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If f(x)=\cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x \cdot \cos 16 x \text { then } f^{\prime}\left(\frac{\pi}{4}\right) \text { is }

Option: 1

\sqrt{2}


Option: 2

\frac{1}{\sqrt{2}}


Option: 3

1


Option: 4

None of these


\begin{aligned} & f(x)=\frac{2 \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x \cdot \cos 16 x}{2 \sin x}=\frac{\sin 32 x}{2^5 \sin x} \\ & \therefore \quad f^{\prime}(x)=\frac{1}{32} \cdot \frac{32 \cos 32 x \cdot \sin x-\cos x \cdot \sin 32 x}{\sin ^2 x} \\ & \quad f^{\prime}\left(\frac{\pi}{4}\right)=\frac{32 \cdot \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \cdot 0}{32 \cdot\left(\frac{1}{\sqrt{2}}\right)^2}=\sqrt{2} \\ & \therefore \quad \end{aligned}

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Posted by

Nehul

Consider f(x)=\left\{\begin{array}{cl}\frac{a+3 \cos x}{x^2}, & x<0 \\ b \tan \left\{\frac{\pi}{[x+3]}\right\}, & x \geq 0\end{array}\right. where [.] represents the greatest integer function. If f(x) is continuous at x=0, then 4 b^2+a is equal to

Option: 1

1


Option: 2

2


Option: 3

0


Option: 4

-1


For x=0 and x=0+h ;[x+3]=3
\mathrm{\begin{aligned} & \therefore \quad f(0)=b \tan \frac{\pi}{3}=b \sqrt{3} \\ & \text { R.H.L. }=\lim _{h \rightarrow 0} b \tan \frac{\pi}{[h+3]}=b \tan \frac{\pi}{3}=b \sqrt{3} \\ & \text { L.H.L. }=\lim _{h \rightarrow 0} \frac{a+3 \cos (0-h)}{(-h)^2}=\lim _{h \rightarrow 0} \frac{a+3 \cos h}{h^2} \\ & =\lim _{h \rightarrow 0} \frac{a+3\left(1-\frac{h^2}{2 !}+\frac{h^4}{4 !}-\ldots\right)}{h^2} \\ & =\lim _{h \rightarrow 0}\left[\frac{(a+3)}{h^2}-\frac{3}{2}+\text { powers of } h\right] \end{aligned} }
For left hand limit to exist, we must have a+3=0
\mathrm{\therefore a=-3 \, and \, \, in\, \, that\, \, case\, \, the\, \, L.H.L. =-\frac{3}{2} }
For continuity, we have
\mathrm{\begin{aligned} & b \sqrt{3}=-\frac{3}{2}=b \sqrt{3} \\ & \Rightarrow a=-3 \text { and } b=-\frac{1}{2} \sqrt{3} \Rightarrow 4 b^2+a=0 \end{aligned} }

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Posted by

seema garhwal

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Let \small x=(9+4 \sqrt{5})^{19} , then x{x} ({⋅} denotes fractional part of x) is equal to 
 

Option: 1

2^{19}


Option: 2

3^{19}


Option: 3

1


Option: 4

None of these


\begin{aligned} & x=(9+4 \sqrt{5})^{19}=[x]+\{x\} \\ & \{x\}+(9-4 \sqrt{5})^{19}=1 \\ & x\{x\}+1=x \\ & x\{x\}=(9+4 \sqrt{5})^{19}-1 \end{aligned}

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chirag

If  C_r stands for { }^{\mathrm{n}} C_r \text {, } the sum of given series \frac{2(n / 2) !(n / 2) !}{n !}\left[C_0^2-2 C_1^2+3 C_2^2-\ldots \ldots+(-1)^n(n+1) C_n^2\right]

where n is an even positive integer, is

Option: 1

0


Option: 2

(-1)^{n / 2}(n+1)


Option: 3

(-1)^n(n+2)


Option: 4

(-1)^{n / 2}(n+2)


We have C_0^2-2 C_1^2+3 C_2^2-\ldots \ldots \ldots+(-1)^n(n+1) C_n^2=\left[C_0^2-C_1^2+C_2^2-\ldots \ldots .+(-1)^n C_n^2\right]-\left[C_1^2-2 C_2^2+3 C_3^2 \ldots \ldots+(-1)^n n \cdot C_n^2\right]

$$ \begin{aligned} & =\frac{2 \cdot \frac{n}{2} ! \frac{n}{2} !}{n !}\left[(-1)^{n / 2} \cdot\left(1+\frac{n}{2}\right) \frac{n !}{\frac{n}{2} ! \frac{n}{2} !}\right]=(-1)^{n / 2}(n+2) \\ & \end{aligned}Therefore the value of given expression

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Ritika Harsh

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The set of points where the function \mathrm{f(x)=|x-1| e^x}   is differentiable is

Option: 1

R


Option: 2

\mathrm{R-|1|}


Option: 3

\mathrm{R-|-1|}


Option: 4

\mathrm{R-|0|}


Since |x-1| is not differentiable at x=1. So.  \mathrm{f(x)=|x-1| c^{\prime}}  is not differentiable at x=1. Hence, the requried set is R-|1|.

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Posted by

Ritika Kankaria

A steel rod of length 400 cm is clamped at the middle. The frequency of the fundamental mode for the longitudinal vibrations of the rod is  2.5K Hz . Find the speed of sound in steel.

 

Option: 1

20km/s


Option: 2

25km/s


Option: 3

15km/s


Option: 4

26km/s


\frac{\lambda }{2}=distance between two connected antinode

\frac{\lambda }{2}=4\\ \lambda=8M

frequency of fundamental mode is 

f=\frac{v}{\lambda }\\2.5\times10^{3}=\frac{v}{8}\\v=8\times2.5\times10^{3}\\v=20km/s

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Posted by

Anam Khan

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If  \vec{A}=a y \hat{x}+b \times \hat{y}  then curl will be-

Option: 1

(a-b) \hat{z}


Option: 2

(b-a) \hat{z}


Option: 3

(a+b) \hat{z}


Option: 4

(a-b) \hat{x}


\vec{A}=a y \hat{x}+b x \hat{y}

{\nabla} \times \vec{A}= curl =\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ a y & b x & 0 \end{array}\right|

\begin{aligned} & \vec{\nabla} \times \vec{A}=\hat{\imath}\left[\frac{\partial}{\partial y}(0)-\frac{\partial}{\partial z} b x\right]-\hat{j}\left[\frac{\partial}{\partial x} 0-\frac{\partial}{\partial z} a y\right] \\ &+\hat{k}\left[\frac{\partial}{\partial x} b x-\frac{\partial}{\partial y} a y\right] \end{aligned}

\\vec{\nabla} \times \vec{A}=i[0]-\hat{j}[0]+\hat{k}[b-a] \\
\vec{\nabla} \times \vec{A}=(b-a) \hat{z} \cdot

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Posted by

himanshu.meshram

In the expansion of \small \left(1+x+x^3+x^4\right)^{10} the coefficient of \small x^4  is 

Option: 1

{ }^{40} \mathrm{C}_4


Option: 2

{ }^{10} \mathrm{C}_4


Option: 3

210


Option: 4

310


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Locus of the midpoint of any focal chord of \mathrm{y}^2=4 \mathrm{ax} is
 

Option: 1

\mathrm{y}^2=\mathrm{a}(\mathrm{x}-2 \mathrm{a})
 


Option: 2

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-2 \mathrm{a})

 


Option: 3

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-\mathrm{a})
 


Option: 4

\text{None of these}


Let the midpoint be \mathrm{P}(\mathrm{h}, \mathrm{k}). Equation of this chord is \mathrm{\mathrm{T}=\mathrm{S}_1}. i.e., \mathrm{\mathrm{yk}-2 \mathrm{a}(\mathrm{x}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. It must pass through (\mathrm{a}, 0)

\mathrm{\Rightarrow 2 \mathrm{a}(\mathrm{a}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. Thus required locus is \mathrm{\mathrm{y}^2=2 \mathrm{ax}-2 \mathrm{a}^2.}

Hence option 2 is correct.

 

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Posted by

SANGALDEEP SINGH

If at t=0, a travelling wave pulse on a string is described by the function y=\frac{6}{x^2+3} . What will be the wave function representing the pulse at time t, if the pulse is propagating along positive x-axis with speed 4m/s?

Option: 1

y=\frac{6}{(x+4 t)^2+3}


Option: 2

y=\frac{6}{(x-4 t)^2+3}


Option: 3

y=\frac{6}{(x- t)^2}


Option: 4

y=\frac{6}{(x- t)^2+13}


y(x, 0)=\frac{6}{x^2+3}

If v is the speed of the wave, then

\\ y(x, t)=y(x-v t, 0) \\ y(x, t)=\frac{6}{(x-v t)^2+3}\\ Here, \quad v=4 \mathrm{~m} / \mathrm{s}.\\ y(x, t)=\frac{6}{(x-4 t)^2+3}

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Posted by

sudhir.kumar

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