Prove that
Question: Prove that $\frac{\cos 4 x \sin 3 x-\cos 2 x \sin x}{\sin 4 x \sin x+\cos 6 x \cos x}=\tan 2 x$ Solution: $=\frac{\cos 4 x \sin 3 x-\cos 2 x \sin x}{\sin 4 x \sin x+\cos 6 x \cos x}$ $=\frac{2 \cos 4 x \sin 3 x-2 \cos 2 x \sin x}{2 \sin 4 x \sin x+2 \cos 6 x \cos x}$ $=\frac{\sin (4 x+3 x)-\sin (4 x-3 x)-\{\sin (2 x+x)-\sin (2 x-x)\}}{\cos (4 x-x)-\cos (4 x+x)+\cos (6 x+x)+\cos (6 x-x)}$ $=\frac{\sin 7 x+\sin x-\sin 3 x+\sin x}{\cos 3 x-\cos 5 x+\cos 7 x+\cos 5 x}$ $=\frac{\sin 7 x-\si...
Read More →if |A|=2 and |B|=4 , then match the relations in column
Question: if |A|=2 and |B|=4, then match the relations in column I with the angle between between A and B in column II. Solution: (a) matches with(ii) (b) matches with (i) (c) matches with (iv) (d) matches with(iii)...
Read More →Prove that
Question: Prove that $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$ Solution: =Sin3x+sin2x-sinx $=(\sin 3 x-\sin x)+\sin 2 x$ $=\left(2 \cos \frac{3 x+x}{2} \sin \frac{3 x-2 x}{2}\right)+\sin 2 x$ $=2 \cos 2 x \sin x+\sin 2 x$ $=2 \cos 2 x \sin x+2 \sin x \cos x$ $=2 \sin x(\cos 2 x+\cos x)$ $=2 \sin x\left(2 \cos \frac{2 x+x}{2} \cos \frac{2 x-x}{2}\right)$ $=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$ Using the formula, $\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{...
Read More →Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
Question: Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals (i) $f(x)=3+(x-2)^{2 / 3}$ on $[1,3]$ (ii) $f(x)=[x]$ for $-1 \leq x \leq 1$, where $[x]$ denotes the greatest integer not exceeding $x$ (iii) $f(x)=\sin \frac{1}{x}$ for $-1 \leq x \leq 1$ (iv) $f(x)=2 x^{2}-5 x+3$ on $[1,3]$ (v) $f(x)=x^{2 / 3}$ on $[-1,1]$ (vi) $f(x)= \begin{cases}-4 x+5, 0 \leq x \leq 1 \\ 2 x-3, 1x \leq 2\end{cases}$ Solution: (i) The given function is $f(x)=3+(x-2)...
Read More →Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
Question: Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals (i) $f(x)=3+(x-2)^{2 / 3}$ on $[1,3]$ (ii) $f(x)=[x]$ for $-1 \leq x \leq 1$, where $[x]$ denotes the greatest integer not exceeding $x$ (iii) $f(x)=\sin \frac{1}{x}$ for $-1 \leq x \leq 1$ (iv) $f(x)=2 x^{2}-5 x+3$ on $[1,3]$ (v) $f(x)=x^{2 / 3}$ on $[-1,1]$ (vi) $f(x)= \begin{cases}-4 x+5, 0 \leq x \leq 1 \\ 2 x-3, 1x \leq 2\end{cases}$ Solution: (i) The given function is $f(x)=3+(x-2)...
Read More →A fighter plane is flying horizontally at an altitude
Question: A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb in order to attack the target? Solution: Given, u = 720 km/h = 200 m/s Let t be the time at which pilot drops the bomb, then Q will be the point vertically above the target T. We also know that the horizontal velocity of the bomb is equal to the velocity of the fighter plane. But the vertical component is zero. When the bomb...
Read More →Prove that
Question: Prove that $\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)$ Solution: $=\frac{\sin x+\sin y}{\sin x-\sin y}$ $=\frac{2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}}$ $=\tan \frac{x+y}{2} \cot \frac{x-y}{2}$ Using the formula, $\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ $\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$...
Read More →Prove that
Question: Prove that $\frac{\cos x+\cos y}{\cos y-\cos x}=\cot \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)$ Solution: $=\frac{\cos x-\cos y}{\cos y-\cos x}$ $=\frac{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}{-2 \sin \frac{x+y}{2} \sin \frac{y-x}{2}}$ $=\frac{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}}$ $=\frac{\cos \frac{x+y}{2} \cos \frac{x-y}{2}}{\sin \frac{x+y}{2} \sin \frac{x-y}{2}}$ $=\cot \frac{x+y}{2} \cot \frac{x-y}{2}$ Using the formula...
Read More →A, B, and C are three non-collinear,
Question: A, B, and C are three non-collinear, non co-planar vectors. What can you say about direction of A(BC)? Solution: The direction of product of vectors B and C is perpendicular to the plane that contains vector B and C which is based on the Right hand grip rule. The direction of vector A(BC) is perpendicular to vector A and is in the plane which contains vectors B and C and is again based on the Right hand grip rule....
Read More →A football is kicked into the air
Question: A football is kicked into the air vertically upwards. What is its (a) acceleration (b) velocity at the highest point Solution: (a) The acceleration of the football will be downward when it is kicked upwards as only gravitational force is acting on it. (b) The football will have its highest velocity when vy= 0....
Read More →For two vectors A and B,
Question: For two vectors A and B, $|A+B|=|A-B|$ is always true when (a) $|A|=|B| \neq 0$ (b) $A \perp B$ (c) $|A|=|B| \neq 0$ and $\mathrm{A}$ and $\mathrm{B}$ are parallel or antiparallel (d)$|A|$ or $|B|$ is zero Solution: The correct answer is (b) $A \perp B$ and (d) when either $|A|$ or $|B|$ is zero...
Read More →Prove that
Question: Prove that $\frac{\sin 2 x-\sin 2 y}{\cos 2 y-\cos 2 x}=\cot (x+y)$ Solution: $=\frac{\sin 2 x-\sin 2 y}{\cos 2 y-\cos 2 x}$ $=\frac{2 \cos \frac{2 x+2 y}{2} \sin \frac{2 x-2 y}{2}}{-2 \sin \frac{2 x+2 y}{2} \sin \frac{2 y-2 x}{2}}$ $=\frac{\cos (x+y) \sin (x-y)}{\sin (x+y) \sin (x-y)}$ $=\frac{\cos (x+y)}{\sin (x+y)}$ $=\cot (x+y)$ Using the formula, $\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$ $\sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$...
Read More →Prove that
Question: Prove that $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)$ Solution: $=(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}$ $=\left(-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right)^{2}+\left(2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right)^{2}$ $=4 \sin ^{2}\left(\frac{x-y}{2}\right)\left(\sin ^{2}\left(\frac{x-y}{2}\right)+\cos ^{2}\left(\frac{x-y}{2}\right)\right)$ $=4 \sin ^{2}\left(\frac{x-y}{2}\right)$ Using the formula, $\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \...
Read More →For a particle performing uniform circular motion,
Question: For a particle performing uniform circular motion, choose the correct statement from the following: (a) magnitude of particle velocity (speed) remains constant (b) particle velocity remains directed perpendicular to radius vector (c) direction of acceleration keeps changing as particle moves (d) angular momentum is constant in magnitude but direction keep changing Solution: The correct answer is (a) magnitude of particle velocity (speed) remains constant, (b) particle velocity remains ...
Read More →Prove that
Question: Prove that $(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x=0$ Solution: $=(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x$ $=\left(2 \sin \frac{3 x+x}{2} \cos \frac{3 x-x}{2}\right) \sin x+\left(-2 \sin \frac{3 x+x}{2} \sin \frac{3 x-x}{2}\right) \cos x$ $=(2 \sin 2 x \cos x) \sin x-(2 \sin 2 x \sin x) \cos x$ $=0 .$ Using the formula, $\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ $\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$...
Read More →Following are four different relations about
Question: Following are four different relations about displacement, velocity, and acceleration for the motion of a particle in general. Choose the incorrect one (s): (a) vav= 1/2 [v(t1) + v(t2)] (b) vav= r(t2)-r(t1)/t2-t2 (c) r = 1/2 [v(t2)-v(t1)](t2-t1) (d) aav= v(t2)-v(t1)/t2-t1 Solution: The correct answer is (a) vav= 1/2 [v(t1) + v(t2)] and (c) r = 1/2 [v(t2)-v(t1)](t2-t1)...
Read More →Find the value of c prescribed by Lagrange's mean value theorem for the function
Question: Find the value ofcprescribed by Lagrange's mean value theorem for the function $f(x)=\sqrt{x^{2}-4}$ defined on $[2,3]$ Solution: We have $f(x)=\sqrt{x^{2}-4}$ Here, $f(x)$ will exist, if $x^{2}-4 \geq 0$ $\Rightarrow x \leq-2$ or $x \geq 2$ Since for each $x \in[2,3]$, the function $f(x)$ attains a unique definite value, $f(x)$ is continuous on $[2,3]$. Also, $f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}}(2 x)=\frac{x}{\sqrt{x^{2}-4}}$ exists for all $x \in(2,3)$. So, $f(x)$ is differentia...
Read More →Find the value of c prescribed by Lagrange's mean value theorem for the function
Question: Find the value ofcprescribed by Lagrange's mean value theorem for the function $f(x)=\sqrt{x^{2}-4}$ defined on $[2,3]$ Solution: We have $f(x)=\sqrt{x^{2}-4}$ Here, $f(x)$ will exist, if $x^{2}-4 \geq 0$ $\Rightarrow x \leq-2$ or $x \geq 2$ Since for each $x \in[2,3]$, the function $f(x)$ attains a unique definite value, $f(x)$ is continuous on $[2,3]$. Also, $f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}}(2 x)=\frac{x}{\sqrt{x^{2}-4}}$ exists for all $x \in(2,3)$. So, $f(x)$ is differentia...
Read More →Prove that
Question: Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$ Solution: L.H.S $\cot 4 x(\sin 5 x+\sin 3 x)$ $=\cot 4 x\left(2 \sin \frac{5 x+3 x}{2} \cos \frac{5 x-3 x}{2}\right)$ $=\cot 4 x(2 \sin 4 x \cos x)$ $=\frac{\cos 4 x}{\sin 4 x}(2 \sin 4 x \cos x)$ $=2 \cos 4 x \cos x$ R.H.S $\cot x(\sin 5 x-\sin 3 x)$ $=\cot x\left(2 \cos \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}\right)$ $=\cot x(2 \cos 4 x \sin x)$ $=\frac{\cos x}{\sin x}(2 \cos 4 x \sin x)$ $=2 \cos 4 x \cos x$ L.H.S=R...
Read More →Prove that
Question: Prove that $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$ Solution: $=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$ $=\frac{2 \sin \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}+2 \sin \frac{9 x+3 x}{2} \cos \frac{9 x-3 x}{2}}{2 \cos \frac{7 x^{2}+5 x}{2} \cos \frac{7 x^{2}-5 x}{2}+2 \cos \frac{9 x^{2}+3 x}{2} \cos \frac{9 x^{2}-3 x}{2}}$ $=\frac{2 \sin 6 x \cos x+2 \sin 6 x \cos 3 x}{2 \cos 6 x \c...
Read More →Two particles are projected in air with speed
Question: Two particles are projected in air with speed v0, at angles 1and 2to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (a) angle of project: q1 q2 (b) time of flight: T1 T2 (c) horizontal range: R1 R2 (d) total energy: U1 U2 Solution: The correct answer is (a) angle of project: q1 q2and (b) time of flight: T1 T2...
Read More →Prove the following
Question: $|A+B|=|A|$. This necessarily implies (a) B = 0 (b) A, B are antiparallel (c) A, B are perpendicular (d) A.B 0 Solution: The correct answer is (a) B = 0...
Read More →If the value of c prescribed in Rolle's theorem for the function
Question: If the value ofcprescribed in Rolle's theorem for the function $f(x)=2 x(x-3)^{n}$ on the interval $[0,2 \sqrt{3}]$ is $\frac{3}{4}$, write the value of $n$ (a positive integer). Solution: We have $f(x)=2 x(x-3)^{n}$ Differentiating the given function with respect tox,we get $f^{\prime}(x)=2\left[x n(x-3)^{n-1}+(x-3)^{n}\right]$ $\Rightarrow f^{\prime}(x)=2(x-3)^{n}\left[\frac{x n}{(x-3)}+1\right]$ $\Rightarrow f^{\prime}(c)=2(c-3)^{n}\left[\frac{c n}{(c-3)}+1\right]$ Given: $f^{\prime...
Read More →If the value of c prescribed in Rolle's theorem for the function
Question: If the value ofcprescribed in Rolle's theorem for the function $f(x)=2 x(x-3)^{n}$ on the interval $[0,2 \sqrt{3}]$ is $\frac{3}{4}$, write the value of $n$ (a positive integer). Solution: We have $f(x)=2 x(x-3)^{n}$ Differentiating the given function with respect tox,we get $f^{\prime}(x)=2\left[x n(x-3)^{n-1}+(x-3)^{n}\right]$ $\Rightarrow f^{\prime}(x)=2(x-3)^{n}\left[\frac{x n}{(x-3)}+1\right]$ $\Rightarrow f^{\prime}(c)=2(c-3)^{n}\left[\frac{c n}{(c-3)}+1\right]$ Given: $f^{\prime...
Read More →Three vectors A, B, and C add up to zero.
Question: Three vectors A, B, and C add up to zero. Find which is false (a) vector (AB)C is not zero unless vectors B, C are parallel (d) vector (AB).C is not zero unless vectors B, C are parallel (c) if vectors A, B, C define a plane, (AB)C is in that plane (d) (AB).C = $|A||B||C|$ such that $\mathrm{C}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}$ Solution: The correct answer is c) if vectors A, B, C define a plane, $(A \times B) C$ is in that plane and d) $(A \times B) . C=$ $|A \| B||C|$ such that $\ma...
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