Practice Start Test

Differentiation

JEE Mathematics · 66 questions · Page 6 of 7 · Click an option or "Show Solution" to reveal answer

Q51
Let g:R→Rg: \mathbf{R} \rightarrow \mathbf{R} be a non constant twice differentiable function such that g′(12)=g′(32)\mathrm{g}^{\prime}\left(\dfrac{1}{2}\right)=\mathrm{g}^{\prime}\left(\dfrac{3}{2}\right). If a real valued function ff is defined as f(x)=12[g(x)+g(2−x)]f(x)=\dfrac{1}{2}[g(x)+g(2-x)], then
A f′′(x)=0f^{\prime \prime}(x)=0 for atleast two xx in (0,2)(0,2)
B f′(32)+f′(12)=1f^{\prime}\left(\dfrac{3}{2}\right)+f^{\prime}\left(\dfrac{1}{2}\right)=1
C f′′(x)=0f^{\prime \prime}(x)=0 for no xx in (0,1)(0,1)
D f′′(x)=0f^{\prime \prime}(x)=0 for exactly one xx in (0,1)(0,1)
Correct Answer
Option A
Solution
f′(x)=g′(x)−g′(2−x)2,f′(32)=g′(32)−g′(12)2=0f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0

Also

f′(12)=g′(12)−g′(32)2=0,f′(12)=0\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0
⇒f′(32)=f′(12)=0\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0
⇒rootsin⁡(12,1)\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)

and

(1,32)\left(1, \frac{3}{2}\right)
⇒f′′(x)\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})

is zero at least twice in

(12,32)\left(\frac{1}{2}, \frac{3}{2}\right)
Q52
If f(x)=∣2cos⁡4x2sin⁡4x3+sin⁡22x3+2cos⁡4x2sin⁡4xsin⁡22x2cos⁡4x3+2sin⁡4xsin⁡22x∣,f(x)=\left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x \end{array}\right|, then 15f′(0)=\dfrac{1}{5} f^{\prime}(0)= is equal to :
A 2
B 1
C 0
D 6
Correct Answer
Option C
Solution
∣2cos⁡4x2sin⁡4x3+sin⁡22x3+2cos⁡4x2sin⁡4xsin⁡22x2cos⁡4x3+2sin⁡24xsin⁡22x∣R2→R2−R1,R3→R3−R1∣2cos⁡4x2sin⁡4x3+sin⁡22x30−303−3∣f(x)=45f′(x)=0\begin{aligned} & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x \end{array}\right| \\ & \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\ & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{array}\right| \\ & \mathrm{f}(\mathrm{x})=45 \\ & \mathrm{f}^{\prime}(\mathrm{x})=0 \\ & \end{aligned}
Q53
If log⁡ey=3sin⁡−1x\log _e y=3 \sin ^{-1} x, then (1−x2)y′′−xy′(1-x^2) y^{\prime \prime}-x y^{\prime} at x=12x=\dfrac{1}{2} is equal to
A 9eπ/29 e^{\pi / 2}
B 9eπ/69 e^{\pi / 6}
C 3eπ/23 e^{\pi / 2}
D 3eπ/63 e^{\pi / 6}
Correct Answer
Option A
Solution
log⁡ey=3sin⁡−1xy=e3sin⁡−1xdydx=e3sin⁡−1x⋅31−x2\begin{aligned} &\log _e y=3 \sin ^{-1} x\\ &\begin{aligned} & y=e^{3 \sin ^{-1} x} \\ & \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}} \end{aligned} \end{aligned}
1−x2dydx=3y\sqrt{1-x^2} \frac{d y}{d x}=3 y

Again differentiate

1−x2⋅y′′−2x21−x2y′=3y′(1−x)2y′′−xy′=3y′(1−x2)\begin{aligned} & \sqrt{1-x^2} \cdot y^{\prime \prime}-\frac{2 x}{2 \sqrt{1-x^2}} y^{\prime}=3 y^{\prime} \\ & (1-x)^2 y^{\prime \prime}-x y^{\prime}=3 y^{\prime}\left(\sqrt{1-x^2}\right) \end{aligned}

So value of

3y′(1−x2)3 y^{\prime}\left(\sqrt{1-x^2}\right)

at

x=12x=\frac{1}{2}
3⋅31−x2esin⁡−1x(1−x2)=9e3π6=9eπ2\begin{aligned} & 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\ & =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}} \end{aligned}
Q54
Let f(x)=ax3+bx2+cx+41f(x)=a x^3+b x^2+c x+41 be such that f(1)=40,f′(1)=2f(1)=40, f^{\prime}(1)=2 and f′′(1)=4f^{\prime \prime}(1)=4. Then a2+b2+c2a^2+b^2+c^2 is equal to:
A 54
B 51
C 73
D 62
Correct Answer
Option B
Solution

Given the polynomial function:

f(x)=ax3+bx2+cx+41f(x) = ax^3 + bx^2 + cx + 41

We are provided the following conditions from the problem: 1.

f(1)=40f(1) = 40

2.

f′(1)=2f^{\prime}(1) = 2

3.

f′′(1)=4f^{\prime \prime}(1) = 4

First, calculate f(1)f(1):

f(1)=a(1)3+b(1)2+c(1)+41=40f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40

Simplifying, we get:

a+b+c+41=40a + b + c + 41 = 40

Therefore:

a+b+c=−1a + b + c = -1

Next, calculate the first derivative f′(x)f^{\prime}(x):

f′(x)=3ax2+2bx+cf^{\prime}(x) = 3ax^2 + 2bx + c

Given f′(1)=2f^{\prime}(1) = 2:

f′(1)=3a(1)2+2b(1)+c=2f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2

Simplifying, we get:

3a+2b+c=23a + 2b + c = 2

Next, calculate the second derivative f′′(x)f^{\prime \prime}(x):

f′′(x)=6ax+2bf^{\prime \prime}(x) = 6ax + 2b

Given f′′(1)=4f^{\prime \prime}(1) = 4:

f′′(1)=6a(1)+2b=4f^{\prime \prime}(1) = 6a(1) + 2b = 4

Simplifying, we get:

6a+2b=46a + 2b = 4

Dividing the entire equation by 2:

3a+b=23a + b = 2

We now have three equations: 1.

a+b+c=−1a + b + c = -1

2.

3a+2b+c=23a + 2b + c = 2

3.

3a+b=23a + b = 2

To solve for aa, bb, and cc, follow these steps: First, subtract the third equation from the second equation:

(3a+2b+c)−(3a+b)=2−2(3a + 2b + c) - (3a + b) = 2 - 2

Which simplifies to:

b+c=0b + c = 0

So,

c=−bc = -b

Substitute c=−bc = -b into the first equation:

(a+b−b=−1)(a + b - b = -1)

Simplifying, we get:

a=−1a = -1

Now substitute a=−1a = -1 into the third equation:

3(−1)+b=23(-1) + b = 2

Which simplifies to:

−3+b=2-3 + b = 2

Therefore:

b=5b = 5

Next, since c=−bc = -b:

c=−5c = -5

Finally, we need to find a2+b2+c2a^2 + b^2 + c^2:

a2+b2+c2=(−1)2+52+(−5)2a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2

Simplifying, we get:

1+25+25=511 + 25 + 25 = 51

Therefore, the answer is: Option B: 51

Q55
Let f(x)=x5+2ex/4f(x)=x^5+2 \mathrm{e}^{x / 4} for all x∈Rx \in \mathbf{R}. Consider a function g(x)g(x) such that (g∘f)(x)=x(g \circ f)(x)=x for all x∈Rx \in \mathbf{R}. Then the value of 8g′(2)8 g^{\prime}(2) is :
A 4
B 2
C 16
D 8
Correct Answer
Option C
Solution

Given that

(g∘f)(x)=x(g \circ f)(x) = x

for all

x∈Rx \in \mathbf{R}

. This means

g(f(x))=xg(f(x)) = x

for all

x∈Rx \in \mathbf{R}

. Differentiating both sides with respect to

xx

, we get:

g′(f(x))⋅f′(x)=1g'(f(x)) \cdot f'(x) = 1

Now, we want to find the value of

8g′(2)8g'(2)

. To do this, we need to find a value of

xx

such that

f(x)=2f(x) = 2

. Let's solve for

xx

:

x5+2ex/4=2x^5 + 2e^{x/4} = 2

By inspection, we see that

x=0x = 0

is a solution. Therefore,

f(0)=2f(0) = 2

. Now, we can substitute this into our differentiated equation:

g′(f(0))⋅f′(0)=1g'(f(0)) \cdot f'(0) = 1
g′(2)⋅f′(0)=1g'(2) \cdot f'(0) = 1

Let's find

f′(0)f'(0)

:

f′(x)=5x4+12ex/4f'(x) = 5x^4 + \frac{1}{2}e^{x/4}
f′(0)=12f'(0) = \frac{1}{2}

Substituting this back into our equation:

g′(2)⋅12=1g'(2) \cdot \frac{1}{2} = 1
g′(2)=2g'(2) = 2

Finally, we can calculate

8g′(2)8g'(2)

:

8g′(2)=8⋅2=168g'(2) = 8 \cdot 2 = \boxed{16}
Q56
If y(θ)=2cos⁡θ+cos⁡2θcos⁡3θ+4cos⁡2θ+5cos⁡θ+2y(\theta)=\dfrac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}, then at θ=π2,y′′+y′+y\theta=\dfrac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y is equal to :
A 12\dfrac{1}{2}
B 1
C 32\dfrac{3}{2}
D 2
Correct Answer
Option D
Solution
y(θ)=2cos⁡θ+cos⁡2θcos⁡3θ+4cos⁡2θ+5cos⁡θ+2=2cos⁡2θ+2cos⁡θ−14cos⁡3θ+8cos⁡2θ+2cos⁡θ−2=2cos⁡2θ+2cos⁡θ−1(2cos⁡2θ+2cos⁡θ−1)(2cos⁡θ+2)=12(1+cos⁡θ)=14cos⁡2θ/2=sec⁡2θ/24y′(θ)=14(2sec⁡θ2⋅sec⁡θ2⋅tan⁡θ2⋅12)=14sec⁡2θ2⋅tan⁡θ2\begin{aligned} & y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\ & y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\ & =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \end{aligned}
y′′(θ)=14(tan⁡θ2)(sec⁡2θ2⋅tan⁡θ2)+14sec⁡2θ2⋅sec⁡2θ2⋅12y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}
 at θ=π2,y(θ)=12,y′(θ)=12,y′′(θ)=1∴y+y′+y′′=2\begin{aligned} & \text{ at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\ & \therefore \quad y+y^{\prime}+y^{\prime \prime}=2 \end{aligned}
Q57
Suppose for a differentiable function h,h(0)=0,h(1)=1h, h(0)=0, h(1)=1 and h′(0)=h′(1)=2h^{\prime}(0)=h^{\prime}(1)=2. If g(x)=h(ex)eh(x)g(x)=h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}, then g′(0)g^{\prime}(0) is equal to:
A 4
B 5
C 3
D 8
Correct Answer
Option A
Solution

To determine

g′(0)g^{\prime}(0)

, we start by applying the chain rule and product rule to find the derivative of the given function

g(x)=h(ex)eh(x)g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}

. The product rule states that if we have two functions

u(x)u(x)

and

v(x)v(x)

, then the derivative of their product is given by:

(u(x)v(x))′=u′(x)v(x)+u(x)v′(x)(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)

Let's denote

u(x)=h(ex)u(x) = h(\mathrm{e}^x)

and

v(x)=eh(x)v(x) = \mathrm{e}^{h(x)}

. First, we need to find

u′(x)u'(x)

and

v′(x)v'(x)

. Using the chain rule, we find:

u′(x)=ddxh(ex)=h′(ex)⋅ddx(ex)=h′(ex)⋅exu'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x

Now, the derivative of

v(x)v(x)

is:

v′(x)=ddx(eh(x))=eh(x)⋅h′(x)v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x)

Using the product rule, we get the derivative of

g(x)g(x)

:

g′(x)=u′(x)v(x)+u(x)v′(x)g'(x) = u'(x) v(x) + u(x) v'(x)

Substituting

u(x)u(x)

,

v(x)v(x)

,

u′(x)u'(x)

, and

v′(x)v'(x)

into the above expression, we get:

g′(x)=(h′(ex)ex)eh(x)+(h(ex))(eh(x)h′(x))g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right)

Next, we need to evaluate this at

x=0x = 0

: First, we know that:

h(0)=0h(0) = 0
h(1)=1h(1) = 1
h′(0)=2h'(0) = 2
h′(1)=2h'(1) = 2

Substituting

x=0x = 0

into the expressions, we get:

u(0)=h(e0)=h(1)=1u(0) = h(\mathrm{e}^0) = h(1) = 1
v(0)=eh(0)=e0=1v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1
u′(0)=h′(e0)e0=h′(1)⋅1=2u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2
v′(0)=eh(0)h′(0)=e0⋅2=2v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2

Therefore, evaluating

g′(0)g'(0)

:

g′(0)=(u′(0)v(0))+(u(0)v′(0))g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right)
g′(0)=(2⋅1)+(1⋅2)g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right)
g′(0)=2+2=4g'(0) = 2 + 2 = 4

Thus, the value of

g′(0)g^{\prime}(0)

is 4, which corresponds to Option A. The correct answer is Option A: 4.

Q58
 If f(x)={x3sin⁡(1x),x≠00,x=0, then \text{ If } f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0 & , x=0 \end{array}\right. \text{, then }
A f′′(0)=0f^{\prime \prime}(0)=0
B f′′(0)=1f^{\prime \prime}(0)=1
C f′′(2π)=24−π22πf^{\prime \prime}\left(\dfrac{2}{\pi}\right)=\dfrac{24-\pi^2}{2 \pi}
D f′′(2π)=12−π22πf^{\prime \prime}\left(\dfrac{2}{\pi}\right)=\dfrac{12-\pi^2}{2 \pi}
Correct Answer
Option C
Solution

Given the function: f(x)={x3sin⁡(1x),x≠00,x=0 f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\dfrac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. we need to find its second derivative at specific points.

First, let’s compute the first derivative f′(x) f^{\prime}(x) : f′(x)=3x2sin⁡(1x)−xcos⁡(1x) f^{\prime}(x) = 3x^2 \sin \left( \dfrac{1}{x} \right) - x \cos \left( \dfrac{1}{x} \right) Next, the second derivative f′′(x) f^{\prime \prime}(x) is: f′′(x)=6xsin⁡(1x)−3xcos⁡(1x)−cos⁡(1x)−1xsin⁡(1x) f^{\prime \prime}(x) = 6x \sin \left(\dfrac{1}{x}\right) - 3x \cos \left(\dfrac{1}{x}\right) - \cos \left(\dfrac{1}{x}\right) - \dfrac{1}{x} \sin \left(\dfrac{1}{x}\right) Therefore, evaluating the second derivative at x=2π x = \dfrac{2}{\pi} : f′′(2π)=6(2π)sin⁡(π2)−3(2π)cos⁡(π2)−cos⁡(π2)−π2sin⁡(π2) f^{\prime \prime} \left(\dfrac{2}{\pi}\right) = 6 \left( \dfrac{2}{\pi} \right) \sin \left(\dfrac{\pi}{2}\right) - 3 \left( \dfrac{2}{\pi} \right) \cos \left(\dfrac{\pi}{2}\right) - \cos \left( \dfrac{\pi}{2} \right) - \dfrac{\pi}{2} \sin \left( \dfrac{\pi}{2} \right) Since sin⁡(π2)=1\sin(\dfrac{\pi}{2}) = 1 and cos⁡(π2)=0\cos(\dfrac{\pi}{2}) = 0, this simplifies to: f′′(2π)=12π−π2=24−π22π f^{\prime \prime} \left( \dfrac{2}{\pi} \right) = \dfrac{12}{\pi} - \dfrac{\pi}{2} = \dfrac{24 - \pi^2}{2\pi} Finally, note that f′(0) f^{\prime}(0) is not defined, as it involves terms like 1x\dfrac{1}{x} when x=0 x = 0 .

Q59
Let f:(−∞,∞)−{0}→Rf:(-\infty, \infty)-\{0\} \rightarrow \mathbb{R} be a differentiable function such that f′(1)=lim⁡a→∞a2f(1a)f^{\prime}(1)=\lim \limits_{a \rightarrow \infty} a^2 f\left(\dfrac{1}{a}\right). Then lim⁡a→∞a(a+1)2tan⁡−1(1a)+a2−2log⁡ea\lim \limits_{a \rightarrow \infty} \dfrac{a(a+1)}{2} \tan ^{-1}\left(\dfrac{1}{a}\right)+a^2-2 \log _e a is equal to
A 52+π8\dfrac{5}{2}+\dfrac{\pi}{8}
B 38+π4\dfrac{3}{8}+\dfrac{\pi}{4}
C 34+π8\dfrac{3}{4}+\dfrac{\pi}{8}
D 32+π4\dfrac{3}{2}+\dfrac{\pi}{4}
Correct Answer
Option A
Solution

Let

f′(1)=kf^{\prime}(1)=k
⇒lim⁡x→0f(x)x2=k(00)\Rightarrow \quad \lim \limits_{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)
lim⁡x→0f′(x)2x=lim⁡x→0f′′(x)2=k⇒f′′(0)=2k\begin{aligned} & \lim \limits_{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\ \Rightarrow & f^{\prime \prime}(0)=2 k \end{aligned}

Given information is not complete.

Q60
Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\sin x \cos y)(f(2 x+2 y)-f(2 x-2 y))=(\cos x \sin y)(f(2 x+2 y)+f(2 x-2 y)), for all x,y∈Rx, y \in \mathbf{R}. If f′(0)=12f^{\prime}(0)=\dfrac{1}{2}, then the value of 24f′′(5π3)24 f^{\prime \prime}\left(\dfrac{5 \pi}{3}\right) is :
A 2
B 3
C −-3
D −-2
Correct Answer
Option C
Solution
sin⁡(x−y)f(2x+2y)=f(2x−2y)sin⁡(x+y)f(2x+2y)sin⁡(x+y)=f(2x−2y)sin⁡(x−y)=k( say )f(2x+2y)=ksin⁡(x+y)f(2x)=5sin⁡x(∵y=0)f(x)=ksin⁡x2f′(x)=k2cos⁡x2f′(0)=12⇒k=1f(x)=sin⁡x2⇒f′(x)=12cos⁡x2f′′(x)=−14sin⁡x224f′′(5π3)=−3\begin{aligned} & \sin (x-y) f(2 x+2 y)=f(2 x-2 y) \sin (x+y) \\ & \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)}=k(\text{ say }) \\ & f(2 x+2 y)=k \sin (x+y) \\ & f(2 x)=5 \sin x \quad(\because y=0) \\ & f(x)=k \sin \frac{x}{2} \\ & f^{\prime}(x)=\frac{k}{2} \cos \frac{x}{2} \\ & f^{\prime}(0)=\frac{1}{2} \Rightarrow k=1 \\ & f(x)=\sin \frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2} \\ & f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2} \\ & 24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=-3 \end{aligned}
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