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InterviewSolution
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.
| 1. |
f the normal at the point `P (at_1, 2at_1)` meets the parabola `y^2=4ax` aguin at `(at_2, 2at_2),` thenA. `t_(1)t_(2)= -1`B. `t_(2)= -t_(1) -(2)/(t_(1))`C. `2t_(1)=t_(2)`D. none of these |
| Answer» Correct Answer - B | |
| 2. |
Let `P(2, 2) and Q(1//2, -1)` be two points on the parabola `y^(2)=2x`, The coordinates of the point R on the parabola `y^(2) =2x` where the tangent to the curve is parallel to the chord PQ, areA. `(2, -1)`B. `(1//8,1//2)`C. `(sqrt(2),1)`D. `(-sqrt(2),1)` |
| Answer» Correct Answer - B | |
| 3. |
The tangent and normal at `P(t)`, for all real positive `t`, to the parabola `y^2= 4ax` meet the axis of the parabola in `T` and `G` respectively, then the angle at which the tangent at `P` to the parabola is inclined to the tangent at `P` to the circle passing through the points `P, T `and `G` isA. `tan^(-1)t^(2)`B. `cot^(-1)t^(2)`C. `tan^(-1)t`D. `cot^(-1)t` |
| Answer» Correct Answer - C | |
| 4. |
If the curves `2x^(2)+3y^(2)=6 and ax^(2)+4y^(2)=4` intersect orthogonally, then a =A. 2B. 1C. 3D. none of these |
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Answer» Correct Answer - A We know that the curves `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(c^(2))+(y^(2))/(d^(2))=1` intersect orthogonally iff `a^(2)-b^(2)=c^(2)-d^(2)` Therefore, the curves `2x^(2)+3y^(2)=6 and ax^(2)+4y^(2) =4` will intersect orthogonally, if `3-2=(4)/(a)-1 rArr 2=(4)/(a) rArr a =2` |
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| 5. |
The curves `ax^(2)+by^(2)=1 and Ax^(2)+B y^(2) =1` intersect orthogonally, thenA. `(1)/(a)+(1)/(A)=(1)/(b) + (1)/(B)`B. `(1)/(a)-(1)/(A)=(1)/(b) - (1)/(B)`C. `(1)/(a)+(1)/(b)=(1)/(B) - (1)/(A)`D. `(1)/(a)+(1)/(b)=(1)/(A) + (1)/(B)` |
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Answer» Correct Answer - B In illustration 4, we have seen that the curves `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(c^(2))+(y^(2))/(d^(2))=1` intersect orthogonally , if `a^(2) -b^(2)=c^(2)-d^(2).` So, given curve will intersect orthogonally, if `(1)/(a)-(1)/(b)=(1)/(A) - (1)/(B)`. |
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| 6. |
Find the condition for the following set of curves to intersectorthogonally:`(x^2)/(a^2)-(y^2)/(b^2)=1`and `x y=c^2``(x^2)/(a^2)+(y^2)/(b^2)=1`and `(x^2)/(A^2)-(y^2)/(B^2)=1.` |
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Answer» 1)`(2x)/a^2+(2y)/b^2-dy/dx=0` `m_1=dy/dx=(-b^2x)/(a^2y)` `(2x)/A^2-(2y)/B^2*dy/dx=0` `m_2=dy/dx=(B^2x)/(A^2y)` `m_1*m_2=-1` `b^2/a^2*x/y*B^2/A^2*x/y=1` `x^2/y^2=(a^2A^2)/(b^2B^2)` `x^2/a^2+y^2/b^2-x^2/A^2+y^2/B^2=0` `x^2(1/a^2-1/A^2)=-y^2(1/b^2+1/B^2)` `x^2/y^2=-(1/b^2+1/B^2)/(1/a^2-1/A^2)` `(a^2A^2)/(b^2B^2)=-(1/b^2+1/B^2)/(1/a^2-1/A^2)` `a^2A^2(1/a^2-1/A^2)=-b^2B^2(1/b^2+1/B^2)` `a^2A^2((A^2-a^2)/(a^2B^2))=-b^2B^2((B^2+b^2)/(b^2B^2))` `a^2-A^2=b^2+B^2` `a^2-b^2=A^2+B^2`. |
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| 7. |
`" For a curve "(("length of normal"))/(("length of tangent"))` is equal toA. subtangentB. subnormalC. slope of tangentD. slope of normal |
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Answer» Correct Answer - C We have, Length of normal `=ysqrt(1+((dy)/(dx))^(2))` Length of tangent`=(ysqrt(1+((dy)/(dx))^(2)))/((dy)/(dx))` ` therefore ("length of normal")/("length of tangent")=(dy)/(dx) =` Slope of tangent |
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| 8. |
Let C be the curve `y^(3) - 3xy + 2 =0`. If H is the set of points on the curve C, where the tangent is horizontal and V is the set of points on the curve C, where the tangent is vertical, then H = … and V = … .A. `H={(x,y):y=0, x inR}, V={(1,1)} `B. `H={(x,y):x=0, y in R}, V={(1,1)} `C. `H=phi , V={(1,1)} `D. `H={(1,1)} , V={(x,y):y=0, x in R} ` |
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Answer» Correct Answer - C We have, ` y^(3)-3xy+2=0 " "...(i)` ` rArr 3y^(2)(dy)/(dx)-3(x(dy)/(dx)+y)=0 ` ` rArr (dy)/(dx)=(y)/(y^(2)-x) ` If the tangent is parallel to x-axis, then ` (dy)/(dx)=0 rArr (y)/(y^(2)-x)=0 rArr y=0 ` But, `y=0 ` does not satisfy equation (i). So, there is no point on the curve where tangent is parallel to x-axis. Therefore, `H= phi.` For the tangent to be parallel to y-axis, we must have `(dx)/(dy)=0 rArr (y^(2)-x)/(y)=0 rArr y^(2)=x ` Putting `x=y^(2)` in (i), we get `y^(3)-3y^(3)+2=0 rArr y^(3)=1 rArr y=1 ` ` therefore x=y^(2) rArr x=1 ` Thus, at (1,1) the tangent is parallel to y-axis. ` therefore V={(1,1)} ` Hence, `H= phi " and " V={(1,1)} ` |
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| 9. |
The curves `x=y^(2) and xy =a^(3)` cut orthogonally at a point, then a =A. `(1)/(3)`B. 3C. 2D. `(1)/(2)` |
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Answer» Correct Answer - D We have, `x=y^(2) " " (i) and , xy =a^(3) " " …(ii) ` These two curves intersect at `P(a, a^(2))`. Now, `x=y^(2) rArr (dy)/(dx)=(1)/(2y) rArr m_(1) = ((dy)/(dx))_(p) = (1)/(2a^(2))` and, `xy=a rArr x(dy)/(dx)+y=0 rArr (dy)/(dx) = (y)/(x) rArr m_(2) = ((dy)/(dx))_(p) = -a` If the two curves intersect orthogonally, then `m_(1)m_(2) = -1 rArr (1)/(2a^(2)) xx -a = -1 rArr a = (1)/(2)` |
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| 10. |
The curves ` x^(3) -3xy^(2) = a and 3x^(2)y -y^(3)=b,` where a and b are constants, cut each other at an angle ofA. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(2)`D. none of these |
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Answer» Correct Answer - C The equation of the two curves are `C_(1) : x^(3) -3xy^(2) =a " " ` …(i) `C_(2): 3x^(2)y -y^(3)=b " " ` …(ii) Differentiating (i) and (ii) w.r.t. x, we get `((dy)/(dx))_(C_(1))=(x^(2)-y^(2))/(2xy) and ((dy)/(dx))_(C_(2)) = -(2xy)/(x^(2)-y^(2))` Clearly, `((dy)/(dx))_(C_(1)) xx((dy)/(dx))_(C_(2))= -1` So, the two curves intersect at right angle. |
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| 11. |
Show that the curves `x = y^2` and `xy = k` cut at right angles; if `8k^2 = 1`A. `2k^(2)-1`B. `4k^(2)=1`C. `6k^(2)=1`D. `8k^(2)=1` |
| Answer» Correct Answer - D | |
| 12. |
The length of subtangent to the curve `x^2 + xy + y^2=7` at the point `(1, -3)` isA. 3B. 5C. 15D. `3//5` |
| Answer» Correct Answer - C | |
| 13. |
The subtangent at any point on the curve `x^(m)y^(n)=a^(m+n)` varies asA. `("abscissae")^(2)`B. `("ordinate")^(2)`C. abscissaD. ordinate |
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Answer» Correct Answer - C We have, `x^(m)y^(n) =a^(m+n)` `rArr m log_(e)x + nlog_(e)y=(m+n) log_(e)a` `rArr (m)/(2) + (n)/(y)(dy)/(dx)=0 " " ` [Differentiating w.r.t. x] `rArr (dy)/(dx) = -(my)/(nx)` `therefore " Length of the subtangent"=|(y)/((dy)/(dx))|=|y xx (nx)/(my)|=(n)/(m) |x| prop x` |
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| 14. |
At any point of a curve `sqrt(("subnormal")/("subtangent")) ` is equal toA. the abscissa of that pointB. the ordinate of that paintC. slope of the tangent at that pointD. slope of the normal at that point |
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Answer» Correct Answer - C We have, Sub normal `=y (dy)/(dx), " Sub-tangent" = (y)/((dy)/(dx))` `rArr sqrt(("sub normal")/("sub tangent"))=sqrt((y(dy)/(dx))/((y)/((dy)/(dx))))=(dy)/(dx) =` Slope of the tangent |
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| 15. |
The normal to the curve `5x^(5)-10x^(3)+x+2y+6 =0` at P(0, -3) meets the curve again at the pointA. (-1, 1), (1, 5)B. (1, -1), (-1, -5)C. (-1, -5),(-1, 1)D. (-1, 5),(1, -1) |
| Answer» Correct Answer - B | |
| 16. |
If the line `a x+b y+c=0`is a normal to the curve `x y=1,`then`a >0,b >0``a >0,b |
| Answer» Correct Answer - B | |
| 17. |
If the line joining the points `(0,3)a n d(5,-2)`is a tangent to the curve `y=C/(x+1)`, then the value of `c`is1 (b) `-2`(c) 4(d) none of theseA. 1B. -2C. 4D. none of these |
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Answer» Correct Answer - C The equation of the line joining (0,3) and (5,-2) is ` x+y-3=0 " "…(i) ` Suppose this line touches the curve `y=(C )/(x+1) " at " (alpha, beta ).` Then, ` ((dy)/(dx))_((alpha"," beta)) =` Slope of the line (i) ` rArr -(C)/((alpha+1)^(2))=-1 " "[ because (dy)/(dx)=- (C)/((x+1)^(2))]` ` rArr C=(alpha+1)^(2) " "...(ii) ` Also, ` (alpha,beta)` lies on the line (i) and the given curve. Therefore, ` alpha+ beta -3=0 " "...(iii) ` and, ` beta=(C)/(alpha+1) " "...(iv) ` Form (ii) and (iv), we get ` beta=alpha+1 " "...(v) ` Solving (iii) and (v), we get ` 2 alpha +1-3 =0 rArr alpha =1 ` Substituting ` alpha=1 ` in (ii), we get C = 4 |
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| 18. |
The slope of the tangent of the curve `y=int_0^x (dx)/(1+x^3)` at the point where `x = 1` isA. ` (1)/(2)`B. 1C. `(1)/(4)`D. non-existent |
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Answer» Correct Answer - A We have, ` y=int_(0)^(x)(1)/(1+t^(3))dt rArr (dy)/(dx)=(1)/(1+x^(3)) rArr ((dy)/(dx))_(x=1)=(1)/(2) ` |
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| 19. |
If m is the slope of the tangent to the curve ` e^(y)=1+x^(2)` , thenA. `|m| gt 1 `B. `m lt 1 `C. `|m| lt 1 `D. `|m| le 1 ` |
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Answer» Correct Answer - D We have, ` e^(y)=1+x^(2) ` ` rArr e^(y) (dy)/(dx)=2x " " [" Differentiating w.r.t. x" ] ` ` rArr (1+x^(2))(dy)/(dx)=2x " " [ because e^(y)=1+x^(2)] ` ` rArr (dy)/(dx) = (2x)/(1+x^(2)) ` ` rArr |m|= (2|x|)/(1+|x|^(2)) ` Now, A.M.` ge ` G.M. ` rArr (1+|x|^(2))/(2) ge sqrt(1xx |x|^(2)) ` ` rArr (1+|x|^(2))/(2) ge |x| ` ` rArr 1+|x|^(2) ge 2|x| ` ` rArr 1 ge (2|x|)/(1+|x|^(2)) rArr 1 ge |m| rArr |m| le 1 ` |
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| 20. |
For the curve `y=ce^(x//a) `, which one of the following is incorrect?A. subtangent is constantB. subnormal varies as the square of the ordinateC. tangent at `(x_(1),y_(1))` on the curve intersects the x-axis at a distance of ` (x_(1)-a) ` from the origin.D. equation of normal at the point where the curve cuts y-axis is ` cy+ax=c ` |
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Answer» Correct Answer - D We have, ` y=ce^(x//a) rArr (dy)/(dx)=(c)/(a)e^(x//a) rArr (dy)/(dx)=(1)/(a)y ` ` therefore (y)/(dy//dx)=a=" Const. " rArr " Subagent " = " Const. " ` Length of the subnormal `= y(dy)/(dx) =y xx (y)/(a) =(y^(2))/(a) prop` (Square of the ordinate) Equation of the tangent at ` (x_(1),y_(1)) ` is ` y-y_(1)=(y_(1))/(a)(x-x_(1)) ` This meets x- axis at a point given by `-y=(y_(1))/(a)(x-x_(1)) rArr x=x_(1)-a. ` The curve meets y- axis at (0,c) ` therefore ((dy)/(dx))_((0","c))=c//a ` So, equation of the normal at (0,c) is ` y-c=-(1)/(c//a)(x-0) rArr ax+cy=c^(2) ` |
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| 21. |
Prove that the curve `y=e^(|x|)` cannot have a unique tangent line at the point `x = 0.` Find the angle between the one-sided tangents to the curve at the point `x = 0.`A. `(pi)/(4) `B. `(pi)/(6) `C. `(pi)/(2) `D. `(pi)/(3) ` |
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Answer» Correct Answer - C We have, `y=e^(|x|)={(e^(-x)"," , x lt0),(e^(x) "," , x ge 0):} rArr (dy)/(dx)={(-e^(-x)"," , x lt0),(e^(x) "," , x gt 0):} ` The slopes of the tangents at x = 0 to the curves ` y=e^(-x), x lt 0 ` and `y= e^(x), x ge 0 " are " m_(1)=-1 " and " m_(2)=1 ` respectively. Clearly , ` m_(1)m_(2)=-1 ` Hence , the required angle is a right angle. |
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| 22. |
The curve `y=a x^3+b x^2+c x+5`touches the x-axis at `P(-2,0)`and cuts the y-axis at the point `Q`where its gradient is 3. Find the equation of the curve completely.A. ` a=(1)/(2),b=-(3)/(4), c=3 `B. ` a=-(1)/(2),b=-(3)/(4), c=3 `C. ` a=(1)/(2),b=(3)/(4), c=3 `D. none of these |
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Answer» Correct Answer - B We have, ` y=ax^(3)+bx^(2)+cx +5 " "...(i) ` ` rArr (dy)/(dx)=3ax^(2) + 2bx +c " "...(ii) ` Since the curve (i) touches x-axis at A(-2,0). Therefore, point A lies on (i) and `((dy)/(dx))_(A)=0.` ` therefore -8a+4b-2c+5=0 " "...(iii) ` and, ` 12a-4b+c=0 " "...(iv) ` The curve (i) cuts y-axis at B(0,5). It is given that ` ((dy)/(dx))_(B)=3 rArr c=3 ` Putting c = 3 in (iii) and (iv) and solving them, we get ` a=-(1)/(2) " and " b=-(3)/(4) ` |
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| 23. |
A curve with equation of the form `y=a x^4+b x^3+c x+d`has zero gradient at the point `(0,1)`and also touches the `x-`axis at the point `(-1,0)`then the value of `x`for which the curve has a negative gradient are:`xgeq-1`b. `x |
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Answer» Correct Answer - C We have , `(dy)/(dx)=0 ` at (0,1) and (-1,0) ` therefore c=0 " and " -4a+3b -c=0 rArr a=(3b)/(4) " and " c=0 " "...(i)` Also, the curve passes through (0,1) and (-1,0) ` therefore d=1 and 0=a-b-c+d rArr a-b-c+1 =0 " "...(ii)` From (i) and (ii), we get `a=3, b=4,c=0 and d=1 ` ` therefore y=3x^(4)+4x^(3)+1 rArr (dy)/(dx)=12x^(3)+12x^(2) ` Now, `(dy)/(dx) lt 0 rArr 12x^(3)+12x^(2) lt 0 rArr x+1 lt 0 rArr x lt -1 ` |
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| 24. |
The normal to the curve `x=a ( cos theta-thetasintheta), y = a( sintheta-theta costheta)` at any point , `theta` , is such thatA. makes a constant angle with x-axisB. is at a constant distance from the originC. passes through the originD. satisfies all the three conditions |
| Answer» Correct Answer - B | |
| 25. |
The tangents to the curve `x=a(theta - sin theta), y=a(1+cos theta)` at the points `theta = (2k+1)pi, k in Z` are parallel to :A. `y=x`B. `y= -x`C. `y=0`D. `x=0` |
| Answer» Correct Answer - C | |
| 26. |
Find the equation of the normal to the curve `x^2+2y^2-4x-6y+8=0`at the point whose abscissa is 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^2+2y^2-4x-6y+8 = 0` Differentiating it w.r.t. `x`, `=>2x+4ydy/dx - 4 -6dy/dx = 0` `=>dy/dx = (4-2x)/(4y-6)` `:. m = (4-2x)/(4y-6)`. Now, we have to find tyhe equation of the normal at a point whose abscissa is `2`. `:. x = 2` `=> m = (4-2(2))/(4y-6) = 0` `:.` Slope of tangent is `0`. `:. n = -1/0` Now, putting `x = 2`, in the equation of the curve. `=>4+2y^2-8-6y+8 = 0` `=>2y^2-6y+4 = 0` `=>2y^2-4y-2y+4 = 0` `=>(2y-2)(y-2) = 0` `=>y = 1 and y = 2` So, we will find equation of normal at points `(2,1)` and `(2,2)`. `:.` Equation of normal to the curve at `(2,1)`. `=>y-1 = -1/0(x-2) => x = 2` Equation of normal to the curve at `(2,2)`. `=>y-2 = -1/0(x-2) => x = 2` `:. x = 2` is the required equation. |
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| 27. |
For the curve `x = t^2 - 1, y = t^2 - t,` the tangent line is perpendicular to `x`-axis, then `t =`(i)`0`(ii)`infty`(iii) `1/(sqrt3)`(iv) `-1/(sqrt3)`A. `t=0`B. `t=oo`C. `t=1//sqrt(3)`D. `t=-1//sqrt(3)` |
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Answer» Correct Answer - A we have, `x=t^(2)-1, y=t^(2)-t` `therefore (dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(2t-1)/(2t)` If the tangent is perpendicular to x-axis, then `(dx)/(dy)=0 rArr (2t)/(2t-1)=0rArr t=0` |
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| 28. |
Find the value of `n in N`such that the curve `(x/a)^n+(y/b)^n=2`touches the straight line `x/a+y/b=2`at the point `(a , b)dot`A. `(b,a)`B. `(a,b)`C. `(1,1)`D. ` ((1)/(b),(1)/(a)) ` |
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Answer» Correct Answer - B We have, `(x^(n))/(a^(n))+(y^(n))/(b^(n)) =2 " "...(i) ` Differentiating with respect to x, we get ` n(x^(n-1))/(a^(n)) +n(y^(n-1))/(b^(n))(dy)/(dx)=0 rArr (dy)/(dx)=-(b^(n))/(a^(n))((x)/(y))^(n-1) ` Suppose the line ` (x)/(a)+(y)/(b)=1 ` touches the curve (i) at `P (alpha ,beta).` The equation of the tangent at `P (alpha, beta )` is ` y- beta =-(b^(n))/(a^(n))((alpha)/(beta))^(n-1) (x-alpha)` ` rArr a^(n) beta^(n-1)y-a^(n)beta^(n)=-b^(n)alpha^(n-1) x+b^(n)alpha^(n)` ` rArr (a^(n)alpha^(n-1))x+(a^(n)beta^(n-1))y=a^(n)beta^(n)+b^(n)alpha^(n) ` ` rArr (a^(n)alpha^(n-1))x+(a^(n)beta^(n-1))y=2a^(n)b^(n) " "[because (alpha,beta) " lies on (i)"] ` Comparing this equation with ` (x)/(a)+(y)/(b)=2 ` , we get ` (ab^(n)alpha^(n-1))/(a^(n)b^(n)) =(ba^(n)beta^(n-1))/(a^(n)b^(n))=(2)/(2) ` ` rArr ((alpha)/(a))^(n-1)=((beta)/(b))^(n-1)=1 ` ` rArr (alpha)/(a)=(beta)/(b)=1 rArr alpha=a,beta =b. ` |
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| 29. |
Prove that `(x/a)^n+(y/b)^n=2`touches the straightline `x/a+y/b=2`for all ` in N`, at the point `(a , b)`. |
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Answer» `n(x/a)^(n-1)*1/a+n(y/b)^(n-1)*1/b*dy/dx=0` `dy/dx=(-n(x/a)^(n-1)*(1/a))/(n(y/b)^(n-1)*(1/b)` `=(-b(x/a)^(n-1))/(a(y/b)^(n-1))` `-b/a((x/a)^(n-1)/(y/b)^(n-1))=-b/a` `y-b=-b/a(x-a)` `y/b-1=-x/a+1` `x/a+y/b=2`. |
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| 30. |
Find the equation of the tangent to the curve `x=sin3t ,y=cos2t`at `t=pi/4dot` |
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Answer» `x_1=sin(3/4pi),x_1=1/G` `y_1=cos(pi/2)=y_1=0` `dy/dx=(dy/dt)/(dx/dx)=(-2sin2t)/(3cos3t)=(-2sin(3/4pi)/(3cos(3/4pi)))` `=(-2)/(3*(1/sqrt2))` `m=(2sqrt2)/3` `y-0=(2sqrt2)/3(x-1/sqrt2)` `y=(2sqrt2x)/3-2/3` `(2sqrt2)/3x-y=2/3`. |
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| 31. |
Find the point on the curve `y=x^2`where the slope of the tangent is equal to the `x-`coordinate of the point. |
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Answer» Let the point be`(x_1,y_1)` `y=x^2` diff. with respect to x `dy/dx=2x_1` `x_1=0,y_1=0`. |
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| 32. |
If the curves `y= a^(x) and y=e^(x)` intersect at and angle `alpha, " then " tan alpha` equalsA. `|(log_(e)a)/(1+log_(e)a)|`B. `|(1+log_(e)a)/(1+log_(e)a)|`C. `|(log_(e)a-1)/(log_(e)a+1)|`D. none of these |
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Answer» Correct Answer - C The equations of the two curves are ` C_(1): y=a^(x) " " …(i) and C_(2) : y=e^(x) " " ` …(ii) At the point of intersection of these two curves, we must have `a^(x)=e^(x) rArr ((a)/(e))^(x)=1 rArr x =0` Putting `x=0` in any one of the two curves, we get y = 1. Thus, the two curves intersect at P(0, 1). Clearly, at the point P(0, 1) `((dy)/(dx))_(C_(1))=log_(e) a and ((dy)/(dx))_(C_(2))=1` `therefore tan alpha = |(((dy)/(dx))_(C_(1))-((dy)/(dx))_(C_(2)))/(1+((dy)/(dx))_(C_(1))xx((dy)/(dx))_(C_(2)))|rArr tan alpha =|(log_(e)a -1)/(log_(e)a+1)|` |
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| 33. |
The normal to the curve `y(x-2)(x-3)=x+6`at the point where the curve intersects the `y-a xi s ,`passes through the point :`(1/2,-1/3)`(2) `(1/2,1/3)`(3) `(-1/2,-1/2)`(4) `((1/(2,1))/2)`A. `(-(1)/(2),-(1)/(2))`B. `((1)/(2),(1)/(2))`C. `((1)/(2),-(1)/(3))`D. `((1)/(2),(1)/(3))` |
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Answer» Correct Answer - B The equation of the curve is `y(x-2)(x-3)=x+6 " " ` …(i) It intersects y-axis at x = 0. Putting x = 0 in (i) , we obtain y = 1. So, the point where (i) cuts y-axis has the coordinates (0, 1). Form (i), we obtain `y=(x+6)/(x^(2)-5x+6)` Differentiating with respect to x, we obtain `(dy)/(dx) = ((x^(2)-5x+6)-(x+6)(2x-5))/((x^(2)-5x+6)^(2))` `therefore ((dy)/(dx))_((0","1))=(6-(-30))/(36) =1` The equation of the normal at (0, 1) is `y-1= -1(x-0) or, x+y-1=0` Clearly, it passes through `((1)/(2), (1)/(2))`. |
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| 34. |
Consider `f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot`A normal to `y=f(x)`at `x=pi/6`also passes through the point:(1) (0, 0)(2) `(0,(2pi)/3)`(3) `(pi/6,0)`(4) `(pi/4,0)`A. (0, 0)B. `(0, (2pi)/(3))`C. `((pi)/(6),0)`D. `((pi)/(4),0)` |
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Answer» Correct Answer - B We have, `y ="tan" ^(-1)sqrt((1+sinx)/(1-sinx))= tan^(-1)(("cos"(x)/(2) + " sin" (x)/(2))/("cos"(x)/(2) - " sin" (x)/(2)))` ` rArr y=tan^(-1) (tan((x)/(4)+(x)/(2)))` ` rArr y= (pi)/(4)+ (x)/(2)` ` rArr (dy)/(dx) = (1)/(2) rArr (1)/(dy//dx)= -2.` When `x=(pi)/(6),y=(pi)/(4)+(pi)/(12)=(pi)/(3).` The equation of the normal at `(pi//6, pi//3)` is ` y-(pi)/(3)= -2(x-(pi)/(6)) or , 2x+y-(2pi)/(3) =0` Clearly, it passes through `(0, (2pi)/(3)).` |
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| 35. |
The tangent to a given cuve is perpendicualr to x-axis, ifA. `(dy)/(dx)=0`B. `(dy)/(dx)=1`C. `(dx)/(dy)=0`D. `(dx)/(dy)=1` |
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Answer» Correct Answer - C If the tangent is perpendicular to x-axis, then `(dy)/(dx) to pm oo rArr (dy)/(dx) =0` |
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| 36. |
The abscissa of the point on the curve `ay^2 = x^3`, the normal at which cuts off equal intercepts from the coordinate axes isA. `(2a)/(9)`B. `(4a)/(9)`C. `-(4a)/(9)`D. `-(2a)/(9)` |
| Answer» Correct Answer - B | |
| 37. |
The abscissa of the point on the curve `ay^2 = x^3`, the normal at which cuts off equal intercepts from the coordinate axes isA. `2a//9`B. `4a//9`C. `-4a//9`D. `-2a//9` |
| Answer» Correct Answer - B | |
| 38. |
The abscissa of a point on the curve `x y=(a+x)^2,`the normal which cuts off numerically equal intercepts from thecoordinate axes, is`-1/(sqrt(2))`(b) `sqrt(2)a`(c) `a/(sqrt(2))`(d) `-sqrt(2)a`A. `(a)/(sqrt(2))`B. aC. `sqrt(2)a `D. `-(a)/(sqrt(2))` |
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Answer» Correct Answer - A::D We have, `xy=(a+x)^(2) " "...(i)` ` rArr y=x+2a+(a^(2))/(x) rArr (dy)/(dx)=1 - (a^(2))/(x^(2)) ` Let `P(x_(1),y_(1)) ` be a point on the curve (i), where the normal cuts off numerically equal intercepts from the coordinate axes. Then, ` (-1)/(((dy)/(dx))_(P))=pm 1 rArr ((dy)/(dx))_(P)=pm 1 rArr 1-(a^(2))/(x_(1)^(2))=pm 1 ` ` rArr 1- (a^(2))/(x_(1)^(2))=1 " or, " 1-(a^(2))/(x_(1)^(2))=-1 ` ` rArr (a^(2))/(x_(1)^(2))=0 " or, " x_(1)=pm (a)/(sqrt(2)) rArr x_(1) = pm (a)/(sqrt(2))` |
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| 39. |
If the sum of thesquares of the intercepts on the axes cut off by tangent to the curve `x^(1/3)+y^(1/3)=a^(1/3), a >0`at `(a/8, a/8)`is 2, then `a=`1 (b) 2(c) 4 (d) 8A. 1B. 2C. 4D. 8 |
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Answer» Correct Answer - C We have, ` x^(1//3) + y^(1//3)= a^(1//3) ` ` rArr (1)/(3)x^(2//3)+(1)/(3)y^(-2//3)(dy)/(dx)=0 ` ` rArr (dy)/(dx)=-(x^(-2//3))/(y^(-2//3))= -(x^(2//3))/(y^(2//3)) rArr ((dy)/(dx))_((a//8 "," a//8))=-1 ` The equation of the tangent at `(a//8 ,a//8)` is given by ` y-(a)/(8)=-1(x-(a)/(8)) rArr x+y-(a)/(4)=0 ` The x and y intercepts of this line on the coordinate axes are each equal to ` a//4. ` So, we have `((a)/(4))^(2)+((a)/(4))^(2)=2 rArr a=4 ` |
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| 40. |
The slope of the tangent to the curve `y=sin^(-1) (sin x) " at " x=(3pi)/(4)` isA. 1B. -1C. 0D. non-existent |
| Answer» Correct Answer - B | |
| 41. |
The equation of the tangent to the curve `y=e^(-|x|)` at the point where the curve cuts the line x = 1, isA. `x+y=e`B. `e(x+y)=1`C. `y+ex=1`D. none of these |
| Answer» Correct Answer - D | |
| 42. |
If the equation of the tangent to the curve `y^2=a x^3+b`at point `(2,3)i sy=4x-5`, then find the values of `aa n db`.A. `a=2, b=7`B. `a=7, b=2`C. `a=2, b= -7`D. `a= -2,b=7` |
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Answer» Correct Answer - C The equation of the curve is `y^(2)=ax^(3)+b.` `therefore 2y (dy)/(dx)=3ax^(2) rArr (dy)/(dx) = (3ax^(2))/(2y) rArr ((dy)/(dx))_((2","3))=2a` The equation of the tangent at (2, 3) is `y-3=2a(x-3) or, 2ax-y-6a+3=0` But, the equation of the tangent at (2, 3) is `y=4x-5.` Comparing these two equations, we get `2a=4 and -6a+3= -5 rArr a=2` As (2, 3) lies on the curve `y^(2)=ax^(3)+b.` `therefore 9=8a+b rArr 9=16+b rArr b = -7` |
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| 43. |
The equation of the normal to the curve `y=x(2-x)` at the point (2, 0) isA. `x-2y=2`B. `x-2y+2=0`C. `2x+y=4`D. `2x+y-4=0` |
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Answer» Correct Answer - A The equation of the curve is `y=x(2-x) or, y=2x-x^(2)` `rArr (dy)/(dx)=2-2x rArr ((dy)/(dx))_((2","0)) = 2-2 xx2 = -2` So, the equation of the normal at (2, 0) is `y-0= -(1)/(-2) (x-2) or, 2y=x-2` |
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| 44. |
The point on the curve `y^(2)=x` where tangent makes `45^(@)` angle with x-axis, isA. `(1//2,1//4)`B. `(1//4,1//2)`C. `(4,2)`D. `(1,1)` |
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Answer» Correct Answer - B Let `(x_(1),y_(1) )` be the required point on the curve `y^(2)=x` Then, `((dy)/(dx))_((x_(1)","y_(1)))=1` `rArr (1)/(2y_(1))=1 " "[because y^(2)=x rArr2y(dy)/(dx)=1 rArr(dy)/(dx)=(1)/(2y)]` `rArr y_(1)=(1)/(2)` Since `(x_(1),y_(1))` lies on `y^(2)=x.` `therefore x_(1)=y_(1)^(2)rArr x_(1)=(1)/(4)` Hence, `(1//4,1//2)` is the required point. |
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| 45. |
If at each point of the curve `y=x^3-a x^2+x+1,`the tangent is inclined at an acute angle with the positive directionof the x-axis, then`a >0`(b) `a |
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Answer» Correct Answer - C We have, ` y= x^(3)-ax^(2)+x+1 " "...(i) ` ` rArr (dy)/(dx) =3x^(2)-2ax+1 ` It is given that at each point on the curve (i), the tangent is inclined at an acute angle with the positive direction of x-axis. ` therefore (dy)/(dx) ge 0 " for all "(x,y)` lying on the curve (i). ` rArr 3x^(2)-2ax+1 ge 0 " for all " x` ` rArr 4a^(2)-12 le 0 rArr a^(2)-3 le 0 rArr - sqrt(3) le a le sqrt(3) rArr |a| le sqrt(3) ` |
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| 46. |
The equation of the normal to the curve `y = sin x` at `(0,0)` isA. `x=0`B. `y=0`C. `x+y=0`D. `x-y=0` |
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Answer» Correct Answer - C We have, `y=sinx rArr (dy)/(dx) = cosx rArr ((dy)/(dx))_((0","0))=cos0=1` So, the equation of the normal at (0, 0), is `y-0=(1)/(1)(x-0) rArr y+x=0` |
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| 47. |
The length of normal at any point to the curve, `y=c cosh(x/c)` isA. `("abscissa")^(2)`B. `("ordinate")^(2)`C. abscissaD. ordinate |
| Answer» Correct Answer - B | |
| 48. |
The angle of intersection of the curves `y=x^(2), 6y=7-x^(3)` at (1, 1), isA. `pi//4`B. `pi//3`C. `pi//2`D. none of these |
| Answer» Correct Answer - C | |
| 49. |
The point at which the tangent to the curve `y=x^(2)-4x` is parallel to x-axis, isA. (0, 4)B. (-2, 4)C. (2,4)D. (2, -4) |
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Answer» Correct Answer - D Let `(x_(1),y_(1))` be the required point. Then, `((dy)/(dx))_((x_(1)","y_(1)))=0rArr 2x_(1)-4=0 rArrx_(1)=2` Since `(x_(1),y_(1))` lies on `y=x^(2)-4x.` `therefore y_(1)=x_(1)-4x_(1) rArr y_(1)=4-8= -4` Hence, the coordinates of the point are (2, -4) |
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| 50. |
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`A. `pm 1`B. `pm 2`C. `pm 3`D. `pm 4` |
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Answer» Correct Answer - A We have, `y=int_(0)^(x)|t|dt, x in R " "...(i)` ` therefore (dy)/(dx)=|x|, x in R ` Let `(x_(1),y_(1))` be a point on the curve `y =int_(0)^(x)|t|dt ` where tangent is parallel to the line `y=2x.` Then, `((dy)/(dx))_((x_(1)","y_(1)))=2 rArr |x_(1)|=2 rArr x_(1)=pm 2 ` Since`(x_(1),y_(1))` lies on (i). Therefore, `y_(1)= int_(0)^(x_(1))|t|d ` When `x=2, y_(1)=int_(0)^(2)|t|dt=int_(0)^(2)t dt =[(t^(2))/(2)]_(0)^(2)=2` When `x=-2, y_(1)=int_(0)^(-2)-tdt=int_(0)^(-2)-t dt =[-(t^(2))/(2)]_(0)^(-2)=-2 ` So, points on the curve are (2,2) and (-2,-2). The equations of tangents at these points are ` y-2=2(x-2) " and " y+2=2(x+2) ` or, ` y=2x-2 ` and `y=2x+2 ` The x-intercepts of these tangents are ` pm 1. ` |
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