NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85384
The equation of normal to the curve \(y=\) \(\sin \left(\frac{\pi x}{4}\right)\) at the point \((2,5)\) is
1 \(x+y=2\)
2 \(x+y=5\)
3 \(y=5\)
4 \(x=2\)
Explanation:
(D) : Given, \(\mathrm{y}=\sin \left(\frac{\pi \mathrm{x}}{4}\right)\) On differentiating both sides, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \mathrm{x}\right)\) Put the value of \(x=2\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \times 2\right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \frac{\pi}{2}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Equation of normal of curve at \((2,5)\) \(\mathrm{x}-2=0\) \(\mathrm{x}=2\)
MHT CET-2020
Application of Derivatives
85385
If the line \(y=4 x-5\) touches the curve \(y^{2}=a x^{3}+b\) at the point \((2,3)\) then
1 \(\mathrm{a}=2, \quad \mathrm{~b}=7\)
2 \(\mathrm{a}=-2, \mathrm{~b}=-7\)
3 \(\mathrm{a}=-2, \mathrm{~b}=7\)
4 \(\mathrm{a}=2, \quad \mathrm{~b}=-7\)
Explanation:
(D) : Given, \(y=4 x-5\) Point \((2,3)\) \(\mathrm{y}^{2}=\mathrm{ax}^{3}+\mathrm{b}\) On differentiating both sides w.r.t. to \(\mathrm{x}\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times 3 \mathrm{x}^{2}+0\) \(\frac{d y}{d x}=\frac{3 a^{2}}{2 y}\) Let, point \((x=2, y=3)\) \(\frac{d y}{d x}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow \frac{d y}{d x}=\frac{12 a}{6}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{a}\) The line \(y=4 x-5\) touches the curve at the point \((2,3)\) So, slope of the tangent \((2,3)\) is 4 . \(\therefore \quad 2 \mathrm{a}=4\) \(\mathrm{a}=2\) Since, \((2,3)\) lie on the curve \(y^{2}=a x^{3}+b\) \((3)^{2}=a(2)^{3}+b\) \(9=8 \mathrm{a}+\mathrm{b} \quad[\because \mathrm{a}=2]\) \(9=8 \times 2+b\) \(\mathrm{b}=-7\) Hence, \(\mathrm{a}=2, \mathrm{~b}=-7\)
MHT CET-2020
Application of Derivatives
85386
The equation of tangent at \(P(-4,-4)\) on the curve \(x^{2}=-4 y\) is
1 \(3 x-y+8=0\)
2 \(2 x+y+4=0\)
3 \(2 x+y-4=0\)
4 \(2 x-y+4=0\)
Explanation:
(D) : Given, Tangent point \((-4,-4)\) Curve, \(x^{2}=-4 y\) On differentiating both sides w.r.t \(x\), \(2 x=-4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\left.\frac{-x}{2} \Rightarrow \frac{d y}{d x}\right|_{(-4,-4)}=\frac{+4}{2}=2\) The equation of the tangent is given \(\left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)\) \((y+4)=2(x+4)\) \((y+4)=2 x+8\) \(2 x-y+4=0\)
MHT CET-2020
Application of Derivatives
85387
The approximate value of the function \(f(x)=x^{3}-3 x+5\) at \(x=1.99\) is
1 7.91
2 7.94
3 6.94
4 6.91
Explanation:
(D) : Given, \(f(x)=x^{3}-3 x+5\) \(f^{\prime}(x)=3 x^{2}-3\) We know that, \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x})+\Delta \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})\) Here, \(\mathrm{x}+\Delta \mathrm{x}=1.99\) \(x=2, \Delta x=-0.01\) \(\therefore \quad \mathrm{f}(1.99)=\mathrm{f}(2)+(-0.01) \mathrm{f}^{\prime}(2)\) \(\mathrm{f}(1.99) =(2)^{3}-3(2)+5-0.01\left[3(2)^{2}-3\right]\) \(=8-6+5-0.01 \times 12-3=7-0.09\) \(\mathrm{f}(1.99) =6.91\) Hence, The approximate value of the function \(\mathrm{f}(1.99\) ) is 6.91
85384
The equation of normal to the curve \(y=\) \(\sin \left(\frac{\pi x}{4}\right)\) at the point \((2,5)\) is
1 \(x+y=2\)
2 \(x+y=5\)
3 \(y=5\)
4 \(x=2\)
Explanation:
(D) : Given, \(\mathrm{y}=\sin \left(\frac{\pi \mathrm{x}}{4}\right)\) On differentiating both sides, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \mathrm{x}\right)\) Put the value of \(x=2\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \times 2\right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \frac{\pi}{2}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Equation of normal of curve at \((2,5)\) \(\mathrm{x}-2=0\) \(\mathrm{x}=2\)
MHT CET-2020
Application of Derivatives
85385
If the line \(y=4 x-5\) touches the curve \(y^{2}=a x^{3}+b\) at the point \((2,3)\) then
1 \(\mathrm{a}=2, \quad \mathrm{~b}=7\)
2 \(\mathrm{a}=-2, \mathrm{~b}=-7\)
3 \(\mathrm{a}=-2, \mathrm{~b}=7\)
4 \(\mathrm{a}=2, \quad \mathrm{~b}=-7\)
Explanation:
(D) : Given, \(y=4 x-5\) Point \((2,3)\) \(\mathrm{y}^{2}=\mathrm{ax}^{3}+\mathrm{b}\) On differentiating both sides w.r.t. to \(\mathrm{x}\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times 3 \mathrm{x}^{2}+0\) \(\frac{d y}{d x}=\frac{3 a^{2}}{2 y}\) Let, point \((x=2, y=3)\) \(\frac{d y}{d x}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow \frac{d y}{d x}=\frac{12 a}{6}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{a}\) The line \(y=4 x-5\) touches the curve at the point \((2,3)\) So, slope of the tangent \((2,3)\) is 4 . \(\therefore \quad 2 \mathrm{a}=4\) \(\mathrm{a}=2\) Since, \((2,3)\) lie on the curve \(y^{2}=a x^{3}+b\) \((3)^{2}=a(2)^{3}+b\) \(9=8 \mathrm{a}+\mathrm{b} \quad[\because \mathrm{a}=2]\) \(9=8 \times 2+b\) \(\mathrm{b}=-7\) Hence, \(\mathrm{a}=2, \mathrm{~b}=-7\)
MHT CET-2020
Application of Derivatives
85386
The equation of tangent at \(P(-4,-4)\) on the curve \(x^{2}=-4 y\) is
1 \(3 x-y+8=0\)
2 \(2 x+y+4=0\)
3 \(2 x+y-4=0\)
4 \(2 x-y+4=0\)
Explanation:
(D) : Given, Tangent point \((-4,-4)\) Curve, \(x^{2}=-4 y\) On differentiating both sides w.r.t \(x\), \(2 x=-4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\left.\frac{-x}{2} \Rightarrow \frac{d y}{d x}\right|_{(-4,-4)}=\frac{+4}{2}=2\) The equation of the tangent is given \(\left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)\) \((y+4)=2(x+4)\) \((y+4)=2 x+8\) \(2 x-y+4=0\)
MHT CET-2020
Application of Derivatives
85387
The approximate value of the function \(f(x)=x^{3}-3 x+5\) at \(x=1.99\) is
1 7.91
2 7.94
3 6.94
4 6.91
Explanation:
(D) : Given, \(f(x)=x^{3}-3 x+5\) \(f^{\prime}(x)=3 x^{2}-3\) We know that, \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x})+\Delta \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})\) Here, \(\mathrm{x}+\Delta \mathrm{x}=1.99\) \(x=2, \Delta x=-0.01\) \(\therefore \quad \mathrm{f}(1.99)=\mathrm{f}(2)+(-0.01) \mathrm{f}^{\prime}(2)\) \(\mathrm{f}(1.99) =(2)^{3}-3(2)+5-0.01\left[3(2)^{2}-3\right]\) \(=8-6+5-0.01 \times 12-3=7-0.09\) \(\mathrm{f}(1.99) =6.91\) Hence, The approximate value of the function \(\mathrm{f}(1.99\) ) is 6.91
85384
The equation of normal to the curve \(y=\) \(\sin \left(\frac{\pi x}{4}\right)\) at the point \((2,5)\) is
1 \(x+y=2\)
2 \(x+y=5\)
3 \(y=5\)
4 \(x=2\)
Explanation:
(D) : Given, \(\mathrm{y}=\sin \left(\frac{\pi \mathrm{x}}{4}\right)\) On differentiating both sides, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \mathrm{x}\right)\) Put the value of \(x=2\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \times 2\right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \frac{\pi}{2}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Equation of normal of curve at \((2,5)\) \(\mathrm{x}-2=0\) \(\mathrm{x}=2\)
MHT CET-2020
Application of Derivatives
85385
If the line \(y=4 x-5\) touches the curve \(y^{2}=a x^{3}+b\) at the point \((2,3)\) then
1 \(\mathrm{a}=2, \quad \mathrm{~b}=7\)
2 \(\mathrm{a}=-2, \mathrm{~b}=-7\)
3 \(\mathrm{a}=-2, \mathrm{~b}=7\)
4 \(\mathrm{a}=2, \quad \mathrm{~b}=-7\)
Explanation:
(D) : Given, \(y=4 x-5\) Point \((2,3)\) \(\mathrm{y}^{2}=\mathrm{ax}^{3}+\mathrm{b}\) On differentiating both sides w.r.t. to \(\mathrm{x}\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times 3 \mathrm{x}^{2}+0\) \(\frac{d y}{d x}=\frac{3 a^{2}}{2 y}\) Let, point \((x=2, y=3)\) \(\frac{d y}{d x}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow \frac{d y}{d x}=\frac{12 a}{6}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{a}\) The line \(y=4 x-5\) touches the curve at the point \((2,3)\) So, slope of the tangent \((2,3)\) is 4 . \(\therefore \quad 2 \mathrm{a}=4\) \(\mathrm{a}=2\) Since, \((2,3)\) lie on the curve \(y^{2}=a x^{3}+b\) \((3)^{2}=a(2)^{3}+b\) \(9=8 \mathrm{a}+\mathrm{b} \quad[\because \mathrm{a}=2]\) \(9=8 \times 2+b\) \(\mathrm{b}=-7\) Hence, \(\mathrm{a}=2, \mathrm{~b}=-7\)
MHT CET-2020
Application of Derivatives
85386
The equation of tangent at \(P(-4,-4)\) on the curve \(x^{2}=-4 y\) is
1 \(3 x-y+8=0\)
2 \(2 x+y+4=0\)
3 \(2 x+y-4=0\)
4 \(2 x-y+4=0\)
Explanation:
(D) : Given, Tangent point \((-4,-4)\) Curve, \(x^{2}=-4 y\) On differentiating both sides w.r.t \(x\), \(2 x=-4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\left.\frac{-x}{2} \Rightarrow \frac{d y}{d x}\right|_{(-4,-4)}=\frac{+4}{2}=2\) The equation of the tangent is given \(\left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)\) \((y+4)=2(x+4)\) \((y+4)=2 x+8\) \(2 x-y+4=0\)
MHT CET-2020
Application of Derivatives
85387
The approximate value of the function \(f(x)=x^{3}-3 x+5\) at \(x=1.99\) is
1 7.91
2 7.94
3 6.94
4 6.91
Explanation:
(D) : Given, \(f(x)=x^{3}-3 x+5\) \(f^{\prime}(x)=3 x^{2}-3\) We know that, \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x})+\Delta \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})\) Here, \(\mathrm{x}+\Delta \mathrm{x}=1.99\) \(x=2, \Delta x=-0.01\) \(\therefore \quad \mathrm{f}(1.99)=\mathrm{f}(2)+(-0.01) \mathrm{f}^{\prime}(2)\) \(\mathrm{f}(1.99) =(2)^{3}-3(2)+5-0.01\left[3(2)^{2}-3\right]\) \(=8-6+5-0.01 \times 12-3=7-0.09\) \(\mathrm{f}(1.99) =6.91\) Hence, The approximate value of the function \(\mathrm{f}(1.99\) ) is 6.91
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85384
The equation of normal to the curve \(y=\) \(\sin \left(\frac{\pi x}{4}\right)\) at the point \((2,5)\) is
1 \(x+y=2\)
2 \(x+y=5\)
3 \(y=5\)
4 \(x=2\)
Explanation:
(D) : Given, \(\mathrm{y}=\sin \left(\frac{\pi \mathrm{x}}{4}\right)\) On differentiating both sides, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \mathrm{x}\right)\) Put the value of \(x=2\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \left(\frac{\pi}{4} \times 2\right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\pi}{4} \cos \frac{\pi}{2}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Equation of normal of curve at \((2,5)\) \(\mathrm{x}-2=0\) \(\mathrm{x}=2\)
MHT CET-2020
Application of Derivatives
85385
If the line \(y=4 x-5\) touches the curve \(y^{2}=a x^{3}+b\) at the point \((2,3)\) then
1 \(\mathrm{a}=2, \quad \mathrm{~b}=7\)
2 \(\mathrm{a}=-2, \mathrm{~b}=-7\)
3 \(\mathrm{a}=-2, \mathrm{~b}=7\)
4 \(\mathrm{a}=2, \quad \mathrm{~b}=-7\)
Explanation:
(D) : Given, \(y=4 x-5\) Point \((2,3)\) \(\mathrm{y}^{2}=\mathrm{ax}^{3}+\mathrm{b}\) On differentiating both sides w.r.t. to \(\mathrm{x}\), we get- \(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a} \times 3 \mathrm{x}^{2}+0\) \(\frac{d y}{d x}=\frac{3 a^{2}}{2 y}\) Let, point \((x=2, y=3)\) \(\frac{d y}{d x}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow \frac{d y}{d x}=\frac{12 a}{6}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{a}\) The line \(y=4 x-5\) touches the curve at the point \((2,3)\) So, slope of the tangent \((2,3)\) is 4 . \(\therefore \quad 2 \mathrm{a}=4\) \(\mathrm{a}=2\) Since, \((2,3)\) lie on the curve \(y^{2}=a x^{3}+b\) \((3)^{2}=a(2)^{3}+b\) \(9=8 \mathrm{a}+\mathrm{b} \quad[\because \mathrm{a}=2]\) \(9=8 \times 2+b\) \(\mathrm{b}=-7\) Hence, \(\mathrm{a}=2, \mathrm{~b}=-7\)
MHT CET-2020
Application of Derivatives
85386
The equation of tangent at \(P(-4,-4)\) on the curve \(x^{2}=-4 y\) is
1 \(3 x-y+8=0\)
2 \(2 x+y+4=0\)
3 \(2 x+y-4=0\)
4 \(2 x-y+4=0\)
Explanation:
(D) : Given, Tangent point \((-4,-4)\) Curve, \(x^{2}=-4 y\) On differentiating both sides w.r.t \(x\), \(2 x=-4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\left.\frac{-x}{2} \Rightarrow \frac{d y}{d x}\right|_{(-4,-4)}=\frac{+4}{2}=2\) The equation of the tangent is given \(\left(y-y_{1}\right)=\frac{d y}{d x}\left(x-x_{1}\right)\) \((y+4)=2(x+4)\) \((y+4)=2 x+8\) \(2 x-y+4=0\)
MHT CET-2020
Application of Derivatives
85387
The approximate value of the function \(f(x)=x^{3}-3 x+5\) at \(x=1.99\) is
1 7.91
2 7.94
3 6.94
4 6.91
Explanation:
(D) : Given, \(f(x)=x^{3}-3 x+5\) \(f^{\prime}(x)=3 x^{2}-3\) We know that, \(\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x})+\Delta \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})\) Here, \(\mathrm{x}+\Delta \mathrm{x}=1.99\) \(x=2, \Delta x=-0.01\) \(\therefore \quad \mathrm{f}(1.99)=\mathrm{f}(2)+(-0.01) \mathrm{f}^{\prime}(2)\) \(\mathrm{f}(1.99) =(2)^{3}-3(2)+5-0.01\left[3(2)^{2}-3\right]\) \(=8-6+5-0.01 \times 12-3=7-0.09\) \(\mathrm{f}(1.99) =6.91\) Hence, The approximate value of the function \(\mathrm{f}(1.99\) ) is 6.91