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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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Posted by

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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Posted by

Ritika Harsh

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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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The equation of normal at the point \left ( 0,3 \right ) of the ellipse \mathrm{9 x^2+5 y^2=45} is 

Option: 1

\mathrm{y-3=0}


Option: 2

\mathrm{y+3=0}


Option: 3

x-axis


Option: 4

y-axis


For \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} ,  equation of normal at point\mathrm{\left(x_1, y_1\right) \text {, is } \frac{\left(x-x_1\right) a^2}{x_1}=\frac{\left(y-y_1\right) b^2}{y_1}}
\mathrm{\text { Here, }\left(x_1, y_1\right)=(0,3) \text { and } a^2=5, b^2=9 \text {, Therefore } \frac{(x-0)}{0} \cdot 5=\frac{(y-3)}{3} \cdot 9 \text { or } x=0 \text { i.e., } y \text {-axis. }}

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Posted by

rishiraj

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The locus of the point of intersection of the perpendicular tangents to the ellipse \mathrm{\frac{x^2}{9}+\frac{y^2}{4}=1} is
 

Option: 1

\mathrm{x^2+y^2=9}


Option: 2

\mathrm{x^2+y^2=4}


Option: 3

\mathrm{x^2+y^2=13}


Option: 4

\mathrm{x^2+y^2=5}


The locus of point of intersection of two perpendicular tangents drawn on the ellipse is \mathrm{x^2+y^2=a^2+b^2} which is called “director circle”.
Given ellipse is \mathrm{\frac{x^2}{9}+\frac{y^2}{4}=1}
\mathrm{\therefore \text { Locus is } x^2+y^2=9+4 \text {, i.e. } x^2+y^2=13}

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Posted by

Kuldeep Maurya

\mathrm{\text { Solve } \lim _{x \rightarrow 0}(1-\sqrt{\cos 4 x}) \cot ^2 x}

Option: 1

1


Option: 2

4


Option: 3

5


Option: 4

2


Note,
\mathrm{1-\sqrt{\cos 4 x}=\frac{1-\cos 4 x}{1+\sqrt{\cos 4 x}}=\frac{1-\left(\cos ^2 2 x-\sin ^2 2 x\right)}{1+\sqrt{\cos 4 x}}=\frac{2 \sin ^2 2 x}{1+\sqrt{\cos 4 x}}=\frac{8 \sin ^2 x \cos ^2 x}{1+\sqrt{\cos 4 x}}}
\mathrm{\text { Consequently, for the limit as } x \rightarrow 0 \text {, we have }}
\mathrm{(1-\sqrt{\cos 4 x}) \cot ^2 x=\frac{8 \sin ^2 x \cos ^2 x}{1+\sqrt{\cos 4 x}} \cdot \frac{\cos ^2 x}{\sin ^2 x}=\frac{8 \cos ^4 x}{1+\sqrt{\cos 4 x}} \rightarrow \frac{8 \cdot 1^4}{1+\sqrt{1}}=\frac{8}{2}=4}

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Posted by

Deependra Verma

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The magnetic field in a plane electromagnetic wave is given by \mathrm{\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \sin \left(0.5 \times 10^3 \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \mathrm{T}}. The electromagnetic wave is:
 

Option: 1

a visible light

 


Option: 2

an infrared rays
 


Option: 3

a microwave
 


Option: 4

a radio wave


\mathrm{\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \sin \left(0.5 \times 10^3 \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \mathrm{T}}

Comparing with the standard equation, we get

\mathrm{\mathrm{B}_{\mathrm{y}}=\mathrm{B}_0 \sin (\mathrm{kx}+\omega \mathrm{t}) }

\mathrm{\mathrm{k}=0.5 \times 10^3 }

\mathrm{ \Rightarrow \lambda=\frac{2 \pi}{0.5 \times 10^3}=0.01256 \mathrm{~m}}

The wavelength range of microwaves is \mathrm{ 10^{-3} \mathrm{~m}\: to\: 0.3 \mathrm{~m}}. The wavelength of this wave lies between \mathrm{ 10^{-3} \mathrm{~m}\: to \: 0.3 \mathrm{~m}}, so the equation represents a microwaves.

Hence option 3 is correct.

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Posted by

Sayak

The locus of the point from which the length of the tangents to the circle \mathrm{x^2+y^2-4=0\: and \: x^2+y^2-8 x+15=0} are equal is given by the equation
 

Option: 1

\mathrm{x=3}
 


Option: 2

\mathrm{y=5}
 


Option: 3

\mathrm{3 y-7=0}
 


Option: 4

\mathrm{8 x-19=0}


Let the point be (\alpha, \beta)

\mathrm{ \alpha^2+\beta^2-4=\alpha^2+\beta^2-8 \alpha+15 \Rightarrow 8 \alpha=19 . }

Hence the locus of \mathrm{ (\alpha, \beta)\: is \: 8 x=19. }

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Posted by

qnaprep

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