Answer» Correct Answer - B `(dx)/(d theta) = a sec theta tan theta, (dy)/(d theta) = b sec^(2) theta .(dy)/(dx) = ((.^(dy)//_(d theta)))/((.^(dx)//_(d theta))) = (b)/(a) " cosec " theta` |
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| 147. |
If `y = sqrt((1 + tan x)/(1 - tan x)) " then " (dy)/(dx)=` ?A. `(1)/(2) sec^(2)x.tan (x + (pi)/(4))`B. `(sec^(2) (x + (pi)/(4)))/(2 sqrt(tan (x + (pi)/(4))))`C. `(sec^(2) ((x)/(4)))/(sqrt(tan (x + (pi)/(4))))`D. none of these |
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Answer» Correct Answer - B `y = {tan (x + (pi)/(4))}^((1)/(2)) rArr (dy)/(dx) = (1)/(2) {tan (x + (pi)/(4))}^(-(1)/(2)).sec^(2) (x + (pi)/(4))` |
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| 148. |
If `y = tan^(-1) {(sqrt(1 + x^(2)) -1)/(x)} " then " (dy)/(dx)=` ?A. `(1)/((1 + x^(2)))`B. `(2)/((1 + x^(2)))`C. `(1)/(2 (1 +x^(2)))`D. none of these |
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Answer» Correct Answer - C Put `x = tan theta`. Then , `y = (1)/(2) tan^(-1) x` |
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| 149. |
If `y = tan^(-1) ((a cos x - b sin x)/(b cos x + a sin x)) " then " (dy)/(dx) =` ?A. `(a)/(b)`B. `(-b)/(a)`C. 1D. `-1` |
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Answer» Correct Answer - D `y = tan^(-1) (((a)/(b) - tan x)/(1 + (a)/(b) tan x)) = tan^(-1) ((tan theta - tan x)/(1 + tan theta tan x))` `= tan^(-1) tan(theta - x) = theta - x = ("tan"^(-1) (a)/(b) -x)` `:. (dy)/(dx) = -1` |
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| 150. |
If `y = tan^(-1) (sec x + tan x) " then " (dy)/(dx)=` ?A. `(1)/(2)`B. `(-1)/(2)`C. 1D. none of these |
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Answer» Correct Answer - A `y = tan^(-1) ((1 + sin x)/(cos x)) = tan^(-1) [({cos (.^(x)//_(2)) + sin (.^(x)//_(2))}^(2))/(cos^(2) (.^(x)//_(2)) - sin^(2) (.^(x)//_(2)))] = tan^(-1) {(cos (.^(x)//_(2)) + sin (.^(x)//_(2)))/(cos (.^(x)//_(2)) - sin (.^(x)//_(2)))}` `= tan^(-1) {(1 + tan (.^(x)//_(2)))/(1 - tan (.^(x)//_(2)))} = tan^(-1) {tan ((pi)/(4) + (x)/(2))} = ((pi)/(4) + (x)/(2))` |
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