Find the points on the curve
Question: Find the points on the curve $y=x^{3}-3 x$, where the tangent to the curve is parallel to the chord joining $(1,-2)$ and $(2,2)$. Solution: Let: $f(x)=x^{3}-3 x$ The tangent to the curve is parallel to the chord joining the points $(1,-2)$ and $(2,2)$. Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$. $\therefore a=1, b=2$ The polynomial function is everywhere continuous and differentiable. So, $f(x)=x^{3}-3 x$ is continuous on $[1,2]$ and differentiable on $(1,2)$. T...
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Question: Prove that $\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}$ Solution: $\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\frac{\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \cdot \tan x}}{\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}}$ $\Rightarrow \frac{\frac{1+\tan x}{1-1 \cdot \tan x}}{\frac{1-\tan x}{1+1 \cdot \tan x}}=\frac{1+\tan x}{1-\tan x} \cdot \frac...
Read More →Find a point on the parabola
Question: Find a point on the parabola $y=(x-3)^{2}$, where the tangent is parallel to the chord joining $(3,0)$ and $(4,1)$. Solution: Let: $f(x)=(x-3)^{2}=x^{2}-6 x+9$ The tangent to the curve is parallel to the chord joining the points $(3,0)$ and $(4,1)$. Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$ $\therefore a=3, b=4$ The polynomial function is everywhere continuous and differentiable. So, $f(x)=x^{2}-6 x+9$ is continuous on $[3,4]$ and differentiable on $(3,4)$. Th...
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Question: Prove that (i) $\cos (n+2) x \cos (n+1) x+\sin (n+2) x \sin (n+1) x=\cos x$ (ii) $\cos \left(\frac{\pi}{4}-\mathrm{x}\right) \cos \left(\frac{\pi}{4}-\mathrm{y}\right)-\sin \left(\frac{\pi}{4}-\mathrm{x}\right) \sin \left(\frac{\pi}{4}-\mathrm{y}\right)=\sin (\mathrm{x}+\mathrm{y})$ Solution: (i) $\cos (n+2) x \cdot \cos (n+1) x+\sin (n+2) x \cdot \sin (n+1) x$ $=\sin ((n+2) x+(n+1) x)($ using $\cos (A-B)=\cos A \cos B+\sin A \sin B)$ $=\cos (n x+2 x-(n x+x))$ $=\cos (n x+2 x-n x-x)$ $...
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Question: Prove that (i) $\sin \left(50^{\circ}+\theta\right) \cos \left(20^{\circ}+\theta\right)-\cos \left(50^{\circ}+\theta\right) \sin \left(20^{\circ}+\theta\right)=\frac{1}{2}$ (ii) $\cos \left(70^{\circ}+\theta\right) \cos \left(10^{\circ}+\theta\right)+\sin \left(70^{\circ}+\theta\right) \sin \left(10^{\circ}+\theta\right)=\frac{1}{2}$ Solution: (i) $\sin \left(50^{\circ}+\theta\right) \cos \left(20^{\circ}+\theta\right)-\cos \left(50^{\circ}+\theta\right) \sin \left(20^{\circ}+\theta\ri...
Read More →Find a point on the curve
Question: Find a point on the curve $y=x^{2}+x$, where the tangent is parallel to the chord joining $(0,0)$ and $(1,2)$. Solution: Let: $f(x)=x^{2}+x$ The tangent to the curve is parallel to the chord joining the points $(0,0)$ and $(1,2)$. Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$. $\therefore a=0, b=1$ The polynomial function is everywhere continuous and differentiable. So, $f(x)=x^{2}+x$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Thus, both the conditions...
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Question: Prove that (i) $\sin 80^{\circ} \cos 20^{\circ}-\cos 80^{\circ} \sin 20^{\circ}=\frac{\sqrt{3}}{2}$ (ii) $\cos 45^{\circ} \cos 15^{\circ}-\sin 45^{\circ} \sin 15^{\circ}=\frac{1}{2}$ (iii) $\cos 75^{\circ} \cos 15^{\circ}+\sin 75^{\circ} \sin 15^{\circ}=\frac{1}{2}$ (iv) $\sin 40^{\circ} \cos 20^{\circ}+\cos 40^{\circ} \sin 20^{\circ}=\frac{\sqrt{3}}{2}$ (v) $\cos 130^{\circ} \cos 40^{\circ}+\sin 130^{\circ} \sin 40^{\circ}=0$ Solution: (i) $\sin 80^{\circ} \cos 20^{\circ}-\cos 80^{\ci...
Read More →Find a point on the parabola
Question: Find a point on the parabola $y=(x-4)^{2}$, where the tangent is parallel to the chord joining $(4,0)$ and $(5,1)$. Solution: Let: $f(x)=(x-4)^{2}=x^{2}-8 x+16$ The tangent to the curve is parallel to the chord joining the points $(4,0)$ and $(5,1)$. Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$. $\therefore a=4, b=5$ The polynomial function is everywhere continuous and differentiable. So, $x^{2}-8 x+16$ is continuous on $[4,5]$ and differentiable on $(4,5)$. Thus,...
Read More →Find the values of all trigonometric functions of 135
Question: Find the values of all trigonometric functions of 135 Solution: $\sin 135^{\circ}=\sin \left(180^{\circ}-45^{\circ}\right)$.............. .(using $\sin \left(180^{\circ}-x\right)=\sin x$ ) $=\sin 45^{\circ} \Rightarrow \frac{1}{\sqrt{2}}$ $\operatorname{Cos} 135^{\circ}=\cos \left(180^{\circ}-45^{\circ}\right)$ ...................$. .\left(\right.$ using $\left.\cos \left(180^{\circ}-x\right)=-\cos x\right)$ $=\cos 45^{\circ} \Rightarrow-\frac{1}{\sqrt{2}}$ $\operatorname{Tan} 135^{\ci...
Read More →Verify the hypothesis and conclusion of Lagrange's man value theorem for the function
Question: Verify the hypothesis and conclusion of Lagrange's man value theorem for the function $f(x)=\frac{1}{4 x-1}, 1 \leq x \leq 4$ Solution: The given function is $f(x)=\frac{1}{4 x-1}$. Since for each $x \in[1,4]$, the function attains a unique definite value, $f(x)$ is continuous on $[1,4]$. Also, $f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}$ exists for all $x \in[1,4]$ Thus, both the conditions of Lagrange's mean value theorem are satisfied. Consequently, there exists some $c \in(1,4)$ such tha...
Read More →Verify the hypothesis and conclusion of Lagrange's man value theorem for the function
Question: Verify the hypothesis and conclusion of Lagrange's man value theorem for the function $f(x)=\frac{1}{4 x-1}, 1 \leq x \leq 4$ Solution: The given function is $f(x)=\frac{1}{4 x-1}$. Since for each $x \in[1,4]$, the function attains a unique definite value, $f(x)$ is continuous on $[1,4]$. Also, $f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}$ exists for all $x \in[1,4]$ Thus, both the conditions of Lagrange's mean value theorem are satisfied. Consequently, there exists some $c \in(1,4)$ such tha...
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Question: Find the value of (i) $\cos 840^{\circ}$ (ii) $\sin 870^{\circ}$ (iii) $\tan \left(-120^{\circ}\right)$ (iv) $\sec \left(-420^{\circ}\right)$ (v) $\operatorname{cosec}\left(-690^{\circ}\right)$ (vi) $\tan \left(225^{\circ}\right)$ (vii) $\cot \left(-315^{\circ}\right)$ (viii) $\sin \left(-1230^{\circ}\right)$ (ix) $\cos \left(495^{\circ}\right)$ Solution: (i) $\operatorname{Cos} 840^{\circ}=\operatorname{Cos}\left(2.360^{\circ}+120^{\circ}\right) \ldots \ldots \ldots \ldots($ using $\o...
Read More →Show that the lagrange's mean value theorem is not applicable to the function
Question: Show that the lagrange's mean value theorem is not applicable to the function $f(x)=\frac{1}{x}$ on $[-1,1]$ Solution: Given: $f(x)=\frac{1}{x}$ Clearly, $f(x)$ does not exist for $x=0$ Thus, the given function is discontinuous on $[-1,1]$. Hence, Lagrange's mean value theorem is not applicable for the given function on $[-1,1]$....
Read More →Discuss the applicability of Lagrange's mean value theorem for the function
Question: Discuss the applicability of Lagrange's mean value theorem for the functionf(x) = |x| on [1, 1] Solution: Given: $f(x)=|x|$ If Lagrange's theorem is applicable for the given function, then $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$. But it is known that $f(x)=|x|$ is not differentiable at $x=0 \in(-1,1)$. Thus, our supposition iswrong.Therefore, Lagrange's theorem is not applicable for the given function....
Read More →Verify Lagrange's mean value theorem for the following
Question: Verify Lagrange's mean value theorem for the following functions on the indicated intervals. In each case find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem (i) $f(x)=x^{2}-1$ on $[2,3]$ (ii) $f(x)=x^{3}-2 x^{2}-x+3$ on $[0,1]$ (iii) $f(x)=x(x-1)$ on $[1,2]$ (iv) $f(x)=x^{2}-3 x+2$ on $[-1,2]$ (v) $f(x)=2 x^{2}-3 x+1$ on $[1,3]$ (vi) $f(x)=x^{2}-2 x+4$ on $[1,5]$ (vii) $f(x)=2 x-x^{2}$ on $[0,1]$ (viii) $f(x)=(x-1)(x-2)(x-3)$ on $[0,4]$ (ix) $f(x)...
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Question: Prove that $4 \sin \frac{\pi}{6} \sin ^{2} \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\operatorname{cosec}^{2} \frac{\pi}{2}=4$ Solution: To prove: $4 \sin \frac{\pi}{6} \sin ^{2} \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\operatorname{cosec}^{2} \frac{\pi}{2}=4$ Taking LHS, $=4 \sin \frac{\pi}{6} \sin ^{2} \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\operatorname{cosec}^{2} \frac{\pi}{2}$ Putting $\pi=180^{\circ}$ $=4 \sin \frac{180}{6} \sin ^{2} \frac...
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Question: Prove that $\sin \frac{\pi}{6} \cos 0+\sin \frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{3} \cos \frac{\pi}{6}=\frac{7}{4}$ Solution: To prove: $\sin \frac{\pi}{6} \cos 0+\sin \frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{3} \cos \frac{\pi}{6}=\frac{7}{4}$ Taking LHS, $=\sin \frac{\pi}{6} \cos 0+\sin \frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{3} \cos \frac{\pi}{6}$ Putting $\pi=180^{\circ}$ $=\sin \frac{180}{6} \cos 0+\sin \frac{180}{4} \cos \frac{180}{4}+\sin \frac{180}{3}...
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Question: Prove that $\tan ^{2} \frac{\pi}{3}+2 \cos ^{2} \frac{\pi}{4}+3 \sec ^{2} \frac{\pi}{6}+4 \cos ^{2} \frac{\pi}{2}=8$ Solution: To prove: $\tan ^{2} \frac{\pi}{3}+2 \cos ^{2} \frac{\pi}{4}+3 \sec ^{2} \frac{\pi}{6}+4 \cos ^{2} \frac{\pi}{2}=8$ Taking LHS, $=\tan ^{2} \frac{\pi}{3}+2 \cos ^{2} \frac{\pi}{4}+3 \sec ^{2} \frac{\pi}{6}+4 \cos ^{2} \frac{\pi}{2}$ Putting $\pi=180^{\circ}$ $=\tan ^{2} \frac{180}{3}+2 \cos ^{2} \frac{180}{4}+3 \sec ^{2} \frac{180}{6}+4 \cos ^{2} \frac{180}{2}$...
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Question: Find the value of $\cos \left(-2220^{\circ}\right)$ Solution: To find: Value of cos 2220 We have, $\cos \left(-2220^{\circ}\right)=\cos 2220^{\circ}$ ${[\because \cos (-\theta)=\cos \theta] }$ $=\cos \left[2160+60^{\circ}\right]$ $=\cos \left[360^{\circ} \times 6+60^{\circ}\right]$ $=\cos 60^{\circ}$ [Clearly, $2220^{\circ}$ is in I $^{\text {st }}$ Quadrant and the multiple of $360^{\circ}$ is even] $=\frac{1}{2}\left[\because \cos 60^{\circ}=\frac{1}{2}\right]$...
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Question: Find the value of $\operatorname{cosec}\left(-750^{\circ}\right)$ Solution: To find: Value of cosec (-750) We have, $\operatorname{cosec}\left(-750^{\circ}\right)=-\operatorname{cosec}\left(750^{\circ}\right)$ ${[\because \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta] }$ $=-\operatorname{cosec}\left[90^{\circ} \times 8+30^{\circ}\right]$ Clearly, $405^{\circ}$ is in $\mathrm{Ist}$ Quadrant and the multiple of $90^{\circ}$ is even $=-\operatorname{cosec} 30^{\circ}$ $=-2\le...
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Question: Find the value of $\cot \left(585^{\circ}\right)$ Solution: To find: Value of $\cot \frac{13 \pi}{4}$ We have $\cot \left(585^{\circ}\right)=\cot \left[90^{\circ} \times 6+45^{\circ}\right]$ $=\cot 45^{\circ}$ [Clearly, $585^{\circ}$ is in III $^{\text {rd }}$ Quadrant and the multiple of $90^{\circ}$ is even] $=1\left[\because \cot 45^{\circ}=1\right]$...
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Question: Find the value of $\tan \left(-300^{\circ}\right)$ Solution: To find: Value of tan (-300) We have, $\tan \left(-300^{\circ}\right)=-\tan \left(300^{\circ}\right)$ ${[\because \tan (-\theta)=-\tan \theta] }$ $=-\tan \left[90^{\circ} \times 3+30^{\circ}\right]$ Clearly, $300^{\circ}$ is in IV $^{\text {th }}$ Quadrant and the multiple of $90^{\circ}$ is odd $=-\cot 30^{\circ}$ $=-\sqrt{3}\left[\because \cot 30^{\circ}=\sqrt{3}\right]$...
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Question: Find the value of $\sec \left(-1470^{\circ}\right)$ Solution: To find: Value of sec (-1470) We have, $\sec \left(-1470^{\circ}\right)=\sec \left(1470^{\circ}\right)$ $[\because \sec (-\theta)=\sec \theta]$ $=\sec \left[90^{\circ} \times 16+30^{\circ}\right]$ Clearly, $1470^{\circ}$ is in Ist Quadrant and the multiple of $90^{\circ}$ is even $=\sec 30^{\circ}$ $=\frac{2}{\sqrt{3}}\left[\because \sec 30^{\circ}=\frac{2}{\sqrt{3}}\right]$...
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Question: Find the value of $\sin 405^{\circ}$ Solution: To find: Value of sin 405 We have, $\sin 405^{\circ}=\sin \left[90^{\circ} \times 4+45^{\circ}\right]$ $=\sin 45^{\circ}$ [Clearly, $405^{\circ}$ is in ${ }^{\text {st }}$ Quadrant and the multiple of $90^{\circ}$ is even] $=\frac{1}{\sqrt{2}}\left[\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right]$...
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Question: Find the value of $\operatorname{cosec}\left(\frac{-41 \pi}{4}\right)$ Solution: To find: Value of $\operatorname{cosec}\left(-\frac{41 \pi}{4}\right)$ We have, $\operatorname{cosec}\left(-\frac{41 \pi}{4}\right)=-\operatorname{cosec} \frac{41 \pi}{4}$ $[\because \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta]$ Putting = 180 $=-\operatorname{cosec} \frac{41 \times 180}{4}$ $=-\operatorname{cosec}\left[41 \times 45^{\circ}\right]$ $=-\operatorname{cosec}\left[1845^{\circ}\ri...
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