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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. |
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. |
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Answer» Correct Answer - `a = - (1)/(2), b = - (3)/(4), c = 3` Given, `y = ax^(3) + bx^(2) + cx + 5` touches X-axis at `P ( - 2, 0 )` which implies that X- axis is tangent at `(-2, 0)` and the curve is also passes through `(-2, 0)`. The curve cuts Y- axis at `(0, 5)` and gradient at this point is given 3, therefore at `(0, 5)` slope of the tangent is 3. Now, `" " (dy)/(dx) = 3ax^(3) + 2 b x + c` Since, X-axis is tangent at `(-2, 0)`. `therefore " " |(dy)/(dx)|_(x = -2 ) = 0` `rArr " " 0 = 3a (-2)^(2) + 2b (-2) + c ` `rArr " " 0 = 12 a - 4 b + c " " ` ... (i) Again, slope of tangent at `(0, 5)` is 3. `therefore " " |(dy)/(dx)| _("("0, 5")")= 3 ` `rArr " " 3= 3a (0)^(2) + 2b (0) + c ` `rArr " " 3= c" " `...(ii) Since, the curve passes through `(-2, 0)` `" " 0 = a(-2)^(3) + b (-2)^(2) + c ( - 2 ) + 5` `rArr " " 0 = - 8a + 4b - 2c + 5 " " ` ... (iii) From Eqs (i) and (ii), `" " 12 a - 4b = -3 " " ` ... (iv) From Eqs. (ii) and(iii), `" " - 8a + 4b = 1 " " `... (v) On adding Eqs. (iv ) and (v), we get ` 4a = -2 rArr a = - 1 //2 ` On putting `a = -1//2` in Eqs. (iv), we get ` " " 12( - 1//2) - 4b =-3 ` `rArr " " - 6 - 4b =-3` `rArr - 3 = 4b ` `rArr " " b = - 3//4` `therefore " " a = -1//2, b = - 3//4 and c = 3` |
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| 2. |
The radius of a circle, having minimum area, whichtouches the curve `y=4-x^2`and the lines `y=|x|`is :`4(sqrt(2)-1)`(2) `4(sqrt(2)+1)`(3) `2(sqrt(2)+1)`(4) `2(sqrt(2)-1)` |
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Answer» C(0,4-r) y-x=0 `|(4-r-0)/sqrt2|=r` `4-r=pmsqrt2r` `4=(sqrt2+1)r` `r=4/(sqrt2+1)*(sqrt2-1)/(sqrt2-1)` `=(4(sqrt2-1))/(2-1)` Option 1 is correct. |
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| 3. |
The point(s) on the parabola `y^2 = 4x` which are closest to the circle`x^2 + y^2 - 24y + 128 = 0` is/are |
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Answer» equation of circle `x^2+y^2-24y+128=0` its center=(0,12) equation of parabola `y^2=4x` where a=1 point P`(at^2,2at)=(t^2,2t)` normal at point (t) `y=-tx+at^3+2at` a=1 `y=-tx+t^3+2t` this equation should be normal for circle as well (0,12) should be satisfied `12=t^3+2t` 3+2t-12=0 (t-2)is a factor after division `(t-2)(t^2+2t+6)=0` but `(t^2+2t+6)!=0` so, (t-2)=0 t=2 putting this value in point P P(4,4) |
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| 4. |
A helicopter flying along the path `y=7+x^((3)/(2))`, A soldier standint at point `((1)/(2),7)` wants to hit the helicopter when it is closest from him, then minimum distance is equal to (a) `(1)/(6)(sqrt2)/(3)` (b) `(1)/(2)` (c) `(1)/(3)sqrt((2)/(3))` (d) `sqrt((5)/(2))`A. `(1)/(2)`B. `(1)/(3)sqrt((7)/(3)`C. `(1)/(6)sqrt((7)/(3))`D. `(sqrt(5))/(6)` |
| Answer» Correct Answer - C | |
| 5. |
Let `A(4,-4)` and B(9,6) be points on the parabola `y^(2)=4x. ` Let C be chosen on the on the arc AOB of the parabola where O is the origin such that the area of `DeltaACB` is maximum. Then the area (in sq. units) of `DeltaACB` is :A. `30 .(1)/(2)`B. `31.(3)/(4)`C. `31.(1)/(4)`D. `32` |
| Answer» Correct Answer - C | |
| 6. |
A swimming pool is to be drained by cleaning. If L represents thenumber of litres of water in the pool `t`seconds after the pool has been plugged off to drain and `L=2000(10-t)^2dot`How fast is the water ruining out at the end of 5 seconds? What is theaverage rate at which the water flows out during the first 5 seconds? |
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Answer» Let L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain, then `L=200(10-t)^(2)` `therefore` Rate at which the water is running out `=-(dL)/(dt)` `(dL)/(dt) = -200.2(10-t).(-1)` `=400(10-t)` Rate at which the water is running out at the end of 5s `=400(10-5)` `=2000L//s `= Final rate Since, initial rate `=-(dL)/(dt)_(t=0) = 4000 L//s` `therefore` Average rate during 5s `=("Initial rate + Final rate")/(2)` `=(4000+2000)/(2)` `=3000L//s` |
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| 7. |
On the curve `x^3=12 y ,`find the interval of values of `x`for which the abscissa changes at a faster rate than the ordinate?A. `(-3,0)`B. `(-oo, -2)UU (2,oo)`C. `(-2,2)`D. `(-3,3)` |
| Answer» Correct Answer - C | |