Q 11 :

Consider the triangles with vertices A(2, 1), B(0, 0) and C(t, 4), t[0, 4]. If the maximum and the minimum perimeters of such triangles are obtained at t=α and t=β respectively, then 6α+21β is equal to _____ .             [2023]



(48)

 



Q 12 :

Consider the triangles with vertices A(2, 1), B(0, 0) and C(t, 4), t[0, 4]. If the maximum and the minimum perimeters of such triangles are obtained at t=α and t=β respectively, then 6α+21β is equal to _____ .             [2023]



(48)

 



Q 13 :

Let (2α,α) be the largest interval in which the function f(t)=|t+1|t2, t<0, is strictly decreasing. Then the local maximum value of the function g(x)=2loge(x-2)+αx2+4x-α, x>2, is ________ .              [2026]



(4)

Drawing graph of f(t) for t<0

g(x)=loge(x-2)-x2+4x+1,  x>2

g'(x)=2x-2-2(x-2),  x>2

g'(x)=1-(x-2)2x-2=-(x-3)(x-1)x-2

  as x>2

maxima occur at x=3

g(3)=2loge1-9+12+1=4



Q 14 :

Let f(x)=x2025-x2000, x[0,1], and the minimum value of the function f(x) in the interval [0,1] be (80)80(n)-81. Then n is equal to         [2026]

  • -80

     

  • -81

     

  • -41

     

  • -40

     

(2)

f(x)=x2025-x2000

f'(x)=0x=(20002025)1/25=α (say)

f(0)=0,  f(1)=0, f(α)=(8081)80.-181=8080·(-81)-81



Q 15 :

The least value of (cos2θ-6sinθ cosθ+3sin2θ+2) is  [2026]

  • 4-10

     

  • 4+10

     

  • 1

     

  • -1

     

(1)

f(θ)=1+cos2θ2-3sin2θ+3(1-cos2θ2)+2

f(θ)=4-3sin2θ-cos2θ

f(θ)[4-10, 4+10]



Q 16 :

 If the solution curve y=f(x) of the differential equation

(x2-4)y'-2xy+2x(4-x2)2=0,  x>2,

passes through the point (3,15) then the local maximum value of f is _____.   [2026]



(16)

(x2-4)y'-2xy=-2x(x2-4)2

ddx(yx2-4)=-2x

y=(-x2+C)(x2-4)

for x=3, y=15C=12

y=(-x2+12)(x2-4)

y'=0x=22

ylocal max=((22)2-4)(-(22)2+12)

= 16



Q 17 :

The critical point and nature for the function f(x,y)=x2-2x+y2+2y-2 is:

  • (1, 1) max.

     

  • (1, - 1) max.

     

  • (1, 1) min.

     

  • (1, - 1) min.

     

(4)

Ans.       (1, - 1) min.

Explanation:

              f(x,y)=x2-2x+y2+2y-2

Partial Derivatives

               f'(x)=2x-2

and         f'(y)=2y+2

Therefore, for critical points,

                f'(x)=0

              2x-2=0

                 x=1

and        f'(y)=0

             2y+2=0

                 y=-1

So, the critical point is (1,-1).

Now, f''(x)=2>0  and  f''(y)=2>0

So at (1,-1).

(fx''×fy'')-fx'fy'=2×2-0=4>0



Q 18 :

What is the maximum area of an equilateral triangle that can be inscribed in a circle of radius  'a' ?

  • 3a24

     

  • a22

     

  • 33a24

     

  • 3a24

     

(3)

Ans.       33a24

Explanation:

Area of an equilateral triangle=34×(Side)2

  Radius, OA=OB=OC=a

Here,              OD=a2

As O is the centroid of ABC,

  BD=OB2-OD2

                =a2-a24=3a2

Now,    BC=2BD=3a

So,      Area=34×(3a)2

                      =334a2



Q 19 :

The function f(x)=2x3-3x2-12x+4, has:

  • two points of local maximum

     

  • two points of local minimum

     

  • one maxima and one minima

     

  • no maxima or minima

     

(3)

Ans.         One maxima and one minima

Explanation:

We have,            f(x)=2x3-3x2-12x+4

                   f'(x)=6x2-6x-12

Now,                  f'(x)=0

     6(x2-x-2)=0

  6(x+1)(x-2)=0

                          x=-1 and  x=2

On the number line for f'(x), we get

Hence, x=-1 is the point of local maxima and x=2 is the point of local minima.

So, f(x) has one maxima and one minima.



Q 20 :

The function f(x)=xx has a stationary point at:

  • x=e

     

  • x=1e

     

  • x=1

     

  • x=e

     

(2)

Ans.     x=1e

Explanation:

We have,   f(x)=xx

Let                 y=xx

and          logy=xlogx

   1y·dydx=x·1x+logx·1

  dydx=(1+logx)xx

  dydx=0

  (1+logx)xx=0

                logx=-1

                logx=loge-1

                      x=e-1

                      x=1e

Hence,  f(x) has a stationary point at x=1e.



Q 21 :

The minimum value of the function y=2x3-21x2+36x-20 is:

  • - 120

     

  • - 126

     

  • - 128

     

  • None of these

     

(3)

Ans.        - 128

Explanation:

                    y=f(x)=2x3-21x2+36x-20

              f'(x)=6x2-42x+36

             f''(x)=12x-42

For maxima/minima point, f(x)=0

             6x2-42x+36=0

             x2-7x+6=0

         (x-6)(x-1)=0x=6, 1

  f''(6)=72-42=30>0

         f''(1)=12-42=-30<0

Hence, the minimum point is 6.

The minimum value is

                 f(6)=2(6)3-21(6)2+36(6)-20

                          =432-756+216-20=-128.

     Minimum value = -128.



Q 22 :

The greatest value of 5sin2x+7cos2x-4sinx cosx will be:

  • 6-5

     

  • 6+5

     

  • -6+5

     

  • -6-5

     

(2)

Ans.        6+5

Explanation:

          y=5sin2x+7cos2x-4sinx cosx

              =(2sinx-cosx)2+sin2x+6cos2x

               =(2sinx-cosx)2+(sin2x+cos2x)+5cos2x

                =(2sinx-cosx)2+1+5cos2x

       ymax=22+12+1+5=6+5



Q 23 :

Which one of the following is correct in respect of the function, f(x)=x3sinx?

  • It has local maximum at x=0

     

  • It has local minimum at x=0

     

  • It has neither maximum nor minimum at x=0

     

  • It has maximum value as 1

     

(2)

Ans.      It has local minimum at x=0

Explanation:

Given,     f(x)=x3sinx

       f'(x)=x3cosx+3x2sinx

Now,     f''(x)=-x3sinx+3x2cosx+3x2cosx+6xsinx

                        =-x3sinx+6x2cosx+6xsinx

To find critical points, f'(x)=0

         x3cosx+3x2sinx=0

         x2(xcosx+3sinx)=0

            x=0

At                x=0,

             f''(x)=0

             f'''(x)=-6x2sinx-x3cosx+12xcosx-6x2sinx+6sinx+6xcosx

                          =-12x2sinx-x3cosx+18xcosx-6x2sinx

At              a=0,

         f'''(x)=0

         f''''(x)=-24xsinx-12x2cosx-3x2cosx+x3sinx+18cosx-18xsinx-12xsinx-6x2cosx

                       =-42xsinx-21x2cosx-30xsinx+x3sinx+18cosx

At     x=0f''''(x)=18>0

    Function is local minimum at x=0.



Q 24 :

If x is real, then the minimum value of x2-8x+17 is:

  • - 1

     

  • 0

     

  • 1

     

  • 2

     

(3)

Ans.        1

Explanation:

Let f(x)=x2-8x+17

            f'(x)=2x-8

So,             f'(x)=0,   gives x=4

Now,         f''(x)=2>0,  x

So, x=4 is the point of local minimum.

Hence, the minimum value of f(x) = at x=4.

                          f(4)=42-8(4)+17=16-32+17=1

    Minimum value = 1



Q 25 :

A right circular cylinder which is open at the top and has a given surface area, will have the greatest volume, if its height h and radius r are related by:

  • 2h=r

     

  • h=4r

     

  • h=2r

     

  • h=r

     

(4)

Ans.      h=r

Explanation:

Surface area, S=2πrh+πr2            (i)

and                 V=πr2h                       (ii)

From Eq. (i),

                       h=S-πr22πr

From Eq. (ii),

                       V=r2(S-πr2)

            dVdr=12(S-3πr2)=0

   S-3πr2=0

               S=3πr2

On putting the value of S in eq. (i), we get

                3πr2=2πrh+πr2

                  r=h



Q 26 :

Let l be the length and b be the breadth of a rectangle such that l+b=k. What is the minimum area of rectangle?

  • 2k2

     

  • k2

     

  • k22

     

  • k24

     

(4)

Ans.        k24

Explanation:

Here,      l+b=k

or                  l=k-b

Area,           A=l×b=(k-b)b

                    A=kb-b2

        dAdb=k-2b

For maximum area,

               dAdb=0

    k-2b=0

             b=k2

             d2Adb2=-2<0             (maximum area)

Now,

  A=(k-b)×b=(k-k2)×k2=k24



Q 27 :

If the function f(x)=x4-62x2+ax+9 attains its local maximum value at x=1, then a is equal to:

  • 120

     

  • 110

     

  • 100

     

  • 90

     

(1)

Ans.        120

Explanation:

Here,     f(x)=x4-62x2+ax+9

         df(x)dx=d(x4-62x2+ax+9)dx=4x3-124x+a

4x3-124x+a=0

Putting x=1, at which attains local maxima,

4-124+a=0

                   a=120