Find the area of the triangle, the equations
Question: Find the area of the triangle, the equations of whose sides are y = x, y = 2x and y 3x = 4. Solution: The given equations are $y=x \ldots$ (i) $y=2 x \ldots$ (ii) and $y-3 x=4 \ldots$ (iii) Let eq. (i), (ii) and (iii) represents the sides $A B, B C$ and $A C$ respectively of $\triangle A B C$ From eq. (i) and (ii), we get $x=0$ and $y=0$ Thus, $A B$ and $B C$ intersect at $(0,0)$ Solving eq. (ii) and (iii), we get $\mathrm{y}=2 \mathrm{x} \ldots$ (ii) and $y-3 x=4 \ldots$ (iii) Putting...
Read More →The value of cos 1° cos 2° cos 3°…
Question: The value of cos 1 cos 2 cos 3 cos 179 isA.1/2B. 0C. 1D. 1 Solution: B. 0 Explanation: According to the question, Since cos90 =0 We get, ⇒cos 1 cos 2 cos 3 cos90 cos 179 = 0 Thus, option (B) 0 is the correct answer....
Read More →The value of (1 – tan215o)/(1 + tan215o) is
Question: The value of (1 tan215o)/(1 + tan215o) is A. 1 B.3 C. 3/2 D. 2 Solution: C. 3/2 Explanation: According to the question, Let = 15⇒2 = 30 Now, since we know that, $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ $\Rightarrow \cos 30^{\circ}=\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}$ Thus, option (C) 3/2 is the correct answer....
Read More →The value of tan 1° tan 2° tan 3°… tan 89° is
Question: The value of tan 1 tan 2 tan 3 tan 89 isA. 0B. 1C. D. Not defined Solution: B. 1 Explanation: According to the question, tan 1 tan 2 tan 3 tan 89 = tan 1 tan 2 tan 45 tan (90-44) tan(90-43)tan (90-1) = tan 1tan 2 tan 45cot 44cot 43cot 1 [∵tan (90-)=cot ] = tan 1 cot 1 tan 2 cot 2tan45 tan 89 cot 89 =1.1.1 = 1 Thus, option (B) 1 is the correct answer....
Read More →Find the intervals in which
Question: Find the intervals in which $f(x)=(x+2) e^{-x}$ is increasing or decreasing ? Solution: we have, $f(x)=(x+2) e^{-x}$ $f^{\prime}(x)=e^{-x}-e^{-x}(x+2)$ $=e^{-x}(1-x-2)$ $=-e^{-x}(x+1)$ Critical points $f^{\prime}(x)=0$ $\Rightarrow-e^{-x}(x+1)=0$ $\Rightarrow x=-1$ Clearly $\mathrm{f}^{\prime}(\mathrm{x})0$ if $\mathrm{x}-1$ $\mathrm{f}^{\prime}(\mathrm{x})0$ if $\mathrm{x}-1$ Hence $f(x)$ increases in $(-\infty,-1)$, decreases in $(-1, \infty)$...
Read More →Which of the following is not correct?
Question: Which of the following is not correct?A. sin = 1/5B. cos = 1C.sec = D. tan = 20 Solution: C.sec = Explanation: According to the question, We know that, a)sin = 1/5 is correct since Sin [-1,1] b) cos = 1 is correct since Cos [-1,1] c)sec = ⇒ (1/cos ) = ⇒cos=2 is incorrect since Cos[-1,1] d) tan = 20 is correct since tan R. Thus, option (C) sec = is the correct answer....
Read More →If tan θ = 1/2 and tan ϕ = 1/3,
Question: If tan = 1/2 and tanϕ= 1/3, then the value of +ϕisA. /6B. C. 0D. /4 Solution: D. /4 Explanation: According to the question, $\tan \theta=\frac{1}{2}$ and $\tan \phi=\frac{1}{3}$ We know that, $\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$ $=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$ $\Rightarrow \tan (\theta+\phi)=\tan \frac{\pi}{4}$ $\Rightarrow \theta+\phi=\frac{\pi}{4}$ Thus, option (D) /4 is the co...
Read More →Find the interval in which
Question: Find the interval in which $f(x)=\log (1+x)-\frac{x}{1+x}$ is increasing or decreasing ? Solution: we have $f(x)=\log (1+x)-\frac{x}{1+x}$ $f^{\prime}(x)=\frac{1}{1+x}-\left(\frac{(1+x)-x}{(1+x)^{2}}\right)$ $=\frac{1}{1+x}-\left(\frac{1}{(1+x)^{2}}\right)$ $=\frac{x}{(1+x)^{2}}$ Critical points $\mathrm{f}^{\prime}(\mathrm{x})=0$ $\Rightarrow \frac{x}{(1+x)^{2}}=0$ $\Rightarrow x=0,-1$ Clearly, $f^{\prime}(x)0$ if $x0$ And $f^{\prime}(x)0$ if $-1x0$ or $x-1$ Hence, $f(x)$ increases in...
Read More →If f(x) = cos2x + sec2x,
Question: If f(x) = cos2x + sec2x, then A. f(x) 1 B. f(x) = 1 C. 2 f(x) 1 D. f(x) 2 [Hint: A.M G.M.] Solution: D. f(x) 2 Explanation: According to the question, We have, f(x) = cos2x + sec2x We know that, A.M G.M.\ $\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq \sqrt{\cos ^{2} x \sec ^{2} x}$ $\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq \sqrt{\cos ^{2} x \frac{1}{\cos ^{2} x}}$ $\Rightarrow \frac{\cos ^{2} x+\sec ^{2} x}{2} \geq 1$ ⇒cos02x + sec2x 2 ⇒f(x) 2 Thus, option (D) f(x) 2 is...
Read More →Find the area of the triangle formed by the lines
Question: Find the area of the triangle formed by the lines x = 0, y = 1 and 2x + y = 2. Solution: The given equations are $x=0 \ldots(\mathrm{i})$ $y=1 \ldots(\mathrm{ii})$ and $2 x+y=2 \ldots$ (iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC From eq. (i) and (ii), we get $x=0$ and $y=1$ Thus, $A B$ and $B C$ intersect at $(0,1)$ Solving eq. (ii) and (iii), we get $y=1 \ldots$ (ii) and $2 x+y=2 \ldots$ (iii) Putting the value of y = 1 in eq. (iii), we ge...
Read More →If sin θ + cosec θ = 2,
Question: If sin + cosec = 2, then sin2 + cosec2 is equal to A. 1 B.4 C. 2 D. None of these Solution: C. 2 Explanation: According to the question, sin + cosec = 2 Squaring LHS and RHS, We get, ⇒(sin + cosec )2= 22 ⇒sin2 + cosec2 + 2 sin cosec = 4 ⇒sin2 + cosec2 + 2 sin (1/sin) = 4 ⇒sin2 + cosec2 + 2 = 4 ⇒sin2 + cosec2 = 2 Thus, option (C) 2 is the correct answer....
Read More →Determine whether
Question: Determine whether $f(x)=x / 2+\sin x$ is increasing or decreasing on $(-\pi / 3, \pi / 3) ?$ Solution: we have, $f(x)=-\frac{x}{2}+\sin x$ $=\mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{2}+\cos \mathrm{x}$ Now, $x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$ $\Rightarrow-\frac{\pi}{3}x\frac{\pi}{3}$ $\Rightarrow \cos \left(-\frac{\pi}{3}\right)\cos x\cos \frac{\pi}{3}$ $\Rightarrow \cos \left(\frac{\pi}{3}\right)\cos x\cos \frac{\pi}{3}$ $\Rightarrow \frac{1}{2}\cos x\frac{1}{2}$ $\Righ...
Read More →Find the general solution of the equation
Question: Find the general solution of the equation (3 1) cos + (3 + 1) sin = 2 [Hint: Put 3 1 = r sin , 3 + 1 = r cos which gives tan = tan((/4) (/6)) = /12] Solution: Let, r sin = 3 1 and r cos = 3 + 1 Therefore, r = {(3 1)2+ (3 + 1)2} = 8 = 22 And,tan = (3 1) / (3 + 1) Therefore,r(sin cos + cos sin ) = 2 ⇒ r sin (+) = 2 ⇒ sin (+) = 1/2 ⇒ sin (+) = sin (/4) ⇒ + = n + ( 1)n(/4), nZ ⇒ = n + ( 1)n(/4) (/12), n Z...
Read More →Find the general solution of the equation
Question: Find the general solution of the equation sin x 3sin2x + sin3x = cos x 3cos2x + cos3x Solution: According to the question, sin x 3sin2x + sin3x = cos x 3cos2x + cos3x Grouping sin x and sin 3x in LHS and, cos x and cos 3x in RHS, We get, sin x + sin3x 3sin2x = cos x + cos3x 3cos2x Applying transformation formula, cos A + cos B = 2cos ((A + B)/2) cos((A B)/2) sin A + sin B = 2sin ((A + B)/2) cos((A B)/2) ⇒ $2 \sin \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)-3 \sin 2 x...
Read More →Show that
Question: Show that $f(x)=\tan ^{-1} x-x$ is a decreasing function on $R$ ? Solution: we have, $f(x)=\tan ^{-1} x-x$ $f^{\prime}(x)=\frac{1}{1+x^{2}}-1$ $=-\frac{x^{2}}{1+x^{2}}$ Now, $\mathrm{X} \in \mathrm{R}$ $\Rightarrow x^{2}0$ and $1+x^{2}0$ $\Rightarrow \frac{x^{2}}{1+x^{2}}0$ $\Rightarrow-\frac{x^{2}}{1+x^{2}}0$ $\Rightarrow f^{\prime}(x)0$ Hence, $f(x)$ is an decreasing function for $R$...
Read More →Find the general solution of the equation
Question: Find the general solution of the equation 5cos2 + 7sin2 6 = 0 Solution: According to the question, 5cos2 + 7sin2 6 = 0 We know that, sin2 = 1 cos2 Therefore,5cos2 + 7(1 cos2) 6 = 0 ⇒5cos2 + 7 7cos2 6 = 0 ⇒-2cos2 + 1 = 0 ⇒cos2 = Therefore,cos = 1/2 Therefore,cos = cos /4 or cos = cos 3/4 Since,solution of cos x = cos is given by x = 2m mZ = n /4, n Z...
Read More →Find the area of the triangle formed by the lines
Question: Find the area of the triangle formed by the lines x + y = 6, x 3y = 2 and 5x 3y + 2 = 0. Solution: The given equations are $x+y=6 \ldots(i)$ $x-3 y=2 \ldots(i i)$ and $5 x-3 y+2=0$ or $5 x-3 y=-2 \ldots$ (iii) Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC Firstly, we solve the equation (i) and (ii) $x+y=6 \ldots(i)$ $x-3 y=2 \ldots(i i)$ Subtracting eq. (ii) from (i), we get $x+y-x+3 y=6-2$ $\Rightarrow 4 y=4$ $\Rightarrow y=1$ Putting the value of...
Read More →Find the value of the expression
Question: Find the value of the expression cos4(/8) + cos4(3/8)+ cos4(5/8)+ cos4(7/8).[Hint: Simplify the expression to $2\left(\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}\right)=2\left[\left(\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}\right)^{2}-2 \cos ^{2} \frac{\pi}{8} \cos ^{2} \frac{3 \pi}{8}\right]$ Solution: According to the question, Let y = cos4(/8) + cos4(3/8)+ cos4(5/8)+ cos4(7/8). ⇒ y = cos4(/8) + cos4(3/8)+ cos4( 3/8)+ cos4( /8). Since we know that, cos ( x) = cos x, we ...
Read More →Show that
Question: Show that $f(x)=\sin x-\cos x$ is an increasing function on $(-\pi / 4, \pi / 4) ?$ Solution: we have, $f(x)=\sin x-\cos x$ $f^{\prime}(x)=\cos x+\sin x$ $=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right)$ $=\sqrt{2}\left(\frac{\sin \pi}{4} \cos x+\frac{\cos \pi}{4} \sin x\right)$ $=\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)$ Now, $x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ $\Rightarrow-\frac{\pi}{4}x\frac{\pi}{4}$ $\Rightarrow 0\frac{\pi}{4}+x\frac{\pi}{2...
Read More →If θ lies in the first quadrant and cos θ = 8/17,
Question: If lies in the first quadrant and cos = 8/17, then find the value of cos(30 + ) + cos (45 ) + cos (120 ) Solution: According to the question, cos = 8/17 sin = (1 cos2) Since, lies in first quadrant, only positive sign can be considered. ⇒sin = (1 64/289) = 15/17 Let, y = cos(30 + ) + cos (45 ) + cos (120 ) We know that, cos(x + y) = cos x cos y sin x sin y Therefore, y = cos30 cos sin30 sin + cos45 cos + sin45sin +cos120 cos + sin120 sin Substituting values of cos30, sin30, cos 120, si...
Read More →If x = sec ϕ – tan ϕ and y = cosec ϕ + cot ϕ,
Question: If x = secϕ tanϕand y = cosecϕ+ cotϕ, then show that xy + x y + 1 = 0. [Hint: Find xy + 1 and then show tan x y = (xy + 1)] Solution: According to the question, x = secϕ tanϕand y =cosecϕ+ cotϕ Given that, LHS = xy + x y + 1 $=(\sec \phi-\tan \phi)(\operatorname{cosec} \phi+\cot \phi)+(\sec \phi-\tan \phi)-(\operatorname{cosec} \phi+\cot \phi)+1$ $=\sec \phi \operatorname{cosec} \phi+\cot \phi \sec \phi-\tan \phi \cot \phi-\tan \phi \operatorname{cosec} \phi$ $+\sec \phi-\tan \phi-(\op...
Read More →State when a function f(x) is said to be increasing
Question: State when a function $f(x)$ is said to be increasing on an interval $[a, b] .$ Test whether the function $f(x)=x^{2}$ $6 x+3$ is increasing on the interval $[4,6]$. Solution: Given:- Function $f(x)=f(x)=x^{2}-6 x+3$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- ...
Read More →If a cos2θ + b sin 2θ = c has α and β as its roots,
Question: If a cos2 + b sin 2 = c has and as its roots, then prove thattan + tan = 2b/(a + c) [Hint: Use the identitiescos 2 = (( 1 tan2)/(1 + tan2) and sin 2 = 2tan /(1 + tan2)] Solution: According to the question, a cos2 + b sin 2 = c and are the roots of the equation. Using the formula of multiple angles, We know that, $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ and $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$ $\therefore \mathrm{a}\left(\frac{1-\tan ^{2} \theta}...
Read More →If a cos2θ + b sin 2θ = c has α and β as its roots,
Question: If a cos2 + b sin 2 = c has and as its roots, then prove thattan + tan = 2b/(a + c) [Hint: Use the identitiescos 2 = (( 1 tan2)/(1 + tan2) and sin 2 = 2tan /(1 + tan2)] Solution: According to the question, a cos2 + b sin 2 = c and are the roots of the equation. Using the formula of multiple angles, We know that, $\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ and $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$ $\therefore \mathrm{a}\left(\frac{1-\tan ^{2} \theta}...
Read More →Find the value of the expression
Question: Find the value of the expression $3\left[\sin ^{4}\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^{4}(3 \pi+\alpha)\right]-2\left[\sin ^{6}\left(\frac{\pi}{2}+\alpha\right)+\sin ^{6}(5 \pi-\alpha)\right]$ Solution: According to the question, Let, y = 3[sin4(3/2 ) + sin4(3 + )] 2[sin6(/2 + ) + sin6(5 )] We know that, sin(3/2 ) = -cos sin(3 + ) = -sin sin(/2 + ) = cos sin(5 ) = sin Therefore, y =3[( cos )4+ ( sin )4] 2[cos6 + sin6] ⇒y = 3 [cos4 + sin4] 2[sin6 + cos6] ⇒y = 3[(sin2 + cos2)2 2sin...
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