Him Academy

Hamirpur

Exam Name: Demo Test Subject: Mathematics

DURATION: 60 mins DATE: 10/06/2025 M. MARKS: 100

Mathematics (Q1 to 25)

Q1.
If the point of minima of the function, 𝑓𝑥
( ) = 1+ 𝑎 𝑥−𝑥
2 3
satisfy the inequality
𝑥 𝑥
2

𝑥 𝑥
2
+

+5
+2

+6
< 0, 𝑎
then ' ' must lie in
the interval :

√ √ √ √
A) (−3 3, 3 3 ) B) (−2 3, −3 3)
√ √ √ √ √ √
C) (2 3, 3 )
3
D) (−3 3, −2 3) ∪ (2 3, 3 3)

Explanation:

Discriminant of 𝑥 𝑥 is 𝐷
2
+ +2 = 1 − 8 = −7 < 0 and coefficient of 𝑥 2
is 1 > 0, hence
𝑥 𝑥 2
+ +2 > 0 ∀𝑥 ∈ 𝑅.
Now, we have,
𝑥 𝑥 2

𝑥 𝑥 <0
+ +2
2
+5 +6

⇒ 𝑥 𝑥 <0 1

⇒ 𝑥 + 5𝑥 + 6 < 0
2
+5 +6

⇒ (𝑥 + 2) (𝑥 + 3) < 0
⇒ −3 < 𝑥 < −2

Now,
𝑓𝑥 ( ) = 1+ 𝑎 𝑥−𝑥
2 3

Differentiating both sides w.r.t , we get 𝑥

𝑓 𝑥 𝑎 − 𝑥√
′
( ) =
2
3
2

√
⇒𝑓 𝑥 (𝑎 − 𝑥) (𝑎
′
( ) = 3 + 3 𝑥)

If sign of 𝑓 𝑥 ′
( ) changes from − ve to + ve, then it is the point of minima.
Hence, 𝑥 = ± √𝑎 are the points of minima.

𝑎
3

If > 0
 −3 < √−𝑎 < −2
√ √
⇒2 3<𝑎<3 3
3

If 𝑎 < 0,

−3 < √𝑎 < −2
√ √
𝑎
3

⇒ −3 3 < √ < −2
√ 3 √ √
∴ 𝑎 ∈ (−3 3, −2 3) ∪ (2 3, 3 3)

Q2.
Find the intervals in which the following function 𝑓𝑥( ) = 20 − 9𝑥 + 6𝑥 − 𝑥 2 3
is
( ) 𝑎 Strictly increasing,
( ) 𝑏 Strictly decreasing.
A) B)

𝑎
( ) (1, 3) ( )(𝑎 −∞ , 1) ∩ (3, ∞)

𝑏 −∞
( )( , 1) ∩ (3, ∞) 𝑏
( ) (1, 3)

C) D)

( )(𝑎 −∞ , 1) ∪ (3, ∞) 𝑎
( ) (1, 3)

𝑏
( ) (1, 3) 𝑏 −∞
( )( , 1) ∪ (3, ∞)

Explanation:

We have:
𝑓𝑥 ( − 9𝑥 + 6𝑥 − 𝑥
) = 20
2 3

𝑓 𝑥
′
( ) = −9 + 12 − 3𝑥
2

= −3 (𝑥 − 4𝑥 + 3)
2

= −3 [𝑥 − 3𝑥 − 𝑥 + 3]
2

= −3[𝑥 (𝑥 − 3) − 1 (𝑥 − 3)

𝑓 ′ (𝑥) = −3 (𝑥 − 1) (𝑥 − 3)
For stationary point, 𝑓 𝑥
′
( ) = 0

−3 (𝑥 − 1) (𝑥 − 3) = 0

⇒ 𝑥 = 1, 𝑥 = 3
Divide 𝑥 = 1, 𝑥 = 3 into following disjoint intervals
( −∞, 1) (1, 3) (3, ∞)
 Interval Test Value sign of 𝑓 𝑥
′
( ) = − (𝑥 − 1) (𝑥 − 3) Nature of 𝑓𝑥
( )

(−∞, 1) 𝑥 = 0 −) (−) (−) = (−) < 0

( Strictly decreasing

(1, 3) 𝑥 = 2 −) (+) (−) = (+) > 0

( Strictly increasing

(3, ∞) 𝑥 = 4 −) (+) (+) = (−) < 0

( Strictly decreasing

Hence 𝑓𝑥 ( ) is strictly increasing in (1, 3) and 𝑓𝑥 ( ) strictly decreasing in

(−∞, 1) ∪ (3, ∞)
Q3.
Let 𝑔𝑥 ( ) = 2 𝑓(𝑥) 𝑓 2
+ (2 − 𝑥) and 𝑓 ′′
𝑥
( ) < 0 for all 𝑥∈ (0, 2). Then, 𝑔𝑥 ( ) increases in

A) ( 12 , 2) B) ( 43 , 2)

C) (0, 2)
D) (0, 4
)
3

Explanation:

𝑔𝑥 ( 𝑓 ( 𝑥 ) 𝑓 − 𝑥 and 𝑓 𝑥 ∀𝑥 ∈
) = 2 + (2 )
′′
( ) < 0 (0, 2)

Now, 𝑓 𝑥 ⇒ 𝑓 𝑥 is a decreasing function on 𝑥 ∈

2
′′ ′
( ) < 0 ( ) (0, 2).

⇒𝑔 𝑥 𝑓 ( )−𝑓 −𝑥
′ 𝑥
( ) =
′ ′
(2 )

Now, 𝑔 𝑥
2
′
( ) > 0

⇒𝑓 (𝑥) 𝑓 ′
−𝑥 >
′
(2 )

⇒ 𝑥 −𝑥
2

< 2

⇒𝑥
2
4
<

So, 𝑔 𝑥 is an increasing function for 𝑥 ∈ (

3

( ) ) 0,
4
3

Q4.
If function 𝑓𝑥 ( ) = cos | 𝑥 − 𝑎𝑥 𝑏 is strictly increasing function on whole number line, then the value of 𝑎 is
| 2 +

A) 𝑏 B) 𝑏
2

C) 𝑎≤ − 2
1
D) 𝑎 >
−3
2

Explanation:

𝑓𝑥
Given, ( ) = cos | 𝑥 − 𝑎𝑥 𝑏
| 2 +

Now, we know that

𝑑
𝑑𝑥 𝑥
{cos | |} = − sin 𝑥 ∵ −𝑥, [ cos ( ) = cos 𝑥 ]

for 𝑥 ∈ 𝑅
Therefore, 𝑓 𝑥 ′
( ) = − sin 𝑥 − 2𝑎
Now, 𝑓𝑥 ( ) is strictly increasing function, therefore
 𝑓 𝑥 ′
( ) > 0

⇒ − sin 𝑥 − 2𝑎 > 0
⇒ 𝑎 < − sin 𝑥 ⇒ 𝑎 ≤ −1
2
1
2

Q5.
Let the function 𝑔 −∞, ∞) → (− 𝜋 , 𝜋 )
: (
2 2
be given by g (u) = 2tan
−1
(eu) − 𝜋 2
Then, g is

A) even and is strictly increasing in (0, ∞ ) B) ∞∞

odd and is strictly decreasing in (− , )

C) odd and is strictly increasing in (− ∞∞ , ) D) neither even nor odd, but is strictly increasing in (− ∞∞
, )

Explanation:

Given, g(( 𝑢 )) = 2tan

−1
(, msup
) −𝜋 2
for 𝑢𝜖 ( −∞,∞)

g (−u) = 2tan
−1
(e−u) − 𝜋 2

⇒ 2 (cot− (eu)) − 𝜋 1

⇒ 2 (· 𝜋 − tan− (eu)) − 𝜋
2
1

=
𝜋 − 2tan− (eu) = −g (u)
1

Therefore, g (-u) = - g( u ) ⇒ g(u) is an odd function.

Since ex is increasing and tan-1x is increasing, hence g(u) is an increasing function.

Q6.
For the function f (x) = (cosx)–x + 1, x ∈ ℝ,
between the following two statements
(S1) f(x) = 0 for only one value of x is [0, π].
(S2) f(x) is decreasing in [0, π
2
] and increasing in
[ 2 , π].
π

A) Both (S1) and (S2) are correct B) Only (S1) is correct

C) Both (S1) and (S2) are incorrect D) Only (S2) is correct

Explanation:
f (x) = cosx –x + 1

f′(x) = – sinx –1

f is decreasing ∀x ∈ R
f (x) = 0

f (0) = 2, f (π) = –π

f is strictly decreasing in [0, π] and f (0). f (π) < 0

⇒ only one solution of f (x) = 0

S1 is correct and S2 is incorrect.

Q7.
In open interval (0,
𝜋)
2

A) cos 𝑥 𝑥 𝑥 + sin < 1

B) cos 𝑥 𝑥 𝑥
+ sin > 1

C) no specific order relation can be ascertained between cos 𝑥 𝑥 𝑥 and

+ sin 1

D) cos 𝑥 𝑥 𝑥 + sin <

1
2

Explanation:

𝑓𝑥 ( 𝑥 𝑥 𝑥−
) = cos + sin 1

𝑓 𝑥 − 𝑥 𝑥 𝑥
′( ) = sin + cos + sin 𝑥
𝑓 𝑥 − 𝑥 𝑥 𝑥
′( ) = sin + cos + sin 𝑥
𝑓 𝑥 is increasing on 𝑥 ∈ (
( ) 0,
𝜋)
𝑥 𝑥 𝑥−
2

∴ cos + sin 1 > 0

⇒ 𝑥 𝑥 𝑥 cos + sin > 1

Q8.
The value of 𝑎 for which the function 𝑓 𝑥 ( ) = sin 𝑥 − 𝑥 − 𝑎𝑥 𝑏 decreases for all real values of 𝑥 is given by
cos +

A) 𝑎≥√ 2 B) 𝑎≥ 1

√
C) 𝑎 < 2 D) 𝑎 < 1

Explanation:

Given:
𝑓𝑥 ( 𝑥 − 𝑥 − 𝑎𝑥 𝑏
) = sin cos +

⇒𝑓 𝑥 ′
𝑥 𝑥−𝑎
( ) = cos + sin

For decreasing function,

𝑓 𝑥≤ ′
( ) 0

⇒ 𝑥 𝑥−𝑎 ≤
cos + sin 0

⇒ 𝑥 𝑥≤𝑎cos + sin
 √
⇒ 2 (√ 1
cos 𝑥 √ 𝑥) ≤ 𝑎
+
1
sin
√
𝑥) ≤ 𝑎
2 2

⇒ 2 sin (
𝜋 +
4

⇒ sin ( 𝜋 4
+ 𝑥) ≤ √𝑎 2

⇒ √𝑎 ≥ 1
𝑥) ≤
2

[∵ −1 ≤ sin ( 𝜋 4
+ 1]

√
⇒𝑎≥ 2

Q9.
The function : 𝑓
A) Has a maxima but no minima. B) Has a minima but no maxima.

C) Has both a maxima and minima. D) Is monotonic.

Explanation:

𝑥→
lim 𝑓𝑥 ( ) = 1 (can be verified)

𝑓 (𝑥) = 𝑒
0

𝑥lim
→∞
Also 𝑓 is increasing for all 𝑥 > 0,

⇒( 𝑑 (can be verified)
)

Q10.
The curve 𝑦 𝑥 𝑎𝑥 𝑏𝑥 𝑐𝑥 touches the 𝑥-axis at the point 𝑃
( ) =
3
+
2
+ +5 −2, 0) and cuts the 𝑦-axis at the point
(

Q , where 𝑦 is equal to . Then the local maximum value of 𝑦 𝑥 is

′
3 ( )

A) 27
B) 29
4 4

C) 37
D) 9
4 2

Explanation:

Given that 𝑦 𝑥 𝑎𝑥 𝑏𝑥 𝑐𝑥 ( ) =
3
+
2
+ +5 pass through (−2, 0)
so 8 𝑎− 𝑏 𝑐 4𝑖 +2 = 5. . . ( )
 𝑥
Since the curve touches -axis at (−2, 0), so its slope would be 0
i.e 𝑦 − ′
( ⇒ ( 𝑎𝑥 𝑏𝑥 𝑐)𝑥 −
2) = 0 3
2
+2 + = 0

𝑎− 𝑏 𝑐 𝑖𝑖
= 2

12 4 + = 0. . . ( )

Also given, for 𝑥 𝑦 𝑥 = 0,

′
( ) = 3

𝑐 𝑖𝑖𝑖
= 3. . . ( )

Solving eq. 𝑖 𝑖𝑖 𝑖𝑖𝑖 , we get,

( ), ( )& ( )

𝑎=− 𝑏 − 1
, =
3

For local maxima 𝑦 𝑥 − 𝑥 − 𝑥

2 4
′ 3 2 3
( ) = +3 = 0

⇒𝑥 𝑥− ⇒ 𝑥− 𝑥
2 2
2
+ 2 = 0 ( 1) ( + 2) = 0

⇒𝑥 and 𝑦= 1
″
(1) < 0

So 𝑦 𝑥 has local maxima at 𝑥

( ) = 1

Hence, 𝑦
27
(1) =
4

Q11.
Let the equation 𝑥 − 𝑥 𝑎 has a unique root in [− 𝜋
sin =
2
,
𝜋 ], then
2

A) 𝑎 ∈ [− 𝜋 2
+ 1, ∞) B) 𝑎 ∈ (−∞ ,
𝜋 − 1]
2

C) 𝑎∈[ 1 − 𝜋 , 𝜋 − 1]
2 2
D) 𝑎∈𝑅−( 1 − 𝜋 , 𝜋 − 1)
2 2

Explanation:

Let 𝑓 𝑥 𝑥− 𝑥 ( ) = sin

⇒𝑓 𝑥 − 𝑥
′
( ) = 1 cos

For 𝑥 ∈ [− 𝑥≤ ⇒− ≤− 𝑥≤ ⇒ ≤𝑓 𝑥 ≤
𝜋 𝜋] − ≤ , , 1 cos 1 1 cos 1 0
′
( ) 2

Thus, 𝑓 𝑥 is increasing in [− 𝜋 𝜋 ] ⇒ 𝑓 𝑥 ∈ [𝑓 (− 𝜋 ) 𝑓 ( 𝜋 )]
2 2

( ) , ( ) ,

⇒𝑓 𝑥 ∈[ − 𝜋 𝜋 − ]
2 2 2 2

( ) 1 , 1

So, 𝑓 𝑥 𝑥 − 𝑥 𝑎 has a unique root in [− 𝜋 𝜋 ] if 𝑎 ∈ [ − 𝜋 𝜋 −

2 2

( ) = sin = , 1 , 1]
2 2 2 2

Q12.
𝑛 th
derivative of ( 𝑥 + 1)
𝑛 is equal to

A) (𝑛− 1)! B) ( 𝑛 + 1)!

C) 𝑛 ! D) 𝑛𝑛 [( + 1)]
𝑛− 1

Explanation:

Given 𝑦 𝑥 = ( + 1)
𝑛 On differentiation w.r.t. 𝑥
 𝑑𝑦
⇒ 𝑑𝑥 = 𝑛𝑥 ( + 1)
𝑛− 1

Again,

𝑑𝑦
⇒ 𝑑𝑥 𝑛 𝑛 − 𝑥 𝑛−
2

= ( 1)( + 1)
2
2

Similarly for 𝑛 order derivative, th

⇒ 𝑑𝑥 𝑑𝑛𝑦𝑛 𝑛 𝑛 − 𝑛 − = ( 1) ( 2) … 3.2. 1( 𝑥 + 1)
0

⇒ 𝑑𝑥 𝑑𝑛𝑦𝑛 𝑛 . = !

Q13.
𝑓 𝑥 𝑥 𝑏𝑥 𝑐
If ( ) =
2
+2 +2
2
and 𝑔𝑥
( ) = −𝑥 − 2𝑐𝑥 + 𝑏
2 2
are such that min 𝑓𝑥 ( ) > max 𝑔𝑥
( ), then the relation
between 𝑏 and is c,

A) no relation B) 0 < 𝑐 <

𝑏
2

C) 𝑐 2
< 2 𝑏 D) 𝑐 2
𝑏
> 2
2

Explanation:

𝑓𝑥 𝑥 ( ) =
2
+2 𝑏𝑥 𝑐 +2
2

Since 𝑎 > 0 graph will be concave up.

Minimum value of 𝑓 𝑥 occurs at 𝑥 − 𝑏𝑎
( ) =

⇒𝑥=− 𝑏 −𝑏
2
(2 )
=

⇒ 𝑓 (𝑥) −𝑏 𝑏 −𝑏 𝑐
2(1)

2 2
= ( ) +2 ( )+2

⇒ 𝑓 (𝑥) 𝑐 −𝑏
min
2 2
min
= 2

𝑔𝑥 ( ) = −𝑥 − 2𝑐𝑥 + 𝑏
2 2

Since 𝑎 < 0 graph will be concave down.

𝑓 𝑥 occurs at 𝑥 − 𝑏𝑎
Maximum value of ( ) =
(−2𝑐)
𝑥 −𝑐
2

⇒ = − 2( −1) =
⇒𝑔 𝑥 − −𝑐 − 𝑐 −𝑐 𝑏
( ) max = ( )
2
2 ( )+
2

⇒𝑔 𝑥 𝑐 𝑏
( ) =
2
+
2

Since 𝑓 𝑥 𝑔𝑥
max

( ) min > ( ) max

⇒ 𝑐 −𝑏2 𝑏 𝑐 2 2
>
2
+
2

⇒𝑐 𝑏 2
> 2
2
Q14.
If ℎ ( 𝑥 ) = 3 𝑓 ( 𝑥 ) 𝑓 ( − 𝑥 )∀
3
2

+ 3
2
x ∈ (−3, 4) where 𝑓 ″ (𝑥) > 0∀x ∈ (−3, 4), then ℎ (𝑥) is:

A) increasing in ( 32 , 4) B) decreasing in (−3, − 32 )

C) increasing in (− 32 , 0) D) decreasing in (0, 3

)
2

Explanation:

Given ℎ ( 𝑥 ) = 3 𝑓 ( 𝑥 ) 𝑓 ( − 𝑥 )∀𝑥 ∈
−3, 4)
2

+ 3
2
(

For checking increasing and decreasing nature of ℎ (𝑥), differentiate it.

3

ℎ′ (𝑥) = 3𝑓 ′ ( 𝑥 ) × 𝑥 − 𝑓 ′ (3 − 𝑥 )2𝑥
2
2 2

= 2𝑥 [𝑓 (
𝑥 ) − 𝑓 ′ (3 − 𝑥 )]
3 3

′ 2
2

∵ 𝑓 ′′ (𝑥) > 0
3

⇒ 𝑓 ′ (𝑥) is an increasing function, so 𝑓 ′ (𝑥 ) > 𝑓 ′ (𝑥 )

⇒𝑥 >𝑥
1 2

For increasing and decreasing nature of ℎ (𝑥), we have to take cases.

1 2

Case-1: When 𝑥 > 0

ℎ (𝑥) = 2 𝑥 [𝑓
( 𝑥 ) − 𝑓 ′ (3 − 𝑥 )]
2
′ ′ 2

ℎ (𝑥) is increasing when ℎ′ (𝑥) > 0

3

∴ 𝑓 ′ ( 𝑥 ) > 𝑓 ′ (3 − 𝑥 )
2
2

𝑥 >3−𝑥
3
2
2

𝑥
3
2
4
> 3

𝑥− 𝑥
3

(2 3) (2 + 3) > 0

𝑥 ∈ (−∞ − ) ∪ ( ∞) ,
3 3
,

since 𝑥 and 𝑥 ∈ −
2 2

> 0 ( 3, 4)

so for 𝑥 ∈ ( ) ℎ 𝑥 is increasing 3
,4 ( )

and for 𝑥 ∈ ( ) ℎ 𝑥 is decreasing.

2
3
0, , ( )

Case- : When 𝑥
2

2 < 0

ℎ 𝑥 is increasing when ℎ 𝑥
( )
′
( ) > 0

𝑓 (𝑥 ) 𝑓 ( −𝑥 )
′
2

<
′
3
2

−𝑥
3

𝑥 2

< 3
2

𝑥
3
2
4
< 3

𝑥− 𝑥
3

(2 3) (2 + 3) < 0

𝑥 ∈(− ) 3
,
3

since 𝑥 and 𝑥 ∈ −
2 2

< 0 ( 3, 4)

so for 𝑥 ∈ ( − ) ℎ 𝑥 2
3
,0 , ( ) is increasing.
Q15.
if what values of , the function𝑥 𝑓𝑥 𝑥 ( ) =
4
− 4𝑥 3
+4 𝑥 2
+ 40 is monotonic decreasing. Then the solution set of 𝑥
is

A) 0 < 𝑥 < 1
B) 1 < 𝑥 < 2

C) 2 < 𝑥 < 3 D) 4 < 𝑥 < 5

Explanation:

𝑓𝑥 𝑥−𝑥 𝑥
( ) =
4
4
3
+4
2
+ 40

𝑓 𝑥 𝑥− 𝑥 𝑥
′
( ) = 4
3
12
2
+8

For monotonic decreasing 𝑓 𝑥 ′

( ) < 0

⇒ 𝑥( 𝑥 − 𝑥 ) 4
2
12 +8 < 0

⇒ 𝑥 (𝑥 − 𝑥 ) 2
3 +2 < 0

⇒𝑥 𝑥− 𝑥− ( 1) ( 2) < 0

​
⇒ 𝑥 ∈ (−∞, 0) ∪ (1, 2)
Q16.
In its domain, 𝑓𝑥
( ) =
sin

cot
−1
−1
𝑥
𝑥 is:

A) an increasing function B) a strictly increasing function

C) a decreasing function D) a strictly decreasing function

Explanation:

Given: 𝑓𝑥 ( ) =
sin
cot
−1
−1
𝑥 , domain of 𝑓 (𝑥) is 𝑥 ∈ [−1, 1]. Differentiating, we get,
𝑥
− 𝑥)−( − 𝑥) (
𝑓 𝑥 (√ )× )
1 1
( cos
1
sin
1
× –
′ 𝑥 2
𝑥 √1– 2

− 𝑥
1–
( ) = 2
1
(cos )

⇒ 𝑓 ′ (𝑥) = 𝑥 𝑥
−1 −1
√
cos +sin

𝑥 2 −1 𝑥 2

⇒ 𝑓 ′ (𝑥) =
1– (cos )

√ π

𝑥 2 − 𝑥1
2

𝑥 𝑥
1– (cos )

(∵ cos
−1 + sin
−1 =
π
)
≤𝑥 𝑥≤
2

Since 0 2
< 1 and 0 ≤ cos− 1
π,
⇒ 𝑓 ′ (𝑥) > 0
So, given function is strictly increasing in its domain.
Q17.
A particle moves along the curve 𝑦 𝑥
=
3
2 in the first quadrant in such a way that its distance from the origin
increases at the rate of 11units per second. The value of 𝑑𝑥
𝑑𝑡 when 𝑥 = 3 is

A) 4 B) 9
2

√
C) 3 3 D) None
2

Explanation:

We have,
𝑦 𝑥 =
3
2

𝑃 𝑥 𝑦 on the curve, from origin is

Now, distance of any point
√ ( , )

𝑆 √𝑥 𝑦 𝑥 𝑥
= 2
+ 2
= 2
+ 3

⇒ 𝑑𝑆
𝑑𝑡 √ ( 𝑥 𝑥 ) ( 𝑑𝑥
𝑑𝑡 )
1 2

𝑥 𝑥 =
2 3
2 +3

(6 + 3 · 3 2 ) 𝑑𝑥
2 +

⇒ 11 = √ 1
𝑑𝑡
√
2 3
2 3 +3

⇒ 𝑑𝑥
𝑑𝑡
22 36 2
= = ×6
𝑑𝑥
33 3

⇒ 𝑑𝑡 = 4 units per section.

Q18.
If the radius af a sphere is measured as 7 𝑐𝑚 with an error of 0. 01 𝑐𝑚, then the approximate error is calculating the
volume is (use 𝜋 =
22
7
)

A) △𝑣 = 6. 16 B) △𝑣 = 6. 56

C) △𝑣 = 7. 18 D) △𝑣 = 6. 26

Explanation:

Given: The radius af a sphere is measured as 7 𝑐𝑚 with an error of 0. 01 𝑐𝑚

Let volume be 𝑣 and error of volume be △𝑣.
We know that, 𝑣 𝜋𝑟 =
4 3

⇒ 𝑑𝑣 𝜋𝑟
3

𝑑𝑟
2
= 4

⇒ 𝑑𝑣
𝑑𝑟
22
= 4× × 49
𝑑𝑣
7

⇒ 𝑑𝑟 = 616

We know that, the approximate error is given by,

△𝑣 𝑑𝑣 ⇒ △𝑣 = 𝑑𝑣 △𝑟
△𝑟 =
𝑑𝑟 𝑑𝑟 ·
 ∴ 𝛥𝑣 = 616 × 0. 01
∴ 𝛥𝑣 = 6. 16

Q19.
If 𝑦 𝑥 − 𝑎𝑥
=
3 2
+ 48 𝑥 +7 is an increasing function for all values of , then 𝑥 𝑎 lies in:
A) ( −14, 14) B) ( −12, 12)

C) (−16, 16) D) ( −21, 21)

Explanation:

Given that
𝑦 𝑥 − 𝑎𝑥 𝑥
=
3 2
+ 48 +7
𝑑𝑦
𝑑𝑥 𝑥 − 𝑎𝑥
= 3
2
2 + 48 > 0

Since, 𝑦 is an increasing function,

Therefore, Discriminant, 𝐷 < 0

We know that discriminant of a quadratic equation 𝑎𝑥 𝑏𝑥 𝑐

2
+ + = 0 is 𝐷 𝑏
=
2
− 4𝑎𝑐
⇒ 4𝑎 − 4 × 3 × 48 < 0
2

⇒ 𝑎 − 144 < 0 2

⇒ 𝑎 ∈ (−12, 12)
Q20.
If 𝑓 𝑥 𝑥 𝑥−
( ) =
3

2 (3 10), 𝑥≥ 0, then 𝑓𝑥
( ) is increasing in

A) ( −∞, −1) ∪ (1, ∞) B) [2, ∞)

C) (−∞, −1) ∪ [2, ∞) D) ( −∞, 0] ∪ (2, ∞)

Explanation:

𝑓 𝑥 𝑥 𝑥− 𝑥≥
( ) =
3

2 (3 10), 0

𝑓 𝑥 𝑥 𝑥−
′
( ) = 𝑥 3
1

2 (3 10) +
3

2 ·3

𝑥 𝑥− 𝑥
2
1
3
= 2
(3 10 + 2 )

𝑓 𝑥 𝑥 𝑥−
2
1
′ 3
( ) = 2 (5 10)

𝑥√ 𝑥 −
2
1
15
= 2
( 2)

Here, 𝑥 ≥ 𝑥− ≥
2

∈ ∞ 0& 2 0, for x [2, )

For a function to be increasing its derivative must be greater than or equal to zero.
∴Thus 𝑓 𝑥 is increasing in ∞ . ( ) [2, )
Q21.
The flower bed is to be in the shape of a circular sector of radius 𝑟 and central angle 𝜃. If the area is fixed &
perimeter is minimum, then value of 𝜃 is
Answer: 2

Explanation:

We know that, the area and length of the sector of a circle of radius 𝑟 and central angle 𝜃𝑐 are
respectively 1
𝑟 𝜃 and 𝑟𝜃
2
.

𝑟𝜃 𝑐
2

Given, area of sector is constant, let

1 2
= .

⇒ 𝜃 = 𝑟𝑐 . . . (𝑖)
2
2
2

Perimeter of the sector is 𝑃 𝑟𝜃 𝑟 = +2

⇒𝑃 = ( 𝜃 + 2) 𝑟
⇒𝑃 = 𝑟 ( 𝑟𝑐 2
+2 )
⇒𝑃 𝑟
2

𝑐
𝑟
2
= +2 .

Now, to find the minimum value of the perimeter, put

𝑑𝑃
𝑑𝑟 = 0.

⇒ 𝑑𝑃
𝑑𝑟 = 2 − 𝑟
𝑐 2
= 0

⇒ 𝑟 = 𝑐.
2

𝑑𝑝 2
𝑐 𝑟 2

𝑟 𝑐 perimeter is minimum.
And 𝑑𝑟 𝑟 𝑟 𝑟 so at
4 4 4 2
= = = > 0 =

𝑖 𝑐 𝑐
2 3 3

Now, from equation ( ) we get 𝜃𝑐 =

𝑟
2
2
=
𝑐
2
= 2.

Q22.
√
The sides of an equilateral triangle are increasing at the rate of 3 cm / sec. Find the rate at which the area of the
triangle is increasing when the side is 4 cm long.
[Enter the value excluding units]

Answer: 6

Explanation:

Let 𝑎 be the side of√triangle.

Given that, 𝑑𝑎
𝑑𝑡 = 3 cm / sec
√
Area of triangle,
√
𝐴 𝑎 =
3 2

⇒ 𝑑𝐴 𝑎 𝑑𝑎𝑑𝑡
4

𝑑𝑡
3
= 2 ·
4
 √ √
( 𝑑𝐴
𝑑𝑡 )𝑎=4 =
4
3
2 (4) · 3

∴ Rate at which area is increasing when side is 4 cm long= 6 cm 2

/ sec

Q23.
𝑎 for which the function
If the set of all values of the parameter
𝑓𝑥 ( 𝑥 − 𝑎 √ 𝑥√ ( 𝑎 𝑎 − )𝑥 increases for all 𝑥 ∈ 𝑅 and has no critical points for all
) = sin 2 8( + 1) sin + 4
2
+8 14

𝑥 ∈ 𝑅, is (−∞ −𝑚 − 𝑛) ∪ ( 𝑛 ∞) then (𝑚 𝑛 ) is (where 𝑚and𝑛 are prime numbers):

, ,
2
+
2

Answer: 29

Explanation:

Given that,
𝑓𝑥 ( ) = sin 2 𝑥− 𝑎 8( + 1) sin 𝑥 + (4 𝑎 2
+8 𝑎 − )𝑥 is an increasing function for all
14

𝑥 ∈ 𝑅.
⇒𝑓 𝑥 ′
( ) > 0

⇒ 2 cos 2𝑥 − 8 (𝑎 + 1) cos 𝑥 + (4𝑎 + 8𝑎 − 14) > 0 2

⇒ 2 (2 cos 𝑥 − 1) − 8 (𝑎 + 1) cos 𝑥 + (4𝑎 + 8𝑎 − 14) > 0

2 2

⇒ 4 cos 𝑥 − 8 (𝑎 + 1) cos 𝑥 + (4𝑎 + 8𝑎 − 16) > 0

2 2

⇒ cos 𝑥 − 2 (𝑎 + 1) cos 𝑥 + (4𝑎 + 2𝑎 − 4) > 0

2 2

⇒ cos 𝑥 − 2 (𝑎 + 1) cos 𝑥 + (𝑎 + 1) − 5 > 0

2 2

√
⇒ {cos 𝑥 − (𝑎 + 1)} − ( 5) > 0
2
2

√ √
⇒ {cos − (𝑎 + 1) + 5} {cos 𝑥 − (𝑎 + 1) − 5} > 0
√ √
⇒ {cos 𝑥 − (𝑎 + 1 − 5)} {cos 𝑥 − (𝑎 + 1 + 5)} > 0
√ √
⇒ cos 𝑥 ∈ (−∞, 𝑎 + 1 5) ∪ (𝑎 + 1 + 5, ∞)
√ √
⇒ cos 𝑥 < 𝑎√+ 1 − 5 or cos 𝑥√> 𝑎 + 1 + 5
⇒ 𝑎 + 1√− 5 > 1 or 𝑎 + √1 + 5 < −1
⇒ 𝑎 > 5 or 𝑎 < −√ 2− 5
√
⇒ 𝑎 ∈ (−∞, −2 − 5) ∪ ( 5, ∞) … (i)
According to problem
𝑎 ∈ (−∞ −𝑚 − √𝑛) ∪ (√𝑛 ∞)
, , … (ii)

Comparing the equations (i) and (ii), we get

𝑚 = 2 and 𝑛 = 5

⇒𝑚 2
+ 𝑛 2
= 4 + 25

⇒𝑚 2
+ 𝑛 2
= 29

Q24.
If 𝑓𝑥 ( ) = 3 𝑥 2
+ 15 𝑥 +5 then find the approximate value of 𝑓 (3. 02).
Answer: 77.66

Explanation:

Given
𝑓𝑥 ( ) = 3 𝑥 2
𝑥
+ 15 +5

Take 𝑥 = 3 then 𝑥 ∆𝑥 + = 3. 02

⇒ ∆𝑥 = 0. 02

When 𝑥 ,𝑦
2
= 3 = 3(3) + 15 (3) + 5

⇒𝑦 = 27 + 45 + 5 = 77

Let 𝑑𝑥 ∆𝑥 = = 0. 02

Now 𝑦 𝑥 𝑥 = 3
2
+ 15 +5
𝑑𝑦
∴ 𝑑𝑥 𝑥 = 6 + 15
𝑑𝑦 ]
∴ 𝑑𝑥 𝑥 =3 = 6 × 3 + 15 = 33

∴ 𝑑𝑦 = 𝑑𝑥𝑑𝑦 𝑑𝑥 = 33 × 0. 02 = 0. 66 = ∆𝑦
∴ 𝑓 (3. 02) = 𝑦 + ∆𝑦 = 77 + 0. 66 = 77. 66
Hence, approximate value of 𝑓 (3. 02) is 77. 66.

Q25.
Let 𝑓𝑥 ( ) = ∣( 𝑥 − (𝑥 1)
2
− 2𝑥 − 3)∣ + 𝑥 − 3, 𝑥 ∈ ℝ. If 𝑚 and 𝑀 are respectively the number of points of local
minimum and local maximum of 𝑓 in the interval (0, 4), then 𝑚 𝑀 is equal to _____.
+

Answer: 3

Explanation:

Given,
𝑓𝑥 𝑥− 𝑥 𝑥− 𝑥−
⎧ 𝑥 − (𝑥 )
( ) = |( 1) ( + 1) ( 3)| + ( 3)

≤𝑥≤
( 3)
2
, 3 4

𝑓 𝑥 ⎨ 𝑥− ( −𝑥 ) ≤𝑥
( ) = ( 3) 2
2
, 1 < 3

⎩⎧ 𝑥 − (𝑥 ) ( 𝑥 3)
2
, 0 < < 1

𝑥−𝑥 𝑥 3
2
6 , 3 < < 4

𝑓 𝑥 ⎨− 𝑥 𝑥
′
( ) = 𝑥 3
2
+6 + 2, 1 < < 3

⎩ 𝑥− − 𝑥 𝑥 3
2
6 , 0 < < 1

𝑓 ′
(3 𝑓+
) > 0 → Minimum ′
(3 ) < 0

𝑓 ′
(1 𝑓 − → Minimum
+
) > 0
′
(1 ) < 0

For 𝑥 ∈ 𝑓 𝑥 at one point → Maximum

(1, 3),
′
( ) = 0

For 𝑥 ∈ 𝑓 𝑥≠(3, 4),

′
( ) 0

For 𝑥 ∈ 𝑓 𝑥 (0, 1),

′
( )+0

So points of minima are , so 𝑚 2 = 2

and point of maxima is 1, so 𝑀 = 1

So, 𝑀 𝑚
+ = 3