Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 11 :

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

(48)

Q 12 :

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

(48)

Q 13 :

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{| t + 1 |}{t^{2}}, t < 0,$ is strictly decreasing. Then the local maximum value of the function $g (x) = 2 \log_{e} (x - 2) + α x^{2} + 4 x - α, x > 2,$ is ________ . [2026]

(4)

$Drawing graph of f (t) for t < 0$

$g (x) = \log_{e} (x - 2) - x^{2} + 4 x + 1, x > 2$

$g' (x) = \frac{2}{x - 2} - 2 (x - 2), x > 2$

$g' (x) = \frac{1 - {(x - 2)}^{2}}{x - 2} = \frac{- (x - 3) (x - 1)}{x - 2}$

$as x > 2$

$maxima occur at x = 3$

$g (3) = 2 \log_{e} 1 - 9 + 12 + 1 = 4$

Q 14 :

Let $f (x) = x^{2025} - x^{2000}$ , $x \in [0, 1]$ , 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]

$- 80$
$- 81$
$- 41$
$- 40$

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(2)

$f (x) = x^{2025} - x^{2000}$

$f' (x) = 0 \Rightarrow x = {(\frac{2000}{2025})}^{1 / 25} = α (say)$

$∴$ $f (0) = 0, f (1) = 0,$ $f (α) = {(\frac{80}{81})}^{80} . \frac{- 1}{81}$ $= 80^{80} \cdot {(- 81)}^{- 81}$

Q 15 :

The least value of $(\cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2)$ is [2026]

$4 - \sqrt{10}$
$4 + \sqrt{10}$
1
-1

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2

3

4

(1)

$f (θ) = \frac{1 + \cos 2 θ}{2} - 3 \sin 2 θ + 3 (\frac{1 - \cos 2 θ}{2}) + 2$

$f (θ) = 4 - 3 \sin 2 θ - \cos 2 θ$

$f (θ) \in [4 - \sqrt{10}, 4 + \sqrt{10}]$

Q 16 :

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

$(x^{2} - 4) y' - 2 x y + 2 x {(4 - x^{2})}^{2} = 0, x > 2,$

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

(16)

$(x^{2} - 4) y' - 2 x y = - 2 x {(x^{2} - 4)}^{2}$

$\Rightarrow \frac{d}{d x} (\frac{y}{x^{2} - 4}) = - 2 x$

$\Rightarrow y = (- x^{2} + C) (x^{2} - 4)$

$for x = 3, y = 15 \Rightarrow C = 12$

$\Rightarrow y = (- x^{2} + 12) (x^{2} - 4)$

$y' = 0 \Rightarrow x = 2 \sqrt{2}$

$y_{local max} = ({(2 \sqrt{2})}^{2} - 4) (- {(2 \sqrt{2})}^{2} + 12)$

$= 16$

Q 17 :

The critical point and nature for the function $f (x, y) = x^{2} - 2 x + y^{2} + 2 y - 2$ is:

(1, 1) max.
(1, - 1) max.
(1, 1) min.
(1, - 1) min.

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(4)

Ans. (1, - 1) min.

Explanation:

$f (x, y) = x^{2} - 2 x + y^{2} + 2 y - 2$

Partial Derivatives

$f' (x) = 2 x - 2$

and $f' (y) = 2 y + 2$

Therefore, for critical points,

$f' (x) = 0$

$2 x - 2 = 0$

$\Rightarrow$ $x = 1$

and $f' (y) = 0$

$2 y + 2 = 0$

$\Rightarrow$ $y = - 1$

So, the critical point is $(1, - 1) .$

Now, $f'' (x) = 2 > 0$ and $f'' (y) = 2 > 0$

So at $(1, - 1) .$

$(f_{x}^{''} \times f_{y}^{''}) - f_{x}^{'} f_{y}^{'} = 2 \times 2 - 0 = 4 > 0$

Q 18 :

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

$\frac{3 a^{2}}{4}$
$\frac{a^{2}}{2}$
$\frac{3 \sqrt{3} a^{2}}{4}$
$\frac{\sqrt{3} a^{2}}{4}$

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4

(3)

Ans. $\frac{3 \sqrt{3} a^{2}}{4}$

Explanation:

$Area of an equilateral triangle = \frac{\sqrt{3}}{4} \times {(Side)}^{2}$

$∴$ $Radius,$ $O A = O B = O C = a$

$Here,$ $O D = \frac{a}{2}$

As $O$ is the centroid of $∆ A B C$ ,

$∴$ $B D = \sqrt{O B^{2} - O D^{2}}$

$= \sqrt{a^{2} - \frac{a^{2}}{4}} = \frac{\sqrt{3} a}{2}$

Now, $B C = 2 B D = \sqrt{3} a$

So, $Area = \frac{\sqrt{3}}{4} \times {(\sqrt{3} a)}^{2}$

$= \frac{3 \sqrt{3}}{4} a^{2}$

Q 19 :

The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4,$ has:

two points of local maximum
two points of local minimum
one maxima and one minima
no maxima or minima

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(3)

Ans. One maxima and one minima

Explanation:

We have, $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$

$∴$ $f' (x) = 6 x^{2} - 6 x - 12$

Now, $f' (x) = 0$

$\Rightarrow$ $6 (x^{2} - x - 2) = 0$

$\Rightarrow$ $6 (x + 1) (x - 2) = 0$

$\Rightarrow$ $x = - 1$ 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 $f (x) = x^{x}$ has a stationary point at:

$x = e$
$x = \frac{1}{e}$
$x = 1$
$x = \sqrt{e}$

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(2)

Ans. $x = \frac{1}{e}$

Explanation:

We have, $f (x) = x^{x}$

Let $y = x^{x}$

and $\log y = x \log x$

$∴$ $\frac{1}{y} \cdot \frac{d y}{d x} = x \cdot \frac{1}{x} + \log x \cdot 1$

$\Rightarrow$ $\frac{d y}{d x} = (1 + \log x) x^{x}$

$∴$ $\frac{d y}{d x} = 0$

$\Rightarrow$ $(1 + \log x) x^{x} = 0$

$\Rightarrow$ $\log x = - 1$

$\Rightarrow$ $\log x = \log e^{- 1}$

$\Rightarrow$ $x = e^{- 1}$

$\Rightarrow$ $x = \frac{1}{e}$

Hence, $f (x)$ has a stationary point at $x = \frac{1}{e}$ .

Q 21 :

The minimum value of the function $y = 2 x^{3} - 21 x^{2} + 36 x - 20$ is:

- 120
- 126
- 128
None of these

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(3)

Ans. - 128

Explanation:

$y = f (x) = 2 x^{3} - 21 x^{2} + 36 x - 20$

$f' (x) = 6 x^{2} - 42 x + 36$

$f'' (x) = 12 x - 42$

For maxima/minima point, $f (x) = 0$

$6 x^{2} - 42 x + 36 = 0$

$\Rightarrow$ $x^{2} - 7 x + 6 = 0$

$\Rightarrow$ $(x - 6) (x - 1) = 0$ $\Rightarrow$ $x = 6, 1$

$∴$ $f'' (6) = 72 - 42 = 30 > 0$

$f'' (1) = 12 - 42 = - 30 < 0$

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 $5 \sin^{2} x + 7 \cos^{2} x - 4 \sin x \cos x$ will be:

$6 - \sqrt{5}$
$6 + \sqrt{5}$
$- 6 + \sqrt{5}$
$- 6 - \sqrt{5}$

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(2)

Ans. $6 + \sqrt{5}$

Explanation:

$y = 5 \sin^{2} x + 7 \cos^{2} x - 4 \sin x \cos x$

$= {(2 \sin x - \cos x)}^{2} + \sin^{2} x + 6 \cos^{2} x$

$= {(2 \sin x - \cos x)}^{2} + (\sin^{2} x + \cos^{2} x) + 5 \cos^{2} x$

$= {(2 \sin x - \cos x)}^{2} + 1 + 5 \cos^{2} x$

$y_{\max} = \sqrt{2^{2} + 1^{2}} + 1 + 5 = 6 + \sqrt{5}$

Q 23 :

Which one of the following is correct in respect of the function, $f (x) = x^{3} \sin x ?$

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

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(2)

Ans. It has local minimum at $x = 0$

Explanation:

Given, $f (x) = x^{3} \sin x$

$∴$ $f' (x) = x^{3} \cos x + 3 x^{2} \sin x$

Now, $f'' (x)$ $= - x^{3} \sin x + 3 x^{2} \cos x + 3 x^{2} \cos x + 6 x \sin x$

$= - x^{3} \sin x + 6 x^{2} \cos x + 6 x \sin x$

To find critical points, $f' (x) = 0$

$x^{3} \cos x + 3 x^{2} \sin x = 0$

$x^{2} (x \cos x + 3 \sin x) = 0$

$∴$ $x = 0$

At $x = 0,$

$f'' (x) = 0$

$f''' (x)$ $= - 6 x^{2} \sin x - x^{3} \cos x + 12 x \cos x$ $- 6 x^{2} \sin x + 6 \sin x + 6 x \cos x$

$= - 12 x^{2} \sin x - x^{3} \cos x + 18 x \cos x$ $- 6 x^{2} \sin x$

At $a = 0,$

$f''' (x) = 0$

$f'''' (x)$ $= - 24 x \sin x - 12 x^{2} \cos x - 3 x^{2} \cos x$ $+ x^{3} \sin x + 18 \cos x - 18 x \sin x$ $- 12 x \sin x - 6 x^{2} \cos x$

$= - 42 x \sin x - 21 x^{2} \cos x - 30 x \sin x$ $+ x^{3} \sin x + 18 \cos x$

At $x = 0$ , $f'''' (x) = 18 > 0$

$\Rightarrow$ Function is local minimum at $x = 0$ .

Q 24 :

If $x$ is real, then the minimum value of $x^{2} - 8 x + 17$ is:

- 1
0
1
2

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4

(3)

Ans. 1

Explanation:

Let $f (x) = x^{2} - 8 x + 17$

$∴$ $f' (x) = 2 x - 8$

So, $f' (x) = 0,$ gives $x = 4$

Now, $f'' (x) = 2 > 0,$ $\forall x$

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

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

$f (4) = 4^{2} - 8 (4) + 17 = 16 - 32 + 17 = 1$

$∴$ 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:

$2 h = r$
$h = 4 r$
$h = 2 r$
$h = r$

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(4)

Ans. $h = r$

Explanation:

Surface area, $S = 2 π r h + π r^{2}$ $\dots (i)$

and $V = π r^{2} h$ $\dots (i i)$

From Eq. (i),

$h = \frac{S - π r^{2}}{2 π r}$

From Eq. (ii),

$V = \frac{r}{2} (S - π r^{2})$

$\Rightarrow$ $\frac{d V}{d r} = \frac{1}{2} (S - 3 π r^{2}) = 0$

$\Rightarrow$ $S - 3 π r^{2} = 0$

$\Rightarrow$ $S = 3 π r^{2}$

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

$3 π r^{2} = 2 π r h + π r^{2}$

$\Rightarrow$ $r = h$

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?

$2 k^{2}$
$k^{2}$
$\frac{k^{2}}{2}$
$\frac{k^{2}}{4}$

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(4)

Ans. $\frac{k^{2}}{4}$

Explanation:

Here, $l + b = k$

or $l = k - b$

Area, $A = l \times b = (k - b) b$

$A = k b - b^{2}$

$∴$ $\frac{d A}{d b} = k - 2 b$

For maximum area,

$\frac{d A}{d b} = 0$

$\Rightarrow$ $k - 2 b = 0$

$\Rightarrow$ $b = \frac{k}{2}$

$\frac{d^{2} A}{d b^{2}} = - 2 < 0$ (maximum area)

Now,

$A = (k - b) \times b = (k - \frac{k}{2}) \times \frac{k}{2} = \frac{k^{2}}{4}$

Q 27 :

If the function $f (x) = x^{4} - 62 x^{2} + a x + 9$ attains its local maximum value at $x = 1$ , then $a$ is equal to:

120
110
100
90

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(1)

Ans. 120

Explanation:

Here, $f (x) = x^{4} - 62 x^{2} + a x + 9$

$\frac{d f (x)}{d x} = \frac{d (x^{4} - 62 x^{2} + a x + 9)}{d x} = 4 x^{3} - 124 x + a$

$4 x^{3} - 124 x + a = 0$

Putting $x = 1$ , at which attains local maxima,

$4 - 124 + a = 0$

$a = 120$

Q 28 :

Let $f (x) =$ ${\begin{matrix} 2 - | x^{2} + 5 x + 6 | & x \neq - 2 \\ b^{2} + 1 & x = - 2 \end{matrix}$

Has relative maximum at $x = - 2$ , then complete set of values $' b'$ can take is:

$| b |$ $\geq 1$
$| b |$ $< 1$
$b > 1$
$b < 1$

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(1)

$b^{2} + 1 \geq 2$

Q 29 :

Let the maximum value of expression $y = \frac{x^{4} - x^{2}}{x^{6} + 2 x^{3} - 1}$ for $x > 1$ is $\frac{p}{q}$ , where $' p'$ and $' q'$ are relatively prime natural numbers, then $p + q =$

(7)

$y = \frac{x - \frac{1}{x}}{x^{3} - \frac{1}{x^{3}} + 2},$

$Let t = x - \frac{1}{x} > 0 for x > 1,$

$y = \frac{t}{t (t^{2} + 3) + 2},$

$x^{3} - \frac{1}{x^{3}} = t (t^{2} + 3)$

$= \frac{1}{t^{2} + \frac{2}{t} + 3}$ $(\begin{matrix} t^{2} + \frac{2}{t} = t^{2} + \frac{1}{t} + \frac{1}{t} \geq 3 \\ ∴ t^{2} + \frac{2}{t} + 3 \geq 6 (AM \geq GM) \end{matrix},$

$y_{\max} = \frac{1}{{(t^{2} + \frac{2}{t} + 3)}_{\min}} = \frac{1}{6},$ $p = 1, q = 6$

Q 30 :

A wire of length 20 m is to be cut into two pieces. A piece of length $ℓ_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $ℓ_{2}$ is made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum then $(π ℓ_{1}) : ℓ_{2}$ is equal to

6 : 1
3 : 1
1 : 6
4 : 1

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(1)

$Since, given ℓ_{1} + ℓ_{2} = 20 \Rightarrow \frac{d ℓ_{2}}{d ℓ_{1}} = - 1$

$Now, A_{1} = {(\frac{ℓ_{1}}{4})}^{2} and A_{2} = π {(\frac{ℓ_{2}}{2 π})}^{2}$

$Let S = 2 A_{1} + 3 A_{2} = \frac{ℓ_{1}^{2}}{8} + \frac{3 ℓ_{2}^{2}}{4 π}$

$For max or min$

$\frac{d S}{d ℓ_{1}} = 0$

$\Rightarrow \frac{2 ℓ_{1}}{8} + \frac{6 ℓ_{2}}{4 π} \cdot \frac{d ℓ_{2}}{d ℓ_{1}} = 0$

$\Rightarrow \frac{ℓ_{1}}{4} = \frac{6 ℓ_{2}}{4 π}$

$\Rightarrow \frac{π ℓ_{1}}{ℓ_{2}} = 6$

Q 31 :

If $m$ and $n$ be the absolute maximum and minimum values of the function $f (x) =$ $| x^{2} - 7 x + 10 |$ $- 5 x + 27, x \in [- 2, 14]$ then $m + n$ is

65
66
68
70

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2

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(2)

$y =$ ${\begin{matrix} x^{2} - 12 x + 37, & x \leq 2 \\ 2 x - x^{2} + 17, & 2 < x < 5 \\ x^{2} - 12 x + 37, & x \geq 5 \end{matrix}}$

$y = a x^{2} + b x + c$

Q 32 :

If the parabola $y = a x^{2} + b x + c$ has vertex at (4, 2) and $a \in [1, 3]$ , then the absolute difference between the extreme values of abc is

3600
144
3456
169

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2

3

4

(3)

$\frac{- b}{2 a} = 4 \Rightarrow b = - 8 a$

$16 a + 4 b + c = 2$

$c = 16 a + 2$

$As a \in [1, 3] \Rightarrow c \in [18, 50]$

Q 33 :

The number of critical points of the function $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}$ in the interval ( $- 2 π, 2 π$ ) is equal to: [2026]

1
3
5
7

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2

3

4

(3)

We have, $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}; x \in (- 2 π, 2 π)$

$f (x) = {\begin{matrix} - \frac{\sin x}{x}, & x \in (- 2 π, - π] \cup [π, 2 π) \\ \frac{\sin x}{x}, & x \in (- π, π) - {0} \\ 1, & x = 0 \end{matrix}$

From graph, clearly f(x) is non-differentiable at $x = - π, π$

$\frac{d}{d x} (\frac{\sin x}{x}) = \frac{x \cos x - \sin x}{x^{2}}$

$= \frac{\cos x (x - \tan x)}{x^{2}}$

Also, $f' (x) = 0$ at $x = - \frac{3 π}{2}, 0, \frac{3 π}{2}$

Hence, the number of critical points is 5.

Q 34 :

Let f(x) be a polynomial of degree 5, and have extrema at x = 1 and x = –1, if $\lim_{x \to 0} (\frac{f (x)}{x^{3}}) = - 5$ , then f(2) – f(–2) is equal to: [2026]

0
50
92
112

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(4)

Degree of f(x) = 5

f(x) has extrema at x = 1 and x = –1 and $\lim_{x \to 0} (\frac{f (x)}{x^{3}}) = - 5$

$∴$ Coefficient of $x^{3}$ is –5

$∴$ f(x) must be of the form

$f (x) = a x^{5} + b x^{4} - 5 x^{3}$

$f' (x) = 5 a x^{4} + 4 b x^{3} - 15 x^{2}$

$= x^{2} (5 a x^{2} + 4 b x - 15)$

Now, x = 1 and x = –1 are roots of $f' (x) = 0$

$∴$ Sum of roots $= \frac{- 4 b}{5 a} = 0 \Rightarrow b = 0$

Products of roots $= \frac{- 15}{5 a} = - 1 \Rightarrow a = 3$

$∴ f (x) = 3 x^{5} - 5 x^{3}$

So, $f (2) - f (- 2) = [3 {(2)}^{5} - 5 {(2)}^{3}] - [3 {(- 2)}^{5} - 5 {(- 2)}^{3}]$

= 192 – 80 = 112

Q 35 :

Consider the system of linear equations in x, y, z;

x + 2y + tz = 0,

6x + y + 5tz = 0,

3x + $t^{2}$ y + f(t)z = 0

where $f : R \to R$ is a differentiable function. If this system has infinitely many solutions for all $t \in R$ , then f [2026]

is a constant function
is strictly increasing on R
is strictly decreasing on R
has two critical points

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2

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(2)

Given system of linear equations are

x + 2y + tz = 0

6x + y + 5tz = 0

3x + $t^{2}$ y + f(t)z = 0

Now, $△ =$ $| \begin{matrix} 1 & 2 & t \\ 6 & 1 & 5 t \\ 3 & t^{2} & f (t) \end{matrix} |$ $= 0$

$\Rightarrow f (t) + 30 t + 6 t^{3} - 3 t - 5 t^{3} - 12 f (t) = 0$

$\Rightarrow 11 f (t) = t^{3} + 27 t$

$\Rightarrow f (t) = \frac{t^{3} + 27 t}{11}$

$\Rightarrow f' (t) = \frac{3 t^{2} + 27}{11} > 0$

Q 36 :

Let $f : R \to R$ be a differentiable function such that $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}$ for all $x, y \in R$ , and $f' (0) = 3$ . Then the minimum value of the function $g (x) = 3 + e^{x} f (x)$ , is: [2026]

$3 (\frac{e + 1}{e})$
$3 (\frac{e - 1}{e})$
$\frac{3 - e}{e}$
3e

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4

(2)

Given,

$f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}$ ... (i)

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= \lim_{h \to 0} \frac{f (\frac{3 x + 3 h}{3}) - f (\frac{3 x + 0}{3})}{h}$

$= \lim_{h \to 0} \frac{\frac{f (3 x) + f (3 h)}{3} - \frac{f (3 x) + f (0)}{3}}{h}$

$= \lim_{h \to 0} \frac{f (3 h) - f (0)}{3 h}$

$\Rightarrow f' (x) = f' (0)$

$\Rightarrow f' (x) = 3$ [ $∵ f' (0) = 3$ ]

Integrating on both sides,

$f (x) = 3 x + λ$ ... (ii)

Put x = y = 0 in (i)

$\Rightarrow f (0) = \frac{f (0) + f (0)}{3}$

$\Rightarrow 3 f (0) = 2 f (0) \Rightarrow f (0) = 0$

Put x = 0 in (ii)

$\Rightarrow 0 = 0 + λ \Rightarrow λ = 0 \Rightarrow f (x) = 3 x$

Now, $g (x) = 3 + e^{x} f (x)$

$\Rightarrow g (x) = 3 + e^{x} \cdot 3 x$

$\Rightarrow g' (x) = 3 e^{x} + 3 x e^{x} = 0$

$\Rightarrow 3 e^{x} (1 + x) = 0 \Rightarrow x = - 1$

$g'' (x) = 3 e^{x} + 3 x e^{x} + 3 e^{x}$

$= 6 e^{x} + 3 x e^{x} = 3 e^{x} (2 + x)$

Minima at x = –1

$∴ g {(x)}_{\min} = g (- 1) = 3 + e^{- 1} 3 (- 1)$

$= 3 - \frac{3}{e} = 3 (\frac{e - 1}{e})$

Topic Question Set

Q 11 :

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

(48)

Q 12 :

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

(48)

Q 13 :

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{| t + 1 |}{t^{2}}, t < 0,$ is strictly decreasing. Then the local maximum value of the function $g (x) = 2 \log_{e} (x - 2) + α x^{2} + 4 x - α, x > 2,$ is ________ . [2026]

Q 14 :

Let $f (x) = x^{2025} - x^{2000}$ , $x \in [0, 1]$ , 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]

Q 15 :

The least value of $(\cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2)$ is [2026]

Q 16 :

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

$(x^{2} - 4) y' - 2 x y + 2 x {(4 - x^{2})}^{2} = 0, x > 2,$

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

Q 17 :

The critical point and nature for the function $f (x, y) = x^{2} - 2 x + y^{2} + 2 y - 2$ is:

Q 18 :

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

Q 19 :

The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4,$ has:

Q 20 :

The function $f (x) = x^{x}$ has a stationary point at:

Q 21 :

The minimum value of the function $y = 2 x^{3} - 21 x^{2} + 36 x - 20$ is:

Q 22 :

The greatest value of $5 \sin^{2} x + 7 \cos^{2} x - 4 \sin x \cos x$ will be:

Q 23 :

Which one of the following is correct in respect of the function, $f (x) = x^{3} \sin x ?$

Q 24 :

If $x$ is real, then the minimum value of $x^{2} - 8 x + 17$ is:

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:

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?

Q 27 :

If the function $f (x) = x^{4} - 62 x^{2} + a x + 9$ attains its local maximum value at $x = 1$ , then $a$ is equal to:

Q 28 :

Let $f (x) =$ ${\begin{matrix} 2 - | x^{2} + 5 x + 6 | & x \neq - 2 \\ b^{2} + 1 & x = - 2 \end{matrix}$

Has relative maximum at $x = - 2$ , then complete set of values $' b'$ can take is:

Q 29 :

Let the maximum value of expression $y = \frac{x^{4} - x^{2}}{x^{6} + 2 x^{3} - 1}$ for $x > 1$ is $\frac{p}{q}$ , where $' p'$ and $' q'$ are relatively prime natural numbers, then $p + q =$

Q 30 :

A wire of length 20 m is to be cut into two pieces. A piece of length $ℓ_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $ℓ_{2}$ is made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum then $(π ℓ_{1}) : ℓ_{2}$ is equal to

Q 31 :

If $m$ and $n$ be the absolute maximum and minimum values of the function $f (x) =$ $| x^{2} - 7 x + 10 |$ $- 5 x + 27, x \in [- 2, 14]$ then $m + n$ is

Q 32 :

If the parabola $y = a x^{2} + b x + c$ has vertex at (4, 2) and $a \in [1, 3]$ , then the absolute difference between the extreme values of abc is

Q 33 :

The number of critical points of the function $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}$ in the interval ( $- 2 π, 2 π$ ) is equal to: [2026]

Q 34 :

Let f(x) be a polynomial of degree 5, and have extrema at x = 1 and x = –1, if $\lim_{x \to 0} (\frac{f (x)}{x^{3}}) = - 5$ , then f(2) – f(–2) is equal to: [2026]

Q 35 :

Consider the system of linear equations in x, y, z;

x + 2y + tz = 0,

6x + y + 5tz = 0,

3x + $t^{2}$ y + f(t)z = 0

where $f : R \to R$ is a differentiable function. If this system has infinitely many solutions for all $t \in R$ , then f [2026]

Q 36 :

Let $f : R \to R$ be a differentiable function such that $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}$ for all $x, y \in R$ , and $f' (0) = 3$ . Then the minimum value of the function $g (x) = 3 + e^{x} f (x)$ , is: [2026]

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Topic Question Set

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]

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]

Q 15 : The least value of (cos2θ-6sinθ cosθ+3sin2θ+2) is [2026]

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]

Q 17 : The critical point and nature for the function f(x,y)=x2-2x+y2+2y-2 is:

Q 18 : What is the maximum area of an equilateral triangle that can be inscribed in a circle of radius 'a' ?

Q 19 : The function f(x)=2x3-3x2-12x+4, has:

Q 20 : The function f(x)=xx has a stationary point at:

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

Q 22 : The greatest value of 5sin2x+7cos2x-4sinx cosx will be:

Q 23 : Which one of the following is correct in respect of the function, f(x)=x3sinx?

Q 24 : If x is real, then the minimum value of x2-8x+17 is:

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:

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?

Q 27 : If the function f(x)=x4-62x2+ax+9 attains its local maximum value at x=1, then a is equal to:

Q 28 : Let f(x)={2-|x2+5x+6|x≠-2b2+1x=-2 Has relative maximum at x=-2, then complete set of values 'b' can take is:

Q 29 : Let the maximum value of expression y=x4-x2x6+2x3-1 for x>1 is pq, where 'p' and 'q' are relatively prime natural numbers, then p+q=

Q 30 : A wire of length 20 m is to be cut into two pieces. A piece of length ℓ1 is bent to make a square of area A1 and the other piece of length ℓ2 is made into a circle of area A2. If 2A1+3A2 is minimum then (πℓ1):ℓ2 is equal to

Q 31 : If m and n be the absolute maximum and minimum values of the function f(x)=|x2-7x+10|-5x+27, x∈[-2,14] then m+n is

Q 32 : If the parabola y=ax2+bx+c has vertex at (4, 2) and a∈[1,3], then the absolute difference between the extreme values of abc is

Q 33 : The number of critical points of the function f(x)={|sinxx|,x≠01,x=0 in the interval (–2π, 2π) is equal to: [2026]

Q 34 : Let f(x) be a polynomial of degree 5, and have extrema at x = 1 and x = –1, if limx→0(f(x)x3)=–5, then f(2) – f(–2) is equal to: [2026]

Q 35 : Consider the system of linear equations in x, y, z; x + 2y + tz = 0, 6x + y + 5tz = 0, 3x + t2y + f(t)z = 0 where f:R→R is a differentiable function. If this system has infinitely many solutions for all t∈R, then f [2026]

Q 36 : Let f:R→R be a differentiable function such that f(x+y3)=f(x)+f(y)3 for all x,y∈R, and f'(0)=3. Then the minimum value of the function g(x)=3+exf(x), is: [2026]

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Q 11 :

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

Q 12 :

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

Q 13 :

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{| t + 1 |}{t^{2}}, t < 0,$ is strictly decreasing. Then the local maximum value of the function $g (x) = 2 \log_{e} (x - 2) + α x^{2} + 4 x - α, x > 2,$ is ________ . [2026]

Q 14 :

Let $f (x) = x^{2025} - x^{2000}$ , $x \in [0, 1]$ , 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]

Q 15 :

The least value of $(\cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2)$ is [2026]

Q 16 :

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

$(x^{2} - 4) y' - 2 x y + 2 x {(4 - x^{2})}^{2} = 0, x > 2,$

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

Q 17 :

The critical point and nature for the function $f (x, y) = x^{2} - 2 x + y^{2} + 2 y - 2$ is:

Q 18 :

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

Q 19 :

The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4,$ has:

Q 20 :

The function $f (x) = x^{x}$ has a stationary point at:

Q 21 :

The minimum value of the function $y = 2 x^{3} - 21 x^{2} + 36 x - 20$ is:

Q 22 :

The greatest value of $5 \sin^{2} x + 7 \cos^{2} x - 4 \sin x \cos x$ will be:

Q 23 :

Which one of the following is correct in respect of the function, $f (x) = x^{3} \sin x ?$

Q 24 :

If $x$ is real, then the minimum value of $x^{2} - 8 x + 17$ is:

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:

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?

Q 27 :

If the function $f (x) = x^{4} - 62 x^{2} + a x + 9$ attains its local maximum value at $x = 1$ , then $a$ is equal to:

Q 28 :

Let $f (x) =$ ${\begin{matrix} 2 - | x^{2} + 5 x + 6 | & x \neq - 2 \\ b^{2} + 1 & x = - 2 \end{matrix}$

Has relative maximum at $x = - 2$ , then complete set of values $' b'$ can take is:

Q 29 :

Let the maximum value of expression $y = \frac{x^{4} - x^{2}}{x^{6} + 2 x^{3} - 1}$ for $x > 1$ is $\frac{p}{q}$ , where $' p'$ and $' q'$ are relatively prime natural numbers, then $p + q =$

Q 30 :

A wire of length 20 m is to be cut into two pieces. A piece of length $ℓ_{1}$ is bent to make a square of area $A_{1}$ and the other piece of length $ℓ_{2}$ is made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum then $(π ℓ_{1}) : ℓ_{2}$ is equal to

Q 31 :

If $m$ and $n$ be the absolute maximum and minimum values of the function $f (x) =$ $| x^{2} - 7 x + 10 |$ $- 5 x + 27, x \in [- 2, 14]$ then $m + n$ is

Q 32 :

If the parabola $y = a x^{2} + b x + c$ has vertex at (4, 2) and $a \in [1, 3]$ , then the absolute difference between the extreme values of abc is

Q 33 :

The number of critical points of the function $f (x) = {\begin{matrix} | \frac{\sin x}{x} |, & x \neq 0 \\ 1, & x = 0 \end{matrix}$ in the interval ( $- 2 π, 2 π$ ) is equal to: [2026]

Q 34 :

Let f(x) be a polynomial of degree 5, and have extrema at x = 1 and x = –1, if $\lim_{x \to 0} (\frac{f (x)}{x^{3}}) = - 5$ , then f(2) – f(–2) is equal to: [2026]

Q 35 :

Consider the system of linear equations in x, y, z;

x + 2y + tz = 0,

6x + y + 5tz = 0,

3x + $t^{2}$ y + f(t)z = 0

where $f : R \to R$ is a differentiable function. If this system has infinitely many solutions for all $t \in R$ , then f [2026]

Q 36 :

Let $f : R \to R$ be a differentiable function such that $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}$ for all $x, y \in R$ , and $f' (0) = 3$ . Then the minimum value of the function $g (x) = 3 + e^{x} f (x)$ , is: [2026]