For the function
Question: For the function $f(x)=x+\frac{1}{x}, x \in[1,3]$, the value of $c$ for the Lagrange's mean value theoremis (a) 1 (b) $\sqrt{3}$ (c) 2 (d) none of these Solution: (b) $\sqrt{3}$ We have $f(x)=x+\frac{1}{x}=\frac{x^{2}+1}{x}$ Clearly, $f(x)$ is continuous on $[1,3]$ and derivable on $(1,3)$. Thus, both the conditions of Lagrange's theorem are satisfied. Consequently, there exists $c \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$ Now, $f(x)=\frac{x^{2}+1}{x...
Read More →For the function
Question: For the function $f(x)=x+\frac{1}{x}, x \in[1,3]$, the value of $c$ for the Lagrange's mean value theoremis (a) 1 (b) $\sqrt{3}$ (c) 2 (d) none of these Solution: (b) $\sqrt{3}$ We have $f(x)=x+\frac{1}{x}=\frac{x^{2}+1}{x}$ Clearly, $f(x)$ is continuous on $[1,3]$ and derivable on $(1,3)$. Thus, both the conditions of Lagrange's theorem are satisfied. Consequently, there exists $c \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$ Now, $f(x)=\frac{x^{2}+1}{x...
Read More →Measure of two quantities along
Question: Measure of two quantities along with the precision of respective measuring instrument is A = 2.5 ms-1 0.5 ms-1, B = 0.10 s 0.01 s. The value of AB will be (a) (0.25 0.08) m (b) (0.25 0.5) m (c) (0.25 0.05) m (d) (0.25 0.135) m Solution: The correct answer is (a) (0.25 0.08) m...
Read More →If θ and Φ lie in the first quadrant such that
Question: If and Φ lie in the first quadrant such that $\sin \theta=\frac{8}{17}$ and $\cos \phi=\frac{12}{13}$ find the values of (i) $\sin (\theta-\Phi)$ (ii) $\cos (\theta-\Phi)$ (iii) $\tan (\theta-\Phi)$ Solution: Given $\sin \theta=\frac{8}{17}$ and $\cos \phi=\frac{12}{13}$ $\cos \theta=\sqrt{\left(1-\sin ^{2} \theta\right)} \Rightarrow \sqrt{\left(1-\left(\frac{8}{17}\right)^{2}\right.}=\sqrt{\left(\frac{289-84}{289}\right)} \Rightarrow \sqrt{\left(\frac{225}{289}\right)}=\frac{15}{17}$ ...
Read More →Which of the following pairs
Question: Which of the following pairs of physical quantities does not havesamedimensional formula? (a) Work and torque (b) Angular momentum and Plancks constant (c) Tension and surface tension (d) Impulse and linear momentum Solution: The correct answer is (a) work and torque...
Read More →The length and breadth of a rectangular
Question: The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is (a) 164 3 cm2 (b) 163.62 2.6 cm2 (c) 163.6 2.6 cm2 (d) 163.62 3 cm2 Solution: The correct answer is (a) 164 3 cm2...
Read More →The numbers 2.745 and 2.735
Question: The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give (a) 2.75 and 2.74 (b) 2.74 and 2.73 (c) 2.75 and 2.73 (d) 2.74 and 2.74 Solution: (d) Key concept: While rounding off measurements, we use the following rules by convention: 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged. 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one. 3. If the digit to be dropped is 5 followed by digits othe...
Read More →If the solve the problem
Question: If $4 a+2 b+c=0$, then the equation $3 a x^{2}+2 b x+c=0$ has at least one real root Iying in the interval (a) (0, 1)(b) (1, 2)(c) (0, 2)(d) none of these Solution: (c) (0, 2) Let $f(x)=a x^{3}+b x^{2}+c x+d$ ......(1) $f(0)=d$ $f(2)=8 a+4 b+2 c+d$ $=2(4 a+2 b+c)+d$ $=d$ $(\because(4 a+2 b+c)=0)$ fis continuous in the closed interval [0, 2] andfis derivable in the open interval (0, 2). Also,f(0) =f(2) By Rolle's Theorem, $f^{\prime}(\alpha)=0 \quad$ for $0\alpha2$ Now, $f^{\prime}(x)=3...
Read More →If the solve the problem
Question: If $4 a+2 b+c=0$, then the equation $3 a x^{2}+2 b x+c=0$ has at least one real root Iying in the interval (a) (0, 1)(b) (1, 2)(c) (0, 2)(d) none of these Solution: (c) (0, 2) Let $f(x)=a x^{3}+b x^{2}+c x+d$ ......(1) $f(0)=d$ $f(2)=8 a+4 b+2 c+d$ $=2(4 a+2 b+c)+d$ $=d$ $(\because(4 a+2 b+c)=0)$ fis continuous in the closed interval [0, 2] andfis derivable in the open interval (0, 2). Also,f(0) =f(2) By Rolle's Theorem, $f^{\prime}(\alpha)=0 \quad$ for $0\alpha2$ Now, $f^{\prime}(x)=3...
Read More →The mass and volume of a body are 4.237 g
Question: The mass and volume of a body are 4.237 g and 2.5 cm3respectively. The density of the material of the body in correct significant figures is: (a) 1.6048 g/cm3 (b) 1.69 g/cm3 (c) 1.7 g/cm3 (d) 1.695 g/cm3 Solution: The correct answer is (c) 1.7 g/cm3...
Read More →Prove that
Question: Prove that $\frac{\sin \left(180^{\circ}+\theta\right) \cos \left(90^{\circ}+\theta\right) \tan \left(270^{\circ}-\theta\right) \cot \left(360^{\circ}-\theta\right)}{\sin \left(360^{\circ}-\theta\right) \cos \left(360^{\circ}+\theta\right) \operatorname{cosec}(-\theta) \sin \left(270^{\circ}+\theta\right)}=1$ Solution: Using $\cos \left(90^{\circ}+\theta\right)=-\sin \theta(I$ quadrant $\cos x$ is positive $\operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta$ $\tan \left(270^{\c...
Read More →The sun of the numbers 436.32, 227.2,
Question: The sun of the numbers 436.32, 227.2, and 0.301 inappropriate significant figures is: (a) 663.821 (b) 664 (c) 663.8 (d) 663.82 Solution: The correct answer is (c) 663.8...
Read More →Prove that
Question: Prove that $\frac{\cos \theta}{\sin \left(90^{\circ}+\theta\right)}+\frac{\sin (-\theta)}{\sin \left(180^{\circ}+\theta\right)}-\frac{\tan \left(90^{\circ}+\theta\right)}{\cot \theta}=3$ Solution: Using $\sin \left(90^{\circ}+\theta\right)=\cos \theta$ and $\sin (-\theta)=\sin \theta, \tan \left(90^{\circ}+\theta\right)=-\cot \theta$ $\operatorname{Sin}\left(180^{\circ}+\theta\right)=-\sin \theta($ III quadrant $\sin x$ is negative $)$ $\frac{\cos \theta}{\sin \left(90^{\circ}+\theta\r...
Read More →The number of significant figures in 0.06900 is:
Question: The number of significant figures in 0.06900 is: (a) 5 (b) 4 (c) 2 (d) 3 Solution: The correct answer is (b) 4 The number of zeroes on the left of the non-zero number is not considered as significant figures but the zeroes that are on the right of the non-zero number are significant figures....
Read More →If the polynomial equation
Question: If the polynomial equation $a_{0} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}=0$ npositive integer, has two different real roots and , then between and , the equation $n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0$ has (a) exactly one root(b) almost one root(c) at least one root(d) no root Solution: (c) at least one root We observe that, $n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0$ is the derivative of the polynomial $a_{n} x^{n}+a_{n-1} x^{n...
Read More →Prove that
Question: Prove that $\frac{\cos (\pi+\theta) \cos (-\theta)}{\cos (\pi-\theta) \cos \left(\frac{\pi}{2}+\theta\right)}=-\cot \theta$ Solution: $\frac{\cos (\pi+\theta) \cdot \cos (-\theta)}{\cos (\pi-\theta) \cdot \cos \left(\frac{\pi}{2}+\theta\right)}=\frac{-\cos \theta \cdot \cos \theta}{-\cos \theta \cdot-\sin \theta}$ $\Rightarrow \frac{\cos \theta}{-\sin \theta}=-\cot \theta$ $\left(\right.$ Using $\cos (\pi-\theta)=-\cos \theta$ and $\left.\cos \left(\frac{\pi}{2}-\theta\right)=-\sin \th...
Read More →Prove that
Question: Prove that $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{0}+\sin 8^{0}}=\tan 37^{0}$ Solution: First we will take out cos8 common from both numerator and denominator, $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\frac{\cos 8^{\circ}\left(1-\tan 8^{\circ}\right)}{\cos 8^{\circ}\left(1+\tan 8^{\circ}\right)} \Rightarrow \frac{\tan 45^{\circ}-\tan 8^{\circ}}{1+\tan 45^{\circ} \cdot \tan 8^{\circ}}=\tan \left(45^{\circ}-8^{\circ}\right) \Rightarrow \tan 37^{\circ}$ ...
Read More →Using Lagrange's mean value theorem, prove that
Question: Using Lagrange's mean value theorem, prove that $(b-a) \sec ^{2} a\tan b-\tan a(b-a) \sec ^{2} b$ where $0ab\frac{\pi}{2}$. Solution: Consider, the function $f(x)=\tan x, x \in[a, b], 0ab\frac{\pi}{2}$ Clearly, $f(x)$ is continuous on $[a, b]$ and derivable on $(a, b)$. Thus, both the conditions of Lagrange's theorem are satisfied. Consequently, $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$. Now, $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x, f(a)=\tan a, f(b)=\ta...
Read More →Using Lagrange's mean value theorem, prove that
Question: Using Lagrange's mean value theorem, prove that $(b-a) \sec ^{2} a\tan b-\tan a(b-a) \sec ^{2} b$ where $0ab\frac{\pi}{2}$. Solution: Consider, the function $f(x)=\tan x, x \in[a, b], 0ab\frac{\pi}{2}$ Clearly, $f(x)$ is continuous on $[a, b]$ and derivable on $(a, b)$. Thus, both the conditions of Lagrange's theorem are satisfied. Consequently, $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$. Now, $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x, f(a)=\tan a, f(b)=\ta...
Read More →Prove that
Question: Prove that $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\tan 37^{\circ}$ Solution: First we will take out cos8 common from both numerator and denominator, $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\frac{\cos 8^{\circ}\left(1-\tan 8^{\circ}\right)}{\cos 8^{\circ}\left(1+\tan 8^{\circ}\right)} \Rightarrow \frac{\tan 45^{\circ}-\tan 8^{\circ}}{1+\tan 45^{\circ} \cdot \tan 8^{\circ}}=\tan \left(45^{\circ}-8^{\circ}\right) \Rightarrow \tan ...
Read More →Prove that
Question: Prove that $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\tan 54^{\circ}$ Solution: First we will take out cos9common from both numerator and denominator, $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\frac{\cos 9\left(1+\tan 9^{\circ}\right)}{\cos 9^{\circ}\left(1-\tan 9^{\circ}\right)} \Rightarrow \frac{\tan 45^{\circ}+\tan 9^{\circ}}{1-\tan 45^{\circ} \cdot \tan 9^{\circ}}=\tan \left(45^{\circ}+9^{\circ}\right) \Rightarrow \tan 54$ $\lef...
Read More →Prove that
Question: Prove that (i) $\cos 15^{\circ}-\sin 15^{\circ}=\frac{1}{\sqrt{2}}$ (ii) $\cot 105^{\circ}-\tan 105^{\circ}=2 \sqrt{3}$ (iii) $\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}=-1$ Solution: (i) $\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$ $\sin 15^{0}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$ $\cos 15^{0}-\sin 15^{0}=\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}}$ $=\frac{\sqrt{3}+1-\sqrt{3}+1}{2 \sqrt{2}}$ $=\frac{2}{2 \sqrt{2}}$ $=\frac{1}{\sqrt{2}}$ (ii...
Read More →Let C be a curve defined parametrically as
Question: Let $C$ be a curve defined parametrically as $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta, 0 \leq \theta \leq \frac{\pi}{2}$. Determine a point $\mathrm{P}$ on $C$, where the tangent to $C$ is parallel to the chord joining the points $(a, 0)$ and $(0, a)$ Solution: As, $x=a \cos ^{3} \theta$ $\Rightarrow \frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$ And, $y=a \sin ^{3} \theta$ $\Rightarrow \frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$ So, $\frac{d y}{d x}=\frac{\left(...
Read More →Find a point on the curve
Question: Find a point on the curve $y=x^{3}+1$ where the tangent is parallel to the chord joining $(1,2)$ and $(3,28)$. Solution: Let: $f(x)=x^{3}+1$ The tangent to the curve is parallel to the chord joining the points $(1,2)$ and $(3,28)$. Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$. $\therefore a=1, b=3$ The polynomial function is everywhere continuous and differentiable. So, $f(x)=x^{3}+1$ is continuous on $[1,3]$ and differentiable on $(1,3)$. Thus, both the conditio...
Read More →Prove that
Question: Prove that (i) $\sin 75^{\circ}=\frac{(\sqrt{6}+\sqrt{2})}{4}$ (ii) $\frac{\cos 135^{\circ}-\cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}=(3-2 \sqrt{2})$ (iii) $\tan 15^{\circ}+\cot 15^{\circ}=4$ Solution: (i) $\sin 75^{\circ}=\sin \left(90^{\circ}-15^{\circ}\right)$ ............(using $\sin (A-B)=\sin A \cos B-\cos A \sin B)$ $=\sin 90^{\circ} \cos 15^{\circ}-\cos 90^{\circ} \sin 15^{\circ}$ $=1 \cdot \cos 15^{\circ}-0 \cdot \sin 15^{\circ}$ $=\cos 15^{\circ}$ $\operatorname{Co...
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