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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.
| 351. |
(a) Find the intervals in which the function `f(x)=log(1+x)-x/(1+x)` is (i) increasing, (ii) decreasing function. (b) Find the intervals in which the function `f(x)= x/(log_(e) x),x gt 0` is increasing or decreasing. |
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Answer» Correct Answer - `(a)(i)[0,infty[" "(ii)]-infty,0]-{-1}` (b) Decreases in `]0,e]-{1}" and inceases in "[e, infty[` |
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| 352. |
Using differentials, find the approximate values of the following: `sqrt26` |
| Answer» Correct Answer - `5.1` | |
| 353. |
Find the intervals in which the function `f(x) = sin x +cos x,x in [0, 2pi]` is (i) strictly increasing, (ii) strictly decreasing. |
| Answer» Correct Answer - (i)`(0,pi/4)cup ((5pi)/(4)2pi)" "(ii)(pi/4,(5pi)/4)` | |
| 354. |
Find the intervals in which the function f given by `f(x)=2x^3-3x^2-36 x+7`is (a) strictly increasing (b) strictly decreasing |
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Answer» Correct Answer - (a) `f(x)` is strictly increas on `[-oo, -2] uu 3, oo]` (b) `f(x)` is strictly decreasing on `[-2, 3]` |
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| 355. |
Find theintervals in which the function `"f "("x")=" x"^3-" "12"x"^2+" "36"x"+17" "`is (a) increasing, (b) decreasing. |
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Answer» Correct Answer - (a) f(x) is increasing on `[-oo, -2] uu [6, oo]` (b) f(x) is decreasing on `[2, 6]` |
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| 356. |
Find the intervals on which each of the following functions is (a) increasing (b) decreasing `f(x) = (x^(3) - 6x^(2) + 9x + 10)` |
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Answer» Correct Answer - (a) f(x) is increasing on `[-oo, 1] uu [3, oo]` (b) f(x) is decreasing on `[-1, 3]` |
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| 357. |
If `f(x)=x^(3)-9x^(2)+24x,` then fA. has minima at x=4B. has maxima at x=4C. has maximum value =16D. has minimum value =-16 |
| Answer» Correct Answer - A | |
| 358. |
Find the intervals on which each of the following functions is (a) increasing (b) decreasing `f(x) = 2x^(3) - 24x + 5` |
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Answer» Correct Answer - (a) `f(x)` is increasing on `[-oo, -2] uu [1, oo]` (b) `f(x)` is decreasing on `[-2, 2]` |
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| 359. |
If `f(x)=x^(3)-9x^(2)+24x,` then fA. has maximum value =16B. has minimum value =10C. has minima at x=2D. has maxima at x=2 |
| Answer» Correct Answer - D | |
| 360. |
If `f(x)=x^(3)-9x^(2)+24x,` then fA. has maximum value =36B. has minimum value =16C. has minima at x=2D. has maxima at x=4 |
| Answer» Correct Answer - B | |
| 361. |
If a rectangle of maximum area is inscibed in the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, then the dimensions areA. `asqrt(2), bsqrt(2)`B. `(a)/(sqrt(2)), (b)/(sqrt(2))`C. `(1)/(2sqrt(a)), (1)/(2sqrt(b))`D. `(1)/(sqrt(2a)), (1)/(sqrt(2b))` |
| Answer» Correct Answer - A | |
| 362. |
Examine the function `f(x)=x^(3)-9x^(2)+24x` for maxima and minima.A. has maximum value =36B. has minimum value =16C. has minima at x=2D. has maxima at x=4 |
| Answer» Correct Answer - C | |
| 363. |
A beam of length `l`is supported at oneend. If `W`is the uniform load perunit length, the bending moment `M`at a distance `x`from the end is givenby `M=1/2l x-1/2W x^2dot`Find the point on thebeam at which the bending moment has the maximum value. |
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Answer» `M = ((lx)/(2) - (Wx^(2))/(2)) rArr (dM)/(dx) = ((l)/(2) - Wx) and (d^(2)M)/(dx^(2)) = -W` For a maxima or minima, we have `(dM)/(dx) = 0` Now, `(dM)/(dx) = 0 rArr (l)/(2) - Wx = 0 rArr x = (l)/(2W)` Also, `(d^(2)M)/(dx^(2)) = - W lt 0` for all values of x `:. x = (l)/(2W)` is a point of maxima So, the required point is at a distance of `(l//2W)` from the supporting end. |
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| 364. |
A metal wire of `36` cm long is bent to form a rectangle. Find its dimensions when its area is maximum.A. 10 cm, 8 cmB. 6 cm, 12 cmC. 14 cm, 4 cmD. 9 cm, 9 cm |
| Answer» Correct Answer - D | |
| 365. |
Examine the following functions for maxima and minima : `f(x)=3x^(3)-9x^(2)-27x+15`A. f has maximum value 66B. f has minimum value 30C. f has maxima at x=3D. f has minima at x=3 |
| Answer» Correct Answer - D | |
| 366. |
If `f(x)=3x^(3)-9x^(2)-27x+15`,t hen the maximum value of `f(x)` isA. f has maximum value 30B. f has minimum value 30C. f has maxima at x=3D. f has minima at x =-1 |
| Answer» Correct Answer - A | |
| 367. |
A wire of length 25m is to be cut into two pieces. One of the wires is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum ? |
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Answer» Let x meters and `(25 -x)` metres be the required lengths. Let a be the side of the square formed and r be the radius of the circle formed. Then, `4a = x and 2pi r = 25 - x or a = (x)/(4) and r = (25 - x)/(2pi)` `:.` area of the square `= ((x^(2))/(16))` sq metres. Area of the circle `= pi ((25 - x)/(2pi))^(2) = ((25 -x)^(2))/(4pi)` Now, the combined area `A = (x^(2))/(16) + ((25 - x)^(2))/(4pi)` `:. (dA)/(dx) = (x)/(8) - ((25 -x))/(2pi) = ((pi + 4) x - 100)/((8pi) , and (d^(2)A)/(dx^(2)) = ((pi + 4))/(8pi)` For a maxima or minima, we have `(dA)/(dx) = 0` Now, `(dA)/(dx) = 0 rArr (pi + 4)x - 100 = 0 rArr x = (100)/((pi + 4))` And, `(d^(2)A)/(dx^(2)) = ((pi + 4))/(8pi) gt 0` for all values of x `:. x = (100)/((pi + 4))` is a point of minimum. `:.` lengths of the pieces are `((100)/(pi + 4))` metres and `((25pi)/(pi + 4))` metres. |
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| 368. |
A wire of length l is cut into two parts. One part is bent into a circle of radius r and other part into a square of side x. The sum of areas of circle and square is least, ifA. `r=x`B. `r=3x`C. `r=(x)/(3)`D. `r=(x)/(2)` |
| Answer» Correct Answer - D | |
| 369. |
A wire of length 2 units is cut into two parts which are bentrespectively to form a square of `s i d e=x`units and a circle of `r a d i u s=r`units. If the sum of the areas of the square andthe circle so formed is minimum, then :(1) `2x=(pi+4)r`(2) `(pi+4)x=pir`(3) `x=2r`(4) `2x=r`A. `2x=(pi+4)r`B. `(4-pi)x=pi r`C. `x=2r`D. `2x =r` |
| Answer» Correct Answer - C | |
| 370. |
If `y = e^(sin sqrtx) " then " (dy)/(dx) =` ?A. `e^(sin sqrtx).cos sqrtx`B. `(e^(sin sqrtx) cos sqrtx)/(2sqrtx)`C. `(e^(sinsqrtx))/(2 sqrtx)`D. none of these |
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Answer» Correct Answer - B `log y = sin sqrtx` |
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| 371. |
Find the equation of the normal to the curve `y = (sin 2x + cot x + 2)^(2) " at " x = (pi)/(2)` |
| Answer» Correct Answer - `24y - 2x + pi - 96 = 0` | |
| 372. |
A wire of length 28 m is to becut into two pieces. One of the pieces is to be made into a square and theother into a circle. What should be the length of the two pieces so that thecombined area of the square and the circle is minimum? |
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Answer» Let r be the radius of the circle and x be the side of square. `:. ` Circumference of circle + perimeter of square = 28 cm ` rArr 2 pi r + 4x =28` `rArr x = (14 - pi r)/2 ` ….(1) and let area of circle + area of square = A `:. A = pi r^(2) + x^(2)` ` = pi r^(2) + ((14 - pi r)/2)^(2)` [From equation (1)] ` = pi r^(2)+ 1/4 (pi^(2)r^(2)- 28 pi r+ 196)` ` rArr (dA)/(dr) = 2 pir+1/4(2pi^(2)r-28pi)` and ` (d^(2)A)/(dr^(2) = 2 pi + 1/4 (2pi^(2))` For maxima/minima ` (dA)/(dr) = 0` ` rArr pir+1/4 (2pi^(2)r-28pi)=0` ` rArr pir+pi^(2)r-14pi =0` ` rArr r(4+pi)=14` ` rArr r=14/(pi+4)` at ` r=14/(pi+4)` ` (d^(2)A)/(pi r^(2)) gt 0` ` rArr ` A is minimum. Now ` 2pi r = 2pi((14)/(pi+4)) = (28pi)/(pi+4)` and ` 4x = 28 - 2pi r ` ` = 28 - (28pi)/(pi+4) = (112)/(pi+4)` Therefore, the length of two pieces of wire ` = (28pi)/(pi+4) cm and (112)/(pi+4)` cm. |
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| 373. |
For all real values of x, the minimum value of `(1-x+x^2)/(1+x+x^2)`is(A) 0 (B) 1 (C) 3 (D) `1/3` |
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Answer» Correct Answer - D Let `y=(1-x+x^(2))/(1+x+x^(2))` Differentiate w.r.t. x ` (dy)/(dx) = ((1+x+x^(2))(-1+2x)-(1-x+x^(2))(1+2x))/((1+x+x^(2))` `-1+2x-x+2x^(2)-x^(2)+2x^(3)` ` = (-1-2x+x+2x^(2)-x^(2)-2x^(3))/((1+x+x^(2))^(2))` ` = (-2+2x^(2)-2x^(2)+2x^(2)0)/((1+x+x^(2))^(2))` ` = (2x^(2)-2)/((1+x+x^(2))^(2)) = (2(x^(2)-1))/((1+x+x^(2))^(2))` Now`(dy)/(dx) = 0 rArr x^(2)=1` ` rArr x = pm 1 ` Now, `(d^(2)y)/(dx^(2))=(2[(1+x+x^(2))^(2)(2x)-(x^(2)-1)(2)(1+x+x^(2))(1+2x)])/((1+x+x^(2))^(4))` `(4(1+x+x^(2))[(1+x+x^(2))x-(x^(2)-1)(1+2x)])/((1+x+x^(2))^(4))` `=(4(x+x^(2)+x^(3)-x^(2)-2x^(3)+1+2x))/((1+x+x^(2))^(3))` ` = (4(1+3x-x^(3)))/((1+x+x^(2))^(3))` at x = 1, `((d^(2)y)/(dx^(2)))_(x = 1) = (4[1+3(1)-1^(3)])/((1+1+1^(2))^(3))= (4(3))/3^(3)=4/9 gt 0` at x =- 1, `((d^(2)y)/(dx^(2)))_(x=-1)=(4[1+3(-1)-(-1)^(3)])/([1+(-1)+(-1)^(2)]^(3))` ` = (4(1-3+1))/((1-1+1)^(3))=4(-1)=-4 lt 0` `:. ` at x = 1, y will be minimum and minimum value ` y=(1-1+1)/(1+1+1)=1/3`. |
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| 374. |
Find a 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» `y=(x-2)^(2)`…(1) `rArr (dy)/(dx) = 2 (x-2)` Slope of chord ` = (4-0)/(4-2) = 2` According to the problem, for parallel lines, ` 2(x-2) = 2` ` rArr x= 3` From equation (1) ` y=(3-2)^(2) = 1` Therefore, required point is (3, 1). |
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| 375. |
The point on the curve `x^2=2y`which is nearest to the point (0, 5) is(A) `(2sqrt(2),4)` (B) `(2sqrt(2),0)` (C) (0, 0) (D) (2, 2)A. `(2sqrt2, 4)`B. `(2 sqrt2, 0)`C. ``(0, 0)`D. `(2, 2)` |
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Answer» Correct Answer - A Let d be the distance of point (x, y) on ` x^(2) = 2y` from the point(0, 5), `d = sqrt((x-0)^(2)+(y-5)^(2)) = sqrt(x^(2)+(y-5)^(2))` ` = sqrt(2y+(y-5)^(2))` ....(1) ` rArr d = sqrt(2y+y^(2)-10y+25)` ` = sqrt(y^(2)-8y+4^(2)+9)=sqrt((y-4)^(2)+9)` d will be minimum when `(y-4)^(2)=0 or y= 4` `:. y = 4 rArr x^(2)=2 xx 4` ` rArr x = pm sqrt8 = pm 2 sqrt2` `:." Points "(2sqrt2,4) and (-2sqrt2,4)` are the points on the curve, at minimum distance from the point (0, 5). |
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| 376. |
The equation of the tangent to the curve `x^2+2y=8` which is the perpendicular to `x-2y+1=0` is |
| Answer» Correct Answer - `2x + y - 6 = 0` | |
| 377. |
The function `x^(100)+sin x-1` is decreasing in the interval:A. `(0,pi/2)`B. `(0, 1)`C. `(pi/2, pi)`D. None of these |
| Answer» Correct Answer - D | |
| 378. |
Determine the points onthe curve `x^2=4y`which are nearest tothe point `(0, 5)`.A. `(2sqrt(3), pm 3)`B. `(pm 2sqrt(3),3)`C. `(-2sqrt(3)pm3)`D. `(pm sqrt(3),3)` |
| Answer» Correct Answer - B | |
| 379. |
Find the points on the curve `y = x^(3) - 3x`, where the tangent to the curve is parallel to the chord joining `(1, -2) and (2, 2)` |
| Answer» Correct Answer - `(sqrt((7)/(3), (-2)/(3) sqrt((7)/(3)))), (-sqrt((7)/(3)), (2)/(3) sqrt((7)/(3)))` | |
| 380. |
if the straight line `xcosalpha+y sin alpha=p` touches the curve `x^my^n=a^(m+n)` prove that `p^(m+n)m^mn^n=(m+n)^(m+n) a^(m+n) sin^nalpha cos^malpha` |
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Answer» Let the point of contact be `(x_(1), y_(1))` The equation of the given curve is `x^(m) y^(n) = a^(m + n)` This, on differentiation, gives `mx^(m-1) y^(n) + x^(m).ny^(n-1) .(dy)/(dx) = 0 rArr (dy)/(dx) = (-my)/(nx)` `:. ((dy)/(dx))_((x_(1)"," y_(1))) = (-my_(1))/(nx_(1))` So, the equation of the tangent at `(x_(1), y_(1)) " is " (y - y_(1))/(x - x_(1)) = (-my_(1))/(nx_(1))` i.e., `my_(1) x + nx_(1) y = (m + n) x_(1)y_(1)`....(i) This is identical with `c cos alpha + y sin alpha = p`...(ii) Comparing coefficients, we get `(my_(1))/(cos alpha) = (nx_(1))/(sin alpha) = ((m + n) x_(1)y_(1))/(p)` `:. x_(1) = (pm)/((m +n) cos alpha) and y_(1) = (pn)/((m + n) sin alpha)` Since `(x_(1)y_(1))` lies on the given curve, `x_(1)^(m) y_(1)^(n) = a^(m +n)` `:. {("pm")/((m + n) cos alpha)}^(m).{("pm")/((m +n) sin alpha)}^(n) = a^(m +n)` or `p^(m +n) .m^(m). n^(n) = (m +n)^(m + n) .a^(m +n) cos^(m) alpha sin^(n) alpha` |
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| 381. |
Curve `b^(2)x^(2) + a^(2)y^(2) = a^(2)b^(2) and m^(2)x^(2)- y^(2)l^(2) = l^(2)m^(2)` intersect each other at right-angle if:A. `a^(2)+b^(2)=l^(2)+m^(2)`B. `a^(2)-b^(2)=l^(2)-m^(2)`C. `a^(2)-b^(2)=l^(2)+m^(2)`D. `a^(2)+b^(2) = l^(2)-m^(2)` |
| Answer» Correct Answer - C | |
| 382. |
The equation of tangent to the curve `x^(2)+y^(2)=5,` where the tangent is parallel to the line 2x-y+1=0 areA. 2x-y+5=0, 2x-y-5=0B. 2x+y+5=0, 2x+y-5=0C. x-2y+5=0, x-2y-5=0D. x+2y+5=0, x=2y-5=0 |
| Answer» Correct Answer - A | |
| 383. |
The value of parameter t so that the line `(4-t)x+ty+(a^(3)-1)=0` is normal to the curve xy = 1 may lie in the intervalA. `(1,4)`B. `(-oo,0)uu(4,oo)`C. `(-4,4)`D. `[3,4]` |
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Answer» Correct Answer - B Slope of line `(4-t)x+ty+(a^(3)-1)=0" is "(t-4)/(t)` For `xy=1, (dy)/(dx)=(-y)/(x)=(-1)/(x^(2))` `therefore" Slope of normal "=-x^(2)=(t-4)/(t)` As `x^(2)gt0,(t-4)/(t)gt0` `therefore" "t in (-oo,0)uu(4,oo)` |
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| 384. |
Find the coordinates of the points on the curve `y=x^2+3x+4,`the tangents at which pass through the origin. |
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Answer» Let the required point be `(x_(1) , y_(1))` Now, `y = x^(2) + 3x + 4 rArr (dy)/(dx) = 2x + 3 rArr ((dy)/(dx))_((x_(1)","y_(1))) = (2x_(1) + 3)` So, the equation of the tangent at `(x_(1), y_(1))` is `(y - y_(1))/(x - x_(1)) = (2x_(1) + 3)` Since the tangent passes through the origin, we have `(-y_(1))/(-x_(1)) = (2x_(1) + 3)` `rArr y_(1) = (2x_(1)^(2) + 3x_(1))` ...(i) But, `(x_(1), y_(1))` lies on the given curve, So, `y_(1) = x_(1)^(2) + 3x_(1) + 4`...(ii) From (i) and (ii), we get `2x_(1)^(2) + 3x_(1) = x_(1)^(2) + 3x_(1) + 4 or x_(1)^(2) = 4 or x_(1) = +-2` Now, `x_(1) = 2 rArr y_(1) = (2^(2) + 3 xx 2 + 4) = 14` And, `x_(1) = -2 rArr y_(1) = [(-2)^(2) + 3 xx (-2) + 4] = 2` Hence, the required points are (2, 14) and `(-2, 1)` |
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| 385. |
Equation of tangent of the curve `y = 1 - e^(x//2)` at that point at which the curve crosses the y-axis, is :A. `x+y=1`B. `2x+y = 1`C. `x=-2y`D. None of these |
| Answer» Correct Answer - C | |
| 386. |
If the line y=4x-5 touches the curve `y^(2)=ax^(3)+b` at the point (2,3), then 7a+2b=0A. 1B. 2C. 0D. 3 |
| Answer» Correct Answer - C | |
| 387. |
If the line x+y=0 touches the curve `2y^(2)=ax^(2)+b` at (1,-1), thenA. a=2, b=0B. a=-2, b=4C. a=0, b=2D. a=4, b=-2 |
| Answer» Correct Answer - A | |
| 388. |
Let the equation of a curve be `x=a(theta+sin theta),y=a(1-cos theta)`. If `theta` changes at a constant rate k then the rate of change of the slope of the tangent to the curve at `theta=pi/3` is(a) `(2k)/sqrt3`(b) `k/sqrt3`(c) k(d) none of theseA. `2k//sqrt3`B. `k//sqrt3`C. kD. none of these |
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Answer» Correct Answer - D `(dy)/(dx)=((dy)/(d theta))/((dx)/(d theta))=(1)/(2)sec^(2).(theta)/(2).(d theta)/(dt)=(k)/(2)sec^(2).(theta)/(2)` `therefore" Required rate"=(k)/(2).sec^(2).(pi)/(6)=(k)/(2).((2)/(sqrt3))^(2)` |
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| 389. |
The normal at the point (1,1) on the curve `2y+x^2=3`is(A) `x + y = 0` (B) `x y = 0` (C) `x + y +1 = 0`(D) `x y = 0`A. x+y=0B. x-y= 0C. x+y+1=0D. x-y=1 |
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Answer» Correct Answer - B Given curve, ` 2y+x^(2)=3` ` rArr 2(dy)/(dx) + 2x - 0 rArr (dy)/(dx) - " "-x` `:. ` slope of tangent at point (1, 1) =- 1 ` rArr" slope of normal at "(1, 1) = (-1)/(((dy)/(dx))_(1, 1)) = (-1)/(-1) = 1` ` :. " equation of normal at "(1, 1) y-1=1*(x-1)` ` rArr x-y=0` |
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| 390. |
Show that the line `d/a+y/b=1`touches the curve `y=b e^(-x/a)`at the point where it crosses the y-axis. |
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Answer» The equation of the curve is `y = be^(-x//a)` ,rbgt It crosses the y-axis at the point where `x = 0` Putting x = 0 in the equation of the curve, we get `y = b` So, the point of contact is (0, b) Now, `y = be^(-x//a) rArr (dy)/(dx) = (-be^(-x//a))/(a)` `:. ((dy)/(dx))_((0","b)) = - (b)/(a)` So, the equaiton of the tangent is `( y- b)/(x - 0) = (-b)/(a) rArr bx + ay = ab rArr (x)/(a) + (y)/(b) = 1` Hence, `(x)/(a) + (y)/(b) = 1` touches the curve `y = be^(-x//a) " at " (0, b)` |
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| 391. |
Find the equations of the tangent and normal to the curve `x = a sin 3t, y = cos 2 t " at " t = (pi)/(4)` |
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Answer» We have `x = sin 3t rArr (dx)/(dt) = 3 cos 3t` and `y = cos 2 t rArr (dy)/(dt) = - 2 sin 2t` `:. (dy)/(dx) = ((.^(dy)//_(dt)))/((.^(dx)//_(dt))) = (-2 sin 2t)/(3 cos 3t)` `rArr [(dy)/(dx)]_(t = (pi)/(4)) = (-2 sin (2 xx (pi)/(4)))/(3 cos (3 xx (pi)/(4))) = (-2 sin (.^(pi)//_(2)))/(-3 cos (.^(pi)//_(4))) = (2 sqrt2)/(3)` When `t = (pi)/(4)`, we have `x = "sin" (3pi)/(4) = sin (pi - (pi)/(4)) = "sin" (pi)/(4) = (1)/(sqrt2) and y = "cos"(2pi)/(4) = "cos"(pi)/(2) = 0` `:.` point of contact of `P ((1)/(sqrt2), 0)` Equation of the tangent at the point P is given by `(y - 0)/(x - (1)/(sqrt2)) = (2 sqrt2)/(3)` `rArr (sqrt2y)/(sqrt2x -1) = (2 sqrt2)/(3)` `rArr 3 sqrt2y = 4x - 2 sqrt2` `rArr 2 sqrt2x - 3y - 2 = 0` Equation of the normal at the point P is given by `(y - 0)/(x - (1)/(sqrt2)) = (-3)/(2 sqrt2) rArr (sqrt2y)/(sqrt2x -1) = (-3)/(2 sqrt2)` `rArr 4y = - 3 sqrt2 x + 3 rArr 3 sqrt2 x + 4y - 3 = 0` |
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| 392. |
The line `y = m x + 1`is a tangent to the curve `y^2=4x`if the value of m is(A) 1 (B) 2 (C) 3 (D) `1/2`A. 1B. 2C. 3D. `1/2` |
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Answer» Correct Answer - A Equation of tangent on the curve is `y=mx+1`. `:." Given curve "y^(2)=4x` ` rArr 2y(dy)/(dx) = 4 rArr (dy)/(dx) = 2/y` If the tangent of the curve is `y=mx + 1" then "2/y = m` ` rArr y = 2/m` put the value of y in equation (1), ` x = y^(2)/4 = 1/4 (2/m)^(2) = 1/m^(2)` `:. ` The point will satisfy the line `y=mx + 1` `:. 2/m = m(1/m^(2))+1 rArr 1/m = 1 rArr m = 1` |
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| 393. |
Find the equation of all lines having slope 2 which are tangents to the curve `y=1/(x-3), x!=3` |
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Answer» Slope of tangent = 2 Equation of curve ` y = 1/(x-3)` `rArr (dy)/(dx) = (-1)/((x-3)^(2))` Slope of tangent at point `(x, y) = (-1)/((x-3)^(2))` `:. (-1)/((x-3)^(2)) = 2` ` rArr (x-3)^(2) = - 1/2` which is impossible. Therefore, there is no tangent of the given curve with slope 2. |
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| 394. |
Find pints on the curve `(x^2)/9+(y^2)/(16)=1`at which thetangents are(i) parallel to x-axis (ii) parallel to y-axis. |
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Answer» Equation of curve `x^(2)/9 +y^(2)/16 = 1`…(1) `rArr (2x)/9+(2y)/16(dy)/(dx) = 0` ` rArr (dy)/(dx) = - (16x)/(9y)` ….(2) (i) Tangent is parallel to x-axis `rArr (dy)/(dx) = 0` ` rArr (-16 x)/(9y) = 0` ` rArr x= 0` put x = 0 in equation (1) ` 0+(y^(2))/16 = 1` ` rArr y^(2) = 16` ` rArr y = pm 4` `:. ` The tangents drawn at points (0, 4) and (0, -4) of the curve are parallel to x-axis. (ii) Tangent is parallel to y-axis ` rArr (dx)/(dy) = 0` ` rArr (-9y)/(16x) = 0` ` rArr y= 0` put y = 0 in equation (1) ` x^(2)/9 + 0 = 1 ` ` rArr x^(2) = 9` ` rArr x = pm 3` `:. ` The tangents drawn at points (3, 0) and (-3, 0) are parallel to y-axis. |
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| 395. |
The slope of the tangent to the curve `x=t^2+3t-8,``y=2t^2-2t-5`at the point `(2, 1)`is(A) `(22)/7` (B) `6/7` (C) `7/6` (D) `(-6)/7`A. `22/7`B. ` 6/7`C. ` 7/6`D. ` (-6)/7` |
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Answer» Correct Answer - B Given, ` x= t^(2) + 3t - 8 and y = 2t^(2) - 2t - 5` …(1) ` rArr (dx)/(dt) = 2t+3 and (dy)/(dt) = 4t - 2` ` :. (dy)/(dx) = (dy)/(dt) xx (dt)/(dx) = (4t - 2)/(2t + 3)` Given point is (2, -1) at x = 2, from equation (1), ` 2 = t^(2) + 3t-8 rArr t^(2) + 3t - 10 = 0` ` rArr (t+5)(t-2)=0 rArr t=2 or t =- 5` at y =- 1 from equation (1), `-1=2t^(2)-2t-5 rArr 2t^(2)-2t - 4 = 0` ` rArr 2(t-2)(t+1)=0 rArr t = 2 or t =- 1` common value of t is 2. `:. ` slope of tangent of curve at point (2, -1) is `((dy)/(dx))_(t=2) = (4 xx 2-2)/(2 xx 2+ 3) = (8-2)/(4+3) = 6/7` . |
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| 396. |
Find the equations of all lines having slope 0 which are tangent to the curve `y=1/(x^2-2x+3)`. |
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Answer» Slope of tangent = 0 Equation of curve `y= 1/(x^(2)-2x+3)` ….(1) ` rArr (dy)/(dx) = (-1)/((x^(2)-2x+3)^(2)) d/(dx) (x^(2)-2x+3)` `=(-2(x-1))/((x^(2)-2x+3)^(2))` Slope of tangent at point `(x, y) = (-2(x-1))/((x^(2)-2x+3)^(2))` `:. (-2(x-1))/((x^(2)-2x+3)^(2))=0` ` rArr x = 1` put x = 1 in equation (1) ` y = 1/(1-2+3) = 1/2` equation of tangent at point ` (1,1/2)` ,brgt ` y - 1/2 = 0 (x-1)` ` rArr y = 1/2` |
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| 397. |
Find the slope of the tangent of the curve (i) `y = (x^(3) -x) " at " x = 2` (ii) `y = (2x^(2) + 3 sin x) " at " x = 0` (iii) `y = (sin 2x + cot x + 2)^(2) " at " x = (pi)/(2)` |
| Answer» Correct Answer - (i) 11 (ii) 3 (iii) `-12` | |
| 398. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `y^(2) = 4ax " at " ((a)/(m^(2)), (2a)/(m))` |
| Answer» Correct Answer - `m^(2) x - my + a = 0, m^(2) x + m^(3) y - 2 am^(2) - a = 0` | |
| 399. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1 " at " (a cos theta, b sin theta)` |
| Answer» Correct Answer - `b x cos theta + ay sin theta = ab, a x sec theta - by "cosec" theta = (a^(2) -b^(2))` | |
| 400. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `y = x^(3) " at " P(1, 1)` |
| Answer» Correct Answer - `y = 3x -2, x + 3y = 4` | |