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##### The range of values of $a$ for which the line $y+x=0$ bisectstwochords drawn fromapoint $left(frac{1+sqrt{2} a}{2}, frac{1-sqrt{2} a}{2}ight)$ to the circle $2 x^{2}+2 y^{2}-(1+sqrt{2} a) x-(1-sqrt{2} a) y=0$ is(A) $(-infty,-2) cup(2, infty)$(B) $(-2,2)$(C) $(2, infty)$(D) none of these

The range of values of $a$ for which the line $y+x=0$ bisectstwochords drawn fromapoint $left(frac{1+sqrt{2} a}{2}, frac{1-sqrt{2} a}{2} ight)$ to the circle $2 x^{2}+2 y^{2}-(1+sqrt{2} a) x-(1-sqrt{2} a) y=0$ is (A) $(-infty,-2) cup(2, infty)$ (B) $(-2,2)$ (C) $(2, infty)$ (D) none of these...

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##### $int_{0}^{pi}|1+2 cos x| d x$ is equal to(A) $frac{pi}{3}-2 sqrt{3}$(B) $frac{pi}{3}-sqrt{3}$(C) $frac{pi}{3}+sqrt{3}$(D) $frac{pi}{3}+2 sqrt{3}$

$int_{0}^{pi}|1+2 cos x| d x$ is equal to (A) $frac{pi}{3}-2 sqrt{3}$ (B) $frac{pi}{3}-sqrt{3}$ (C) $frac{pi}{3}+sqrt{3}$ (D) $frac{pi}{3}+2 sqrt{3}$...

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##### Assertion: The solution of the equation $x sin heta d heta+$ $left(x^{3}-2 x^{2} cos heta+cos hetaight) d x=0$ is $2 cos heta=x+c x e^{-x 2}$Reason: Integrating factor $=frac{e^{x^{2}}}{x}$

Assertion: The solution of the equation $x sin heta d heta+$ $left(x^{3}-2 x^{2} cos heta+cos heta ight) d x=0$ is $2 cos heta=x+c x e^{-x 2}$ Reason: Integrating factor $=frac{e^{x^{2}}}{x}$...

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##### The locus of the centres of the circles which touch the two circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4 a x$ externally is(A) $12 x^{2}-4 y^{2}-24 a x+9 a^{2}=0$(B) $12 x^{2}+4 y^{2}-24 a x+9 a^{2}=0$(C) $12 x^{2}-4 y^{2}+24 a x+9 a^{2}=0$(D) none of these

The locus of the centres of the circles which touch the two circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4 a x$ externally is (A) $12 x^{2}-4 y^{2}-24 a x+9 a^{2}=0$ (B) $12 x^{2}+4 y^{2}-24 a x+9 a^{2}=0$ (C) $12 x^{2}-4 y^{2}+24 a x+9 a^{2}=0$ (D) none of these...

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##### If $f(x)$ is differentiable on $(-1,1)$ and $f^{prime}(0)=1$, then $f^{prime}(x)$ is equal to(A) $frac{1}{sqrt{1-x^{2}}}$(B) $-frac{1}{sqrt{1-x^{2}}}$(C) $frac{1}{sqrt{1+x^{2}}}$(D) $-frac{1}{sqrt{1+x^{2}}}$

If $f(x)$ is differentiable on $(-1,1)$ and $f^{prime}(0)=1$, then $f^{prime}(x)$ is equal to (A) $frac{1}{sqrt{1-x^{2}}}$ (B) $-frac{1}{sqrt{1-x^{2}}}$ (C) $frac{1}{sqrt{1+x^{2}}}$ (D) $-frac{1}{sqrt{1+x^{2}}}$...

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##### If a circle passes through the points of intersection of the coordinate axes with the lines $lambda x-y+1=0$ and $x-2 y+3=0$, then the value of $lambda$ is(A) 2(B) 1(C) $-1$(D) $-2$

If a circle passes through the points of intersection of the coordinate axes with the lines $lambda x-y+1=0$ and $x-2 y+3=0$, then the value of $lambda$ is (A) 2 (B) 1 (C) $-1$ (D) $-2$...

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##### $a$ and $b$ are mutually perpendicular unit vectors. If $r$ is a vector satisfying $r cdot a=0, r cdot b=1$ and $[r a b]=1$, then $r$ is(A) $a imes b+b$(B) $a+(a imes b)$(C) $b+(a imes b)$(D) $a imes ar{b}+a$

$a$ and $b$ are mutually perpendicular unit vectors. If $r$ is a vector satisfying $r cdot a=0, r cdot b=1$ and $[r a b]=1$, then $r$ is (A) $a imes b+b$ (B) $a+(a imes b)$ (C) $b+(a imes b)$ (D) $a imes ar{b}+a$...

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##### A circle touches the line $y=x$ at $a$ point $P$ such that $O P=4 sqrt{2}$, where $O$ is the origin. The circle contains the point $(-10,2)$ in its interior and the length of its chord on the line $x+y=0$ is $6 sqrt{2}$. The equation of the circle is(A) $x^{2}+y^{2}+18 x-2 y+32=0$(B) $x^{2}+y^{2}-18 x-2 y+32=0$(C) $x^{2}+y^{2}+18 x+2 y+32=0$(D) none of these

A circle touches the line $y=x$ at $a$ point $P$ such that $O P=4 sqrt{2}$, where $O$ is the origin. The circle contains the point $(-10,2)$ in its interior and the length of its chord on the line $x+y=0$ is $6 sqrt{2}$. The equation of the circle is (A) $x^{2}+y^{2}+18 x-2 y+32=0$ (B) $x^{2}+y^{2}-...

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##### The function $f(x)$ is equal to(A) $cos ^{-1} x$(B) $sin ^{-1} x$(C) $an ^{-1} x$(D) $sec ^{-1} x$

The function $f(x)$ is equal to (A) $cos ^{-1} x$ (B) $sin ^{-1} x$ (C) $ an ^{-1} x$ (D) $sec ^{-1} x$...

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##### The order and degree of the differential equation $left(1+3 frac{d y}{d x}ight)^{2 / 3}=4 frac{d^{3} y}{d x^{3}}$ are $quad$ (A) $left(1, frac{2}{3}ight)$(B) $(3,1)$(C) $(3,3)$(D) $(1,2)$

The order and degree of the differential equation $left(1+3 frac{d y}{d x} ight)^{2 / 3}=4 frac{d^{3} y}{d x^{3}}$ are $quad$ (A) $left(1, frac{2}{3} ight)$ (B) $(3,1)$ (C) $(3,3)$ (D) $(1,2)$...

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