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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If `g(x)` is a curve which is obtained by the reflection of `f(x) = (e^x - e^(-x)) / 2` then by the line `y = x`thenA. g(x) has more than one tangent parallel to x-axisB. g(x) has more than one tangent parallel to y-axisC. g(x) has no extremumD. |
| Answer» Correct Answer - D | |
| 2. |
If `f:[1,10]->[1,10]` is a non-decreasing function and `g:[1,10]->[1,10]` is a non-decreasing function. Let `h(x)=f(g(x))` with `h(1)=1,` then `h(2)`A. lies in (1,2)B. is more than 2C. is equal to 1D. is not defined |
| Answer» Correct Answer - C | |
| 3. |
`if ab gt 0 " and " 3a + 5b +15 c=0` then which of the following statement is INCORRECT ?A. there exist exactly one root of equation `ax^(4) +bx^(2) +c=0` in (-1,0)B. there exist exactly one root equation `ax^(4)+bx^(2) +c=0` in (0,1)C. there exist exactly two root of equation `ax^(4) +bx^(2) +bx^(2) +c=0` in (-1,1)D. number of roots of equation `ax^(4) +bx^(2) c=0` can be two in (-1,0) |
| Answer» Correct Answer - D | |
| 4. |
If `f(x)=|ax-b|+c|x|` is stricly increasing at atleast one point of non differentiability of the function where `a > 0, b > 0, c > 0` then (A) `c gt a` (B) `a gt c` (C) `b gt a+c` (D) a=bA. `c gt a`B. `a gt c`C. `b gt a +c`D. `a =b` |
| Answer» Correct Answer - A | |
| 5. |
Find angle `theta`, 0 < `theta` < `pi/2` , which increase twice as fast as sine |
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Answer» Let `theta` increases twice as fast at its sine. `rArr theta=2sintheta` Now, on differentiating both sides w.r.t., we get `(d(theta))/(dt) =2.costheta.(d(theta))/(dt) rArr 1=2costheta` `rArr 1/2 = costheta rArr costheta = cospi/3`. So, the required angle is `pi/3` |
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| 6. |
The volume of a cube is increasing at a rate of 7 `cm^(2)//sec.` How fast is the suface area increasing when the length of an edge is 4cm? |
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Answer» Let at some time t, the length of edge is x cm. `v= x^(3) " "rArr " "(dv)/(dt)= 3x^(2) (dx)/(dt) "("but .(dv)/(dt) =7")"` `rArr " "(dx)/(dt)= (7)/(3x^(2)) cm//sec` `"Now" " "S=6x^(2)` `(dS)/(dt)= 12x .(dx)/(dt) " "rArr " "(dS)/(dt)=12x.(7)/(3x^(2)) =(28)/(x)` `" when "" "x=4 cm. (dS)/(dt)=7 cm^(2)//sec.` |
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| 7. |
Find the equation of the normal to the curve `x^3+y^3=8x y`at the point where it meets the curve `y^2=4x`other than the origin.A. `y=x`B. `x=-x+4`C. `y=2x`D. `y=-2x` |
| Answer» Correct Answer - A | |
| 8. |
The tangent to the curve `y=e^(2x)` at the point (0,1) meets X-axis atA. `(0,1)`B. `(-1/2,0)`C. `(2,0)`D. `(0,2)` |
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Answer» Correct Answer - B the equation of curve is `y=e^(2x)` Since, it passes through the point (0,1). `therefore (dy)/(dx)=e^(2x).2=2.e^(2x)` `rArr (dy)/(dx)_(0,1) = 2.e^(2.0)=2`= Slope of the tangent to the curve. `therefore` Equation of tangent is `y-1=2(x-0)` `rArr y=2x+1` Since, tangent to the curve `y=e^(2x)` at the point (0,1) meets X-axis i.e., y=0. `therefore 0=2x+1 rArr x=-1/2` So, the required point is `(-1/2,0)`. |
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| 9. |
How fast the area of a circle increases when its radius is 5cm. (i) respect to radius (ii) with respect to diameter. |
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Answer» (i)`A =pir^(2) ," "(dA)/(dr) =2pir` `:. (dA)/(dr)]_(r=5) " " =10 pi cm^(2)//cm.` `(ii) A =(pi)/(4) D^(2) , (dA)/(dD) =(pi)/(2)D` `:. (dA)/(dD)]_(D=10)=(pi)/(2). 10=5pi cm^(2)//cm.` |
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| 10. |
A spherical balloon is filled with 4500p cubic meters ofhelium gas. If a leak in the balloon causes the gas to escape at the rate of `72pi`cubic meters per minute, then the rate (in metersper minute) at which the radius of the balloon decreases 49 minutes after theleakage began is(1) `9/7`(2) `7/9`(3) `2/9`(4) `9/2` |
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Answer» `V_i = 4500 pi m^3` `(d V_i)/(dt) = -72 pi m^3/(min)` `V_(49 min) = 4500 pi - (72 pi xx 49) = 4500 pi - 3528 pi` `= 972 pi m^3 ` `V= 4/3 pi r^3` at 49 min,`972 = 4/3 xx pi xx r^3` `r^3 = 3 xx 972/4` `r= 3^2 = 9 m` `(dv)/dt = 4/3 pi 3 r^2 (dr)/(dt)` `- 72 pi = 4/3 pi xx 3 xx 81 (dr)/(dt)` `(dr)/(dt) = -72/(4 xx 81) = -2/9` option 2 is correct |
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| 11. |
A spherical balloon is filled with 4500p cubic meters ofhelium gas. If a leak in the balloon causes the gas to escape at the rate of `72pi`cubic meters per minute, then the rate (in metersper minute) at which the radius of the balloon decreases 49 minutes after theleakage began is(1) `9/7`(2) `7/9`(3) `2/9`(4) `9/2`A. `(9)/(7)`B. `(7)/(9)`C. `(2)/(9)`D. `(9)/(2)` |
| Answer» Correct Answer - C | |
| 12. |
Find the slope of the normal to the curve `x=1-asintheta,y=bcos^2theta`at `theta=pi/2`. |
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Answer» Let `m` is the slope of tangent of the curve and `n` is the slope of normal to the curve. Then, `m*n = -1=> n = -1/m`. Now, equation of the curve, `x = 1-asintheta , y = bcos^2theta` `:. m = dy/dx = (dy/(d theta))/(dx/( d theta)) = (b(2cos theta)(-sintheta))/(-acostheta)` `=> m = (2b)/a sin theta` `:. n = -1/m = -a(2bsintheta)` At, `theta = pi/2`, `n = -a/(2b)` So, solope of normal to the given curve at `theta = pi/2` is `-a/(2b).` |
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| 13. |
The slope of the normal to the curve `y=2x^2+3`sin x at `x = 0`is(A) 3 (B) `1/3` (C)`-3` (D) `-1/3` |
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Answer» Slope of normal to a curve is given as, `m = -1/(dy/dx)` Here, equation of the curve is, `y = 2x^2 + 3sinx` `=>dy/dx = 4x + 3cosx` `:. dy/dx|_(x = 0) = 4(0)+3cos 0 = 3` `:.` Slope of the normal ` = -1/3.` |
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| 14. |
Find the point on the curve `y=x/(1+x^2)`, where the tangent to the curve has the greatest slope. |
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Answer» slope=`dy/dx=d/dx(x/(1+x^2))` `=((1+x^2)*1-x(2x))/(1+x^2)^2` `=(1-x^2)/(1+x^2)^2` `dy/dx` is maximum when`(x/(1+x^2))^2` is minimum `x=0,y=0` |
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| 15. |
Find the equation of all lines having slope ` 1`that are tangents to the curve `y=1/(x-1), x!=1`. |
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Answer» Equation of the curve, `y = 1/(x-1)` `:.` Slope of tangent,`( dy/dx) = -1/(x-1)^2` Slope of tangent to this curve is given `-1`. `:. -1/(x-1)^2= -1` `:. x = 0 and x = 2` `=>y = -1 and y = 1` Therefore, equation of tangents to the given curve, `(y+1)/(x-0) = -1 and (y-1)/(x-2) = -1` `=>x+y+1 = 0 and x+y-3 = 0`, which are the required equations. |
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| 16. |
Find equation of normal to the curve `y = |x^2-|x||` at `x =-2` In the neighborhood of `x =-2, y=x^2+x`. Hence the point of contact is (-2, 2). |
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Answer» In the neighborhood of `x=-2 ,y =x^(2) +x.` Hence the point of contact is `(-2,2)` `(dy)/(dx)= 2x+1 " "rArr " "(dy)/(dx)|_(x=-2) =-3` So the slope of normal at `(-2,2)` is Hence equation of normal is `(1)/(3) (x+2) =y-2 " "rArr" "3y =x+8` |
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| 17. |
Find the equation of all lines having slope 2 and being tangent to the curve `y+2/(x-3)=0`. |
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Answer» Eqaution of the curve is, `y +2/(x-3) = 0` Differentiating it w.r.t. `x`, `=>dy/dx - 2/(x-3)^2 = 0` `=> dy/dx = 2/(x-3)^2` Now, it is given that slope of the tangent is `2`. `:. dy/dx = 2` `=>2/(x-3)^2 = 2` `=>(x-3)^2 = 1` `=>x-3 = +-1` `=>x = 4 and x = 2` When `x = 4`, `y+2/(4-3) = 0` `=>y = -2` So, equation of tangent with this point,`y+2 = 2(x-4) => 2x - y = 10` When `x = 2`, `y+2/(2-3) = 0` `=>y = 2` So, equation of tangent with this point,`y-2 = 2(x-2) => 2x - y = 2.` |
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| 18. |
The tangent to the curve `y=x^2-5x+5.` parallel to the line `2y=4x+1,` also passes through the point :A. `((1)/(4), (7)/(2))`B. `((7)/(2), (1)/(4))`C. `(-(1)/(8), 7)`D. `((1)/(8), -7)` |
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Answer» Correct Answer - D The given curve is ` y = x ^(2) - 5x + 5 " " ` … (i) Now, slope of tangent at any point `(x, y)` on the curve is `" " (dy)/(dx) = 2x - 5" "`… (ii) `" " `[ on differentiating Eq. (i) w.r.t. x] `because` It is given that tangent is parallel to line ` " " 2y = 4x + 1 ` So, ` (dy)/(dx) = 2 [ because` slope of line ` 2y = 4x + 1 ` is 2`]` `rArr 2x - 5 = 2 rArr 2x = 7 rArr x = (7)/(2)` On putting `x= (7)/(2)` in Eq. (i), we get `" " y = (49)/( 4) - ( 35)/( 2) + 5 = ( 69)/( 4) -( 35)/( 2) = - (1)/(4)` Now, equation of tanent to the curve (i) at point `((7)/(2), - (1)/(4))` and having slope 2 , is `" " y + (1)/(4) = 2(x -(7)/(2)) rArr y + (1)/(4) = 2x - 7` `rArr " " y = 2x - (29)/(4)" "` ...(iii) On checking all the options, we get the point `((1)/(8), -7)` satisfy the line (iii). |
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| 19. |
Find the required point be `P(x_1, y_1)dot`The tangent to the curve `sqrt(x)+sqrt(y)=4`at which tangent is equally inclined to the axes. |
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Answer» We have, `sqrt(x) + sqrt(y)=4` `rArr x^(1//2)+y^(1//2)=4` `rArr 1/2.1/(x^(1//2))+1/2.(1/y^(1//2)).(dy)/(dx)=0` `(dy)/(dx) = -1/2.x^(-1//2)2.y^(1//2)` `=-sqrt(y/x)` Since, tangent is equally inclined to the axes. `therefore (dy)/(dx) = +-1` `rArr y/x = 1 rArr y=x` From Eq. (i), `sqrt(y)+sqrt(y)=4` `rArr 2sqrt(y)=4` `therefore y=4` and x=4 When y=4, then x=4 So, the required conditions are (4,4). |
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| 20. |
Find the point on the curve `y = (x - 2)^(2)` at which the tangent is parallel to the chord joining the points (2,0) and (4,4). |
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Answer» Equation of the curve is, `y = (x-2)^2` `:. dy/dx = 2(x-2)` `:.` Slope of the tangent`(m_1) = dy/dx = 2(x-2)` Now, slope of the chord joining points `(2,0)` and `(4,4) = (4-0)/(4-2) = 2` As, chord is parallel to the tangent, so their slopes will be equal. `:. 2(x-2) = 2` `=>x = 3` `:. y = (3-2)^2 = 1` So, the required point is `(3,1)`. |
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| 21. |
The point(s) on thecurve `y^3+ 3x^2=12 y`where the tangent isvertical, is(are) ??`(+-4/(sqrt(3)), -2)`(b) `(+- sqrt((11)/3, ) 1)``(0, 0)`(d) `(+-4/(sqrt(3)), 2)`A. `(pm (4)/(sqrt3), -2)`B. `(pm sqrt((11)/(3)), 0)`C. `(0, 0`)D. ` (pm (4)/(sqrt3), 2)` |
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Answer» Correct Answer - D Given, `" " y^(3) + 3x ^(2) = 12 y " " `…(i) On differentiating w.r.t. x, we get `rArr " " 3y^(2) (dy)/(dx) + 6x = 12 (dy)/(dx)` `rArr " " (dy)/(dx) = ( 6x)/(12 - 3y^(2))` `rArr " " (dx)/(dy) = ( 12 - 3y^(2))/(6x)` For vertical tangent, `(dx)/(dy) =0` `rArr 12 - 3y^(2) =0 rArr y = pm 2` On putting, `y = 2` in Eq. (i), we get ` x = pm ( 4)/(sqrt3)` and again putting `y = - 2 ` in Eq. (i), we get ` 3x^(2) = - 16`, no real solution . So, the required point is `(pm (4)/(sqrt(3)), 2)`. |
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| 22. |
The curve `y-e^(xy)+x=0` has a vertical tangent at the point :A. (1,1)B. (0,1)C. (1,0)D. no point |
| Answer» Correct Answer - C | |
| 23. |
if the tangent ta the curve `x=a(1+sintheta) , y=a(1+costheta)` at `theta=pi/3` makes an angle `x`=axis then `alpha`A. `(pi)/(3)`B. `(2pi)/(3)`C. `(pi)/(6)`D. `(5pi)/(6)` |
| Answer» Correct Answer - D | |
| 24. |
Find the point on the curve `y=x^3-11 x+5`at which the tangent is `y = x - 11`. |
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Answer» Equation of the tangent is , `y = x -11` Comparing it with `y = mx+c`, `m = 1` So, slope of the tangent is `1`. Now, equation of the curve is, `y = x^3-11x+5` `:. 1 = dy/dx = 3x^2-11` `=>3x^2 -11 = 1` `=>3x^2 = 12` `=>x^2 = 4` `=> x= +-2` When `x = 2, y = (2)^3-11(2)+5 = -9` If we put `x = 2, y = -9`, it satisfy the equation, `y = x-11`. So, it is the required point. When `x = -2, y = (-2)^3-11(-2)+5 = 19` If we put `x = -2, y = 19`, it does not satisfy the equation, `y = x-11`. So, the required point on the curve is `(2,-9)`. |
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| 25. |
Find the equations of the tangent and normal to the curve `x^(2/3)+y^(2/3)=2`at (1, 1) |
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Answer» Equation of the given curve, `x^(2/3)+y^(2/3) = 2` Differentiating it w.r.t. x, `2/3x^(-1/3)+2/3y^(-1/3)dy/dx = 0` `=>dy/dx = -(x/y)^(-1/3) = -(y/x)^(1/3)` `:. ` Slope of the tangent at point `(1,1) = m_1 = dy/dx = -(1)^(1/3) = -1` Let slope of the normal is `m_2`. then, `m_1*m_2 = -1 => -1*m_2 = -1 => m_2 = 1` Now, equation of the tangent, `y-1 = -1(x-1) => x+y-2 = 0` Equation of the normal, `y-1 = x-1 => x-y = 0.` |
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| 26. |
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? |
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Answer» `A-H,G-F,E-D,B-C` volume of box`=y^2x` `=y^2(9-y/2)` `=9y^2-y^3/2` `V(y)=V'(y)=0` `V''(y)<0` `18y-(3y^2)/2=0` `3(y)(6-y/2)=0` `y=0 or y=12` `V''(y)=18-6y/2` `=18-36` `=-18<0` `9-y/2=9-12/2` `=3` |
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| 27. |
A spherical iron ball 10cm in radius is coated with a layer of ice ofuniform thickness that melts at a rate of `50c m^3//m in`. When the thickness of ice is 5cm, then findthe rate at which the thickness of ice decreases.A. `(5)/( 6pi) cm//min`B. `(1)/(54pi) cm//min`C. `(1)/(18pi) cm//min`D. `(1)/(36pi)cm//min` |
| Answer» Correct Answer - C | |
| 28. |
A spherical ball of salt is dissolving in water in such a manner thatthe rate of decrease of volume at any instant is proportional to the surface.Prove that the radius is decreasing at a constant rate. |
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Answer» We have, rate of decrease of the volume of spherical ball of salt at any instant is `propto` surface. Let the radius of the spherical ball of the salt be r. `therefore` Volume of the ball (V) `=4/3pir^(3)` and surface area (S) = `4pir^(2)` `therefore (dV)/(dt) propto S rArr d/(dt) (4/3pir^(3)) propto 4pir^(2)` `rArr 4/3.pi.3r^(2).(dr)/(dt) propto 4pir^(2) rArr (dr)/(dt) propto (4pir^(2))/(4pir^(2))` `(dr)/(dt) = k.1` [where, k is the proportionally constant] `rArr (dr)/(dt) = k` Hence, the radius of ball is decreasing at a constant rate. |
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| 29. |
For the fucntion f(x) `=x cot^(-1) x , x ge 0`A. there is atleast one `x in (0,1)` for which `cot^(-1) x =(x)/(1+x^(2))`B. for atleast one x in the interval `(0,oo) f(x+(2)/(pi)) -f(x) lt 1`C. number of solution of the equation f(x) =sec x is 1D. f(x) is strictly decreasing in the interval `(0,oo)` |
| Answer» Correct Answer - B::D | |
| 30. |
The angle between curves `x^(2)+4y^(2) =32 " and "x^(2)-y^(2)=12` isA. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(2)` |
| Answer» Correct Answer - D | |
| 31. |
Find the shortest distance between the line `x - y +1 = 0` and the curve `y^2 = x.`A. `(3sqrt(2))/(8)`B. `(2sqrt(3))/(8)`C. `(3sqrt(2))/(5)`D. `(sqrt(3))/(4)` |
| Answer» Correct Answer - A | |
| 32. |
The shortest distance between curves `(x^(2))/(32)+(y^(2))/(18) =1 " and "(x+(7)/(4))^(2)+y^(2)=1`A. 15B. `(11)/(2)`C. `(15)/(4)`D. `(11)/(4)` |
| Answer» Correct Answer - D | |
| 33. |
The two curves `x^3-3xy^2+2=0` and `3x^2y-y^3-2=0`A. `pi/4`B. `pi/3`C. `pi/2`D. `pi/6` |
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Answer» Correct Answer - C Equation of two curves are given by `x^(3)-3xy^(2)+2=0` and `3x^(2)y-y^(3)-2=0` [on differentiating w.r.t. x] `rArr 3x^(2)-3[x.2y(dy)/(dx)+y^(2).1]+0=0` and `3y^(2)(dy)/(dx) = 3x^(2)(dy)/(dx)+6xy` `rArr (dy)/(dx)=(3x^(2)-3y^(2))/(6xy)` and `(dy)/(dx)=(6xy)/(3y^(2)-3x^(2))` `rArr (dy)/(dx)= (3(x^(2))-y^(2))/(6xy)` and `(dy)/(dx) = (-6xy)/(3(x^(2)-y^(2))` `rArr m_(1)=(x^(2)-y^(2))/(2xy)` and `m_(1)=(x^(2)-y^(2))/(2xy)` and `m_(2)=(-2xy)/(x^(2)-y^(2))` `therefore m_(1)m_(2)=(x^(2)-y^(2))/(2xy) -(2xy)/(x^(2)-y^(2))=-1` Hence, both the curves are intersecting at right angles i.e., making `pi/2` with each other. |
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| 34. |
A bird is sitting on the top of a vertical pole 20 mhigh and its elevation from a point O on the ground is `45o`. It flies off horizontally straight away from thepoint O. After one second, the elevation of the bird from O is reduced to `30o`. Then the speed (in m/s) of the bird is(1) `40(sqrt(2)-1)`(2)`40(sqrt(3)-2)`(3) `20""sqrt(2)`(4) `20(sqrt(3)-1)` |
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Answer» in `/_ AOB` `tan 45^@ = (AB)/(0B)` `1= 20/(OB)` `OB= 20m` in`/_ COD` `tan 30^@ = (CD)/(OD)` `1/sqrt3 = 20/(20+x)` `20 + x = 20 sqrt3` `x= 20(sqrt3 - 1)` Answer |
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| 35. |
Which of the following statements is`//`are correct?A. x+ sinx is increasing functionB. sec x is neither increasing nor decreasing functionC. x + sin x is decreasing functionD. sec x is an increasing function |
| Answer» Correct Answer - A::B | |
| 36. |
For the curve `x=t^(2) +3t -8 ,y=2t^(2)-2t -5` at point (2,-1)A. length of subtangent is `7//6`B. slope of tangent `=6//7`C. length of tangent `=sqrt((85))//6`D. none of these |
| Answer» Correct Answer - A::B::C | |
| 37. |
The intercepts on x-axis made by tangents to thecurve, `y=int_0^x|t|dt , x in R ,`which are parallel to the line `y""=""2x`, are equalto(1) `+-2`(2) `+-3`(3) `+-4`(4) `+-1` |
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Answer» `y = int_0^x |t| dt` `dy/dx = |x|= 2` `x +- 2` when `x=2` `y = int_0^2 |t|dt` `= |t|^2/2 = 2` when `x=-2` `y= - int_0^-2 |t| dt` `= - [|t|^2/2]` `y= -2` now,`y-2 = 2(x-2)` `y-2 = 2x - 4` `y= 2x -2` `y-2 = -2(x+2)` `y-2 = -2x - 4` `y = -2x-2` so, by solving `2x-2= 0` `x=1` `0= -2x-2` `x=-1` `x= +- 1` Answer is option 4 |
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| 38. |
which of the following inequalities are valid-A. `|tan^(-1) x -tan^(-1)y| le|x-y|AA, x in R`B. `|tan^(-1)x-tan^(-1)y|ge|x-y|`C. `|sin x -sin y| le|x-y|`D. `|sin x-sin y| ge |x-y|` |
| Answer» Correct Answer - A::C | |
| 39. |
Equation of the normal to the curve `y=-sqrt(x)+2` at the point (1,1)A. `2x -y-1=0`B. `2x-y+1=0`C. `2x+y-3=0`D. none of these |
| Answer» Correct Answer - A | |
| 40. |
Using differentials find the approximate value of `tan46^0,`if it is being given that `1^0=0. 01745`radians.A. `3`B. `1.035`C. `1.033`D. `1.135` |
| Answer» Correct Answer - B | |
| 41. |
Find the equation of the tangent to the curve `(1+x^2)y=2-x ,`where it crosses the x-axis.A. `x+5y=2`B. `x-5y=2`C. `5x-y=2`D. `5x+y=2` |
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Answer» Correct Answer - A We have, equation of the curve `y(1+x^(2))=2-x` `therefore y.(0+2x)+(1=x^(2)).(dy)/(dx)=0-1` [on differentiaing w.r.t. x] `rArr 2xy+(1+x^(2))(dy)/(dx)=-1` `rArr (dy)/(dx) = (-1-2xy)/(1+x^(2))` Since, the given curve passes through X-axis i.e., y=0 [uisng Eq. (i)] `therefore 0(1+x^(2))=2-x` `rArr x=2` So, the curve passes through the point (2,0). `therefore (dy)/(dx)_(2,0)=(-1 -2 xx 0)/(1+2^(2))=-1/5`= Slope of the curve `therefore` Slope of the tangent to the curve `=-1/5` `therefore` Equation of tangent of the curve passing through (2,0) is `y-0=-1/5(x-2)` `rArr y+x/5=2/5` `rArr 5y+x=2` |
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| 42. |
If `y = x^4−12` and if `x` changes from 2 to 1.99. what is the appoinmate change in `y.` |
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Answer» `y=x^4-12` `dy=4x^3dx` `/_y cong (4x^3)*/_x` `Z->1.99` `/_x=1.99-2=-0.01` `/_y cong (4.2^3)*(-0.01)` `-32(-0.01)=-0.32`. |
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| 43. |
A balloon, which always remains spherical, has a variable diameter `3/2(2x+1)`.Find the rate of change of its volume with respect to x. |
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Answer» `d=3/2(2z+1)` `dv/dx=?,v=4/3pix^3=pi/6d^3` `(d(pi/6)d^3)/dx=pi/6(d(d^3))/dx` `=pi/6(d(d^3))/d (d(d))/dx` `=pi/63d^2(d(3/2)(2x+1))/dx` `=(pi/2 3/2d^2)(d2x+1)/dx` `=((3pid^2)/4)2` `=(3pi)/2d^2` `=3/2pi(3/2(2x+1))^2` |
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| 44. |
Consider the curves `y=x^(2)+2` and `y=10-x^(2)` . Let `theta` be the angle between both the curves at point of intersection, then find `|tan theta|` (a) `(8)/(15)` (b) `(5)/(17)` (c) `(3)/(17)` (d) `(8)/(17)`A. `(7)/(17)`B. `(8)/(15)`C. `(4)/(9)`D. `(8)/(17)` |
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Answer» Correct Answer - B Given equation of curves are `" " y = 10 - x ^(2)" " `…(i) and `" " y = 2 + x^(2)" " ` … (ii) For point of intersection, consider `" " 10- x^(2) = 2 + x ^(2)` `rArr " " 2x^(2) = 8` `rArr " "x ^(2) = 4` `rArr " " x= pm 2 ` Clearly, when `x =2` , then `y =6` (using Eq. (i)) and when `x = - 2` , then `y = 6` Thus, the point of intersection are `(2, 6) and (-2, 6)`. Let `m_1` be the slope of tangent to the curve (i) and `m_2` be the slope of tangent to the curve (ii) For curve (i) `(dy)/(dx) = -2 x and ` for curve (ii) `(dy)/(dx)= 2x` `therefore ` At ` (2, 6)`, slope `m_1 =-4 and m_2=-4` and in that case `|tan theta| = |(m_2 - m_1)/(1+ m_1m_2)| = | (-4 -4)/( 1-16)| = (8)/(15)` At `(- 2, 6)`, slopes `m_1 = 4 and m_2 = - 4 ` and in that case `|tan theta| = |(m_2 - m_1)/(1+m_1m_2)| = |(-4-4)/(1-16)|= (8)/(15)` |
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| 45. |
If the curves `y^2=6x`, `9x^2+by^2=16` intersect each other at right angles then the value of b is: (1) 6 (2) `7/2` (3) `4` (4) `9/2`A. `6`B. `(7)/(2)`C. `4`D. `(9)/(2)` |
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Answer» Correct Answer - D We have, `y^(2) = 6x ` `rArr " " 2y (dy)/(dx) = 6 rArr (dy)/(dx) = (3)/(y)` Slope of tangent at `(x_1, y_1)` is `m_1 = (3)/(y_1)` Also, ` 9x^(2) + by^(2) = 16` `rArr " " 18 x + 2by (dy)/(dx) = 0 rArr (dy)/(dx) = (-9x)/(by)` Slope of tangent at `(x_1, y_1)` is `m_2` = `(-9x_1)/(by_1)` Since, these are intersection at right angle. `therefore m_1m_2 = -1 rArr = (27x_1)/( by_1^(2)) = 1 ` `rArr " " ( 27 x_1)/( 6bx _1) = 1 " " [ because y_1^(2) = 6x_1]` `rArr " " b = (9)/(2)` |
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| 46. |
The function `f(x)=x^(x)` has a stationary point atA. x=eB. x`=1/e`C. x=1D. `x=sqrt( E)` |
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Answer» Correct Answer - B We have, `f(x)= x^(x)` Let `y=x^(x)` and log `y=x logx` `therefore 1/y.(dy)/(dx)=x.1.x+log x.1` `rArr (dy)/(dx)=(1+logx).x^(x)` `therefore (dy)/(dx)=0` `therefore (dy)/(dx)=0` `rArr (1+logx).x^(2)=0` `rArr log x=-1` `rArr logx=loge^(-1)` `rArr x=e^(-1)` `rArr x=1/e` Hence, f(x) has a stationary point at `x=1/e` |
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| 47. |
Prove that the curves `y^2=4xa n dx^2+y^2-6x+1=0`touch each other at the point (1,2). |
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Answer» We have, `y^(2)=4x` and `x^(2)+y^(2)-6x+1=0` Since, both the curves touch each other at (1,2) i.e., curves are passing through (1,2). `therefore 2y.(dy)/(dx)=4` and `(dy)/(dx) = 4/(2y)` and `(dy)/(dx)= (6-2x)/(2y)` `rArr (dy)/(dx)_(1.2) = 4/4=1` and `(dy)/(dx)_(1/2) = ((6-2).1)/(2.2) = 4/4=1` `rArr m_(1)=1` and `m_(2)=1` Thus, we see that slope of both the curves are equal to each other i.e, `m_(1)=m_(2)=1` at the point (1.2). Hence, both the curves touch each other. |
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| 48. |
Show that the maximumvalue of `(1/x)^x`is `e^(1//e)`.A. eB. `e^( e)`C. `e^(1//e)`D. `(1/e)^(1//e)` |
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Answer» Correct Answer - C Let `y=(1/x)^(x)` `rArr logy=x.log1/x` `therefore 1/y(dy)/(dx)=x.1/(1/x). (-1/x^(2))+ log 1/x.1` `=-1+log1/x` `therefore (dy)/(dx)= (log 1/x-1).(1/x)^(x)` Now, `(dy)/(dx)=0` `rArr log1/x=1=loge` `rArr 1/x=e` `x=1/e` Hence, the maximum value of `f(1/e)=(e)^(1//e)` |
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| 49. |
The curves `y=4x^(2)+2x-8` and `y=x^(3)-x+13` touch each other at the point |
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Answer» The curves `y=4x^(2)+2x-8` and `y=x^(3)-x+13` touch each other at the pont `(3,34)` Give, equation of curves are `y=4x^(2)+2x-8` and `y=x^(3)-x+13` `therefore (dy)/(dx)=8x+2` ltbgt and `(dy)/(dx)=3x^(2)-1` So, the slope of both curves should be same `therefore 3x^(2)-9x+x-3=0` `rArr 3x(x-3)+1(x-3)=0` `rArr (3x+1)(x-3)=0` `therefore x=-1/3` and x=3. For x`=-1/3`, `y=4.(-1/3)^(2)+2.(-1/3)-8` `=4/9-2/3-8=(4-6-72)/(9)` `=-74/9` and for x=3, y=`4.(3)^(2)+2.(3)-8= 36+6-8=34` So, the required points are `(3,34)` and `(-1/3, -79/9)`. |
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| 50. |
The equation of normal to the curve y=tanx at (0,0) is ……………….. |
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Answer» The equation of normal to the curve y= tan x at (0,0) is x+y=0. `y=tanx rArr (dy)/(dx)= sec^(2)x` `rArr (dy)/(dx)_(0,0)= sec^(2)0=1` and `-1/((dy)/(dx))=-1/1` `therefore` Equation of normal to the curve `y=tanx` at `(0,0)` is `y-0=-1(x-0)` `rArr y+x=0` |
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